tag:blogger.com,1999:blog-58034171811218208382024-02-01T19:10:26.258-08:00Maths: Geometry, Functions and Graphs* Problems and Solutions * Geometry Calculator * Analytic Geometry * Graphs Calculator * Math FunctionsKTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.comBlogger26125tag:blogger.com,1999:blog-5803417181121820838.post-39754217620519928782016-04-02T12:09:00.001-07:002016-04-07T00:32:42.805-07:00Problem: a, b, c are lengths of side of a triangleLet <span style="color: #cc0000;">a, b, c</span> are lengths of three sides of a triangle, and:<br />
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<br />
Show that: <span style="color: red;">P < 0</span><br />
<br />
<b><i>Solution:</i></b><br />
<br />
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<!--StartFragment--><span style="font-family: "cambria"; font-size: 12.0pt;">We can assume that </span><span style="color: #e36c0a; font-family: "cambria"; font-size: 14.0pt;">c</span><span style="font-family: "cambria"; font-size: 12.0pt;"> this the longest side of the triangle ABC (it does not make different if the longest side is </span><span style="color: #e36c0a; font-family: "cambria"; font-size: 14.0pt;">a </span><span style="font-family: "cambria"; font-size: 12.0pt;">or </span><span style="color: #e36c0a; font-family: "cambria"; font-size: 14.0pt;">b</span><span style="font-family: "cambria"; font-size: 12.0pt;">), draw the height CH with H is on AB => H is
lying between points A and B (see the pic below)</span><br />
<span style="font-family: "cambria"; font-size: 12.0pt;"><br /></span>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg_Dc8uTe9Ij-O94Ezo6B5baehuoPl-O4r3yVMmMjFxiHtukme0Q0MH5ZbhpoSISp8zxgeGCFJ5C27iivhDu4jATUr1d-6DKl6X21rHMwyaR8A-4Xh_sN32zC_WHGuznRnP6NxoAoMsSmBK/s1600/toan1.2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="220" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg_Dc8uTe9Ij-O94Ezo6B5baehuoPl-O4r3yVMmMjFxiHtukme0Q0MH5ZbhpoSISp8zxgeGCFJ5C27iivhDu4jATUr1d-6DKl6X21rHMwyaR8A-4Xh_sN32zC_WHGuznRnP6NxoAoMsSmBK/s400/toan1.2.jpg" width="400" /></a></div>
<span style="font-family: "cambria"; font-size: 12.0pt;">then: </span><br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgUgVqWzm3K5-MgLmS8azjFt0naY3oabL10gT0_gut8UtOJbgZZ_IJPNP7wcfe53q3nwCjlA64kUJf-ut6fUiDItH04IgUnO8_hmdj6RcATxK9WzKQfkq5QZUEFYtcTgTedzaDtjf4d8ZrU/s1600/toan1.3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="29" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgUgVqWzm3K5-MgLmS8azjFt0naY3oabL10gT0_gut8UtOJbgZZ_IJPNP7wcfe53q3nwCjlA64kUJf-ut6fUiDItH04IgUnO8_hmdj6RcATxK9WzKQfkq5QZUEFYtcTgTedzaDtjf4d8ZrU/s320/toan1.3.jpg" width="320" /></a></div>
<span style="font-family: "cambria"; font-size: 12.0pt;"><br /></span>
<span style="font-family: "cambria"; font-size: 12.0pt;">let change the expression P as following:</span><br />
<span style="font-family: "cambria"; font-size: 12.0pt;"><br /></span>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEitZLxqN-eCJCA29nY9XYYUbWo0fgvrsJZcBAeGvffGcW-H3IP94tpnMaOIhKV3b4rGD0jvkLCe1PiJNKO9nwVrzO76p1F5XYMI7OIxH0abWLoTukueGpff4xSYQtnXw9HfFOLctzC_hHY6/s1600/toan1.4.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="220" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEitZLxqN-eCJCA29nY9XYYUbWo0fgvrsJZcBAeGvffGcW-H3IP94tpnMaOIhKV3b4rGD0jvkLCe1PiJNKO9nwVrzO76p1F5XYMI7OIxH0abWLoTukueGpff4xSYQtnXw9HfFOLctzC_hHY6/s400/toan1.4.jpg" width="400" /></a></div>
<div class="separator" style="clear: both; text-align: center;">
<br /></div>
<br />KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-81962354977831298582015-02-24T21:47:00.002-08:002015-02-24T21:47:15.437-08:00Geometry problem (2)<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">Given triangle ABC, angle BAC < 90 degrees. Two squares with side AB, AC are drawn outside of the triangle. M is a midpoint of BC. Show that DM ⊥ ME and DM = ME</span><br />
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<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-55380832865748136972015-02-24T21:43:00.003-08:002015-02-24T21:43:41.103-08:00Geometry Problems (1): Prove that points are laying on one circle!<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;">Given arbitrary five-wings star, the circumscribed circles of its wings are drawn. Show that intersections points A, B, C, D, E are laying on one circle!</span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhqO8x1UeV5Xrrg5Jf9ibZ5-Ex5yJmwWwiFav5E-W9w1xFRPp9gAFJ5VzgFu4EHqM8tKLqJdsD_7LCT01CnTTflRUiYVwx-W7guEhgF8HRv8fVW6GcMKwm44pC8d3HiD2HynA5jA6kkk9GV/s1600/1238277_1479994118884400_4358907612190420884_n.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhqO8x1UeV5Xrrg5Jf9ibZ5-Ex5yJmwWwiFav5E-W9w1xFRPp9gAFJ5VzgFu4EHqM8tKLqJdsD_7LCT01CnTTflRUiYVwx-W7guEhgF8HRv8fVW6GcMKwm44pC8d3HiD2HynA5jA6kkk9GV/s1600/1238277_1479994118884400_4358907612190420884_n.jpg" height="400" width="356" /></a></div>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 20px;"><br /></span>KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-8483005475593238182014-11-12T23:18:00.000-08:002015-02-24T21:08:50.688-08:00Geometry Pad-2D: Select an object to work withIn many cases you need to select an object (point, line segment, circle, triangle, polygon, angle) to work with. <i>For example, you want to draw a circumscribed circle of triangle ABC, in this case you need to select triangle, and use triangle tools to draw its circumscribed circle</i>.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTM-n_OlwVqY56N-TA3zsNUDxoDA7narOHn3gaqNi3Kg5vsl0rdzJD4BhMoifna4fyH0F3WcRBprf5auTG7xdESa5XEW2TDd2icDJ0FNgjOZy6DKjl7ENX4SmPC_Kkg_xqCz4sBvuVppsc/s1600/IMG_0183.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTM-n_OlwVqY56N-TA3zsNUDxoDA7narOHn3gaqNi3Kg5vsl0rdzJD4BhMoifna4fyH0F3WcRBprf5auTG7xdESa5XEW2TDd2icDJ0FNgjOZy6DKjl7ENX4SmPC_Kkg_xqCz4sBvuVppsc/s320/IMG_0183.jpg" width="320" /></a></div>
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There are some ways to select an object in the construction showing in screen:<br />
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<b>Select by tapping on screen</b><br />
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<ul>
<li><i>To select a point</i>, you can simply tap nearly it on screen</li>
<li><i>To select a line segment</i>: tap nearly its center point</li>
<li><i>To select a circle</i>: tap on it</li>
<li><i>To select a triangle, polygon</i>: tap somewhere in the "centre" of a shape</li>
<li><i>To select an angle</i>: select its vertex first, then tap somewhere inside its, nearly but not too close to the vertex </li>
</ul>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuM6bhSJI-aptvlBx4_Gj4L4RyTrdfhBD8skIL8YhVr4qtFHX67-9J0Bcl_mroCQ7IL6THUBzNkTBFNTSsAGEHYAI0ybQ9vr62_oUBAtVVNbxM9WPHGbV8Y-WWE8PZViG51zqFHqY1fS6g/s1600/IMG_0187.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuM6bhSJI-aptvlBx4_Gj4L4RyTrdfhBD8skIL8YhVr4qtFHX67-9J0Bcl_mroCQ7IL6THUBzNkTBFNTSsAGEHYAI0ybQ9vr62_oUBAtVVNbxM9WPHGbV8Y-WWE8PZViG51zqFHqY1fS6g/s400/IMG_0187.jpg" width="300" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>select an object by tapping on screen</i></td></tr>
</tbody></table>
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<i>Limitation</i>: when your construction is complex enough, this may make some confuse when you tap on the screen. For example, when two or more line segments have the same center point or there are some angles with the same vertex,...<br />
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<i>Note</i>: the selected object by default will be highlighted light-green filling<br />
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<b>Use "bottom" tools </b><br />
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This tools will appeared when you select some point. It is to help you select objects related to the selected point. For example, when you select point "B", the "line segment" selection tool will show you all line segments comes from/to point "B".<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMxDCoT2jmATlpjPyWw8qP14yIuz50Q3aJpvxwVwoTUSigSTcfvabqj_4x_k8fW_BSFX8yYCbEpKU01WgvmxhH9Dpg8WiwfmDZTFnng7vnHSdkTx0yhFcc12ZNPWi-gDIYXKpgmBCHurES/s1600/IMG_0185.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMxDCoT2jmATlpjPyWw8qP14yIuz50Q3aJpvxwVwoTUSigSTcfvabqj_4x_k8fW_BSFX8yYCbEpKU01WgvmxhH9Dpg8WiwfmDZTFnng7vnHSdkTx0yhFcc12ZNPWi-gDIYXKpgmBCHurES/s400/IMG_0185.jpg" width="300" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>bottom selection tools</i></td></tr>
</tbody></table>
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<i>Limitation</i>: this method shows you all possible variations for choosing an objects, sometime there are too many variations and it makes you difficult to detect which one is your object!<br />
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<b>Use main selection tools</b><br />
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This is the most careful method to select an object, it was designed for line segments, triangles, angles and polygons. Steps to choose an object are:<br />
<br />
<ol>
<li>Choose selection tool</li>
<li>Choose type of object</li>
<li>Tap on all points (vertices) of an object which you want to select</li>
</ol>
<div>
<i>Note</i>: You should tap on the points in order you needed. For example, when you tap points: "A", "B", "C", angle ABC will be selected; but tapping on "B", "A", "C" will select angle "BAC". In the case of polygon: quadrilaterals ABCD and ACBD are different each of other!</div>
<div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhKXQKNuvIUPa-2bUtTKSWzhr6lWuwkR_kXHYRAsWZFlI715r3ci6Le2M87N6jhM_1P9ib9hSb_mIOd8vm9TGB4x4SFE_tBWFPiHIUh1-VvhfDo22XYc72ohY4uwwh1My2ilUiEhJx0P2No/s1600/IMG_0186.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhKXQKNuvIUPa-2bUtTKSWzhr6lWuwkR_kXHYRAsWZFlI715r3ci6Le2M87N6jhM_1P9ib9hSb_mIOd8vm9TGB4x4SFE_tBWFPiHIUh1-VvhfDo22XYc72ohY4uwwh1My2ilUiEhJx0P2No/s400/IMG_0186.jpg" width="300" /></a></div>
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<span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13.3333330154419px; line-height: 12.3199996948242px;">2014.11.13</span><br />
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-72436483717051649912014-11-11T01:38:00.003-08:002014-11-12T23:25:30.771-08:00Geometry Pad-2D: Working with triangle [1]Suppose that we have some problem like: "<i>Given triangle ABC with the height AH, draw line HD // AC, D is on side AC</i>...."<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9Ph9TekNzN3U2p-HNSSpJ_VKJsAghdFPsJGjvG6Bi9D8BDDb73CHM9M1vhcZeNic1w7WQowD0ELL6D8zq7YiEIf4cdEkxjarxPNenj8OJc1JKKusIqdf2SDX49PxoR4k8X9-lZACIdYnw/s1600/IMG_0165.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh9Ph9TekNzN3U2p-HNSSpJ_VKJsAghdFPsJGjvG6Bi9D8BDDb73CHM9M1vhcZeNic1w7WQowD0ELL6D8zq7YiEIf4cdEkxjarxPNenj8OJc1JKKusIqdf2SDX49PxoR4k8X9-lZACIdYnw/s1600/IMG_0165.jpg" height="213" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>pic 1</i></td></tr>
</tbody></table>
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With the "Geometry Pad-2D" app, you can draw the picture easily by following steps:<br />
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<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEio6_K5gyb2CVAE4B-pOELJ3GtFjemZlf0qExJq47v-BYGUqUCP6b7OAtKF8TbWEFAjYMYH3ewzTAh6btKco8MDhrYJVYyu1Lhjbvpl5FtyCZWlxyi9OAuGHOz21yLqx3_LSNxawDm3fvRP/s1600/IMG_0174.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEio6_K5gyb2CVAE4B-pOELJ3GtFjemZlf0qExJq47v-BYGUqUCP6b7OAtKF8TbWEFAjYMYH3ewzTAh6btKco8MDhrYJVYyu1Lhjbvpl5FtyCZWlxyi9OAuGHOz21yLqx3_LSNxawDm3fvRP/s1600/IMG_0174.jpg" height="320" width="240" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>Step 1: Draw triangle ABC</i></td></tr>
</tbody></table>
<b>Step 1: Draw triangle ABC</b><br />
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a) Tap on the "+"<span style="text-align: center;"> tool of the TOP toolsbox</span><br />
<span style="text-align: center;">b) Tap on the "triangle" tool of the RIGHT toolsbox</span><br />
<span style="text-align: center;">c) Select type of triangle</span><br />
<span style="text-align: center;">d) Move triangle into right position </span><br />
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<i>Note</i>: You also can change the drawn triangle by tap on any vertex or side then move it, change it using points or line segments tools<br />
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<b>Step 2: Draw height AH</b><br />
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There are different ways to draw the heights of triangle: use points tools, use lines tools or use triangle tools. The following shows how to use points tools to draw height AH.<br />
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a) Select vertex A<br />
b) Select "draw perpendicular segment" tool<br />
c) Select showed point to draw line to...<br />
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<i>Note</i>: new the point on BC is automatically named as "D", we will change name of the point later.<br />
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<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; text-align: left;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhHvc4yHtyfGgd6Qyxd58Anb1HxS_PpBFiI-3GJewg4wCN-70SKghJTw1MQ__NlAj5NuY6ejPzv0wdTRANm5B_YtBfbXDf0_t1sp4xNK9RA1fquvxvAZgHrbnYdL_zA5qsKiNVvxTkLH0EY/s1600/IMG_0161.jpg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhHvc4yHtyfGgd6Qyxd58Anb1HxS_PpBFiI-3GJewg4wCN-70SKghJTw1MQ__NlAj5NuY6ejPzv0wdTRANm5B_YtBfbXDf0_t1sp4xNK9RA1fquvxvAZgHrbnYdL_zA5qsKiNVvxTkLH0EY/s1600/IMG_0161.jpg" height="320" width="240" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>Step 2: Draw height from A</i></td></tr>
</tbody></table>
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<b>Step 3: Draw line segment parallel to AB</b><br />
<span id="goog_793060860"></span><span id="goog_793060861"></span><br />
a) Select line AB, then select point D<br />
b) Select "draw perpendicular..." of the Right toolsbox<br />
c) Tap on small brown circle appeared on AC<br />
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<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="clear: right; float: right; margin-bottom: 1em; text-align: right;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhxumsPFPk3IY4xdjGNjypbxoTy6dX6bUXzN-_Zm-8QNyH3mLmq-jabEijpzCjad2yq8CuMJ9jS_WD1olWkiY7guqGP_F3jVb940JUeomBmls7hLHOcn8H5cI_Yzf-RV4W4fZwawT62pqX6/s1600/IMG_0176.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhxumsPFPk3IY4xdjGNjypbxoTy6dX6bUXzN-_Zm-8QNyH3mLmq-jabEijpzCjad2yq8CuMJ9jS_WD1olWkiY7guqGP_F3jVb940JUeomBmls7hLHOcn8H5cI_Yzf-RV4W4fZwawT62pqX6/s1600/IMG_0176.jpg" height="320" width="240" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>Step 3: Draw DE // AB</i></td></tr>
</tbody></table>
<i>Note</i>: by default, new added line segments have "dotted" style<br />
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<b>Step 4: Naming, styles and colours</b><br />
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a) Now you can use points tools (select any point to show point tools on the right side of screen) to change name(s) of points. In our case change "D" -> "H" and "E" -> "D"<br />
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<i>(The naming tool is the first one of the points toolsbox)</i><br />
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b) Select line segment to show lines toolsbox on the right side of screen. Use change line-style, change line-colour tools to make a picture look as you want<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgmT3qLQqHs-xPFjkWFSMic552AqT6E6ovmG5B11jWAkuf3Qb1ZTtJbFGFfys3Q9Oa8KXGeXjPrTgTeSUbgUaaycWsukquvK-xGn3ZP3s6CFltoF8y5spC5xEoeBWaeMqkszrAc1uvhdNnQ/s1600/IMG_0180.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgmT3qLQqHs-xPFjkWFSMic552AqT6E6ovmG5B11jWAkuf3Qb1ZTtJbFGFfys3Q9Oa8KXGeXjPrTgTeSUbgUaaycWsukquvK-xGn3ZP3s6CFltoF8y5spC5xEoeBWaeMqkszrAc1uvhdNnQ/s1600/IMG_0180.jpg" height="320" width="240" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>line segments toolsbox</i></td></tr>
</tbody></table>
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c) Select angle AHC and mark it with, the right angle mark will be drawn<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaMwj3LY2H3OYYNYYtNc81tONwR5FjPOv5lB7wpQCdIAwWOAfLoWJExtNg7Co_P3wBlyhFdVOL55tAsmV6XzmUkvCB-jtEu1hvm72923weLok2Db4jXpTg9yNFskTuvJ7aq3baxVK48nfv/s1600/IMG_0182.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaMwj3LY2H3OYYNYYtNc81tONwR5FjPOv5lB7wpQCdIAwWOAfLoWJExtNg7Co_P3wBlyhFdVOL55tAsmV6XzmUkvCB-jtEu1hvm72923weLok2Db4jXpTg9yNFskTuvJ7aq3baxVK48nfv/s1600/IMG_0182.jpg" height="320" width="240" /></a></td></tr>
<tr><td class="tr-caption" style="font-size: 12.6666669845581px;"><i>mark angle</i></td></tr>
</tbody></table>
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<span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13.3333330154419px; line-height: 12.3199996948242px;">2014.11.10</span><br />
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-84408395894340884652014-10-02T23:51:00.003-07:002014-10-27T03:47:55.833-07:00Geometry 2D-Pad: How to add new object? <div style="text-align: right;">
<a href="http://itunes.apple.com/app/id874778157">Free download</a></div>
<br />
This is to describe the first way to add new object to a geometry construction using Geometry 2D-Pad (v2.0) on your iPad (with iOS 7++). In our example, the new object is a triangle.<br />
<br />
1) Tap the "add" tool on the top screen menu (+)<br />
2) Tap the triangle tool on the right toolset (it appeared after 1st action)<br />
3) The new triangle with the "moving circle" (pink) will appeared in drawing screen<br />
4) Drag and drop the new shape with the pink circle at the right position<br />
<br />
See the picture:<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2MWtzjguefEMEKrQRWf8EBQSgHarr630agdJPv__QLllq86Ur13oy64dtRnNuWAoJxc5q6cY7pZTTly41-DpF0RmRKjXRZNMfow9WA5N9Sb46elMxlmhC5c7iNGLNBHDh69E7Fg9XUBcB/s1600/help02.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2MWtzjguefEMEKrQRWf8EBQSgHarr630agdJPv__QLllq86Ur13oy64dtRnNuWAoJxc5q6cY7pZTTly41-DpF0RmRKjXRZNMfow9WA5N9Sb46elMxlmhC5c7iNGLNBHDh69E7Fg9XUBcB/s1600/help02.png" height="400" width="300" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>Step to add new triangle</i></td></tr>
</tbody></table>
<br />
The triangle ABC has been added. Now you can choose it and use "working with triangle toolset" to add the altitude AH as bellow:<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpmebSnG8nq1eWn3tAGSUmXeDahyphenhyphenovPf0bxwY7Vxjjiq_lcI_Dws0MOIQm-j6M-GeLi238-TnqRGVuR8A8G13RoKyik-v6D3SJatxg4ZwNFGKXqfdRpOhJ0Csq44W-eKPK4RDoTBhYJvX4/s1600/help02a.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpmebSnG8nq1eWn3tAGSUmXeDahyphenhyphenovPf0bxwY7Vxjjiq_lcI_Dws0MOIQm-j6M-GeLi238-TnqRGVuR8A8G13RoKyik-v6D3SJatxg4ZwNFGKXqfdRpOhJ0Csq44W-eKPK4RDoTBhYJvX4/s1600/help02a.png" height="400" width="300" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>Choose triangle ABC and draw altitude AH</i></td></tr>
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<b>Notes</b>: before moving the new triangle, you could see some interesting pictures on the bottom side of drawing area, what is it? This is a new interesting feature, which allows you to choose a type of new triangle: equilateral, right or isosceles. Let try it and see the results! <br />
<br />
It is so easy, isn't it?<br />
<br />
<div style="text-align: right;">
<a href="http://itunes.apple.com/app/id874778157">Free download</a></div>
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<span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13.3333330154419px; line-height: 12.3199996948242px;">2014.10.03</span><br />
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<br />KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-69874541950470349172014-06-10T23:24:00.001-07:002014-06-10T23:31:42.602-07:00How to draw an inscribed and a circumscribed circles of a triangle?<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh-MhTGZKvhKnGPKEVHwWJSqpnNXaJm17uazTh38zMPTDtUfsDMdVuEBBgEUv7AMfgrGIelE0eDGfJyo8j6krPDjAcs2OeXsfbbkzYKZFG169Go613gHnOFBLgDRqXi-DIgruOJnadQcwch/s1600/IMG_7694.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh-MhTGZKvhKnGPKEVHwWJSqpnNXaJm17uazTh38zMPTDtUfsDMdVuEBBgEUv7AMfgrGIelE0eDGfJyo8j6krPDjAcs2OeXsfbbkzYKZFG169Go613gHnOFBLgDRqXi-DIgruOJnadQcwch/s1600/IMG_7694.jpg" height="320" width="318" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">(O1) is an inscribed, (O2) is a circumscribed circle of triangle ABC</td></tr>
</tbody></table>
<br />
This post is to show how to use Geometry Drawer tools to draw an inscribed and a circumscribed circle of a given triangle. On a paper, this task is a combination of several activities to detect a centre points of required circles. With Geometry Drawer Tools it made easy:<br />
<br />
1) Select a triangle ABC with selection tools or by selecting point A, then long press on on side BC<br />
2) When the triangle is selected, the circles tools will be appeared.<br />
3) Select required tool and drag it into drawing area, then drop it<br />
<br />
That is all! See the picture bellow:<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9IcH7ahHqj5mamHi_Axi__yoCeIxHL069i1pJW0Fh5LYn8LX9OTO05x6j4ReQs_a-yHtfD6m2i4TQkXbgDV3_VROokKN3rmvSfuPw5DNbXawJNeinBL-7L8TExtj8t74Ghm2tOLj86j6q/s1600/IMG_7693.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9IcH7ahHqj5mamHi_Axi__yoCeIxHL069i1pJW0Fh5LYn8LX9OTO05x6j4ReQs_a-yHtfD6m2i4TQkXbgDV3_VROokKN3rmvSfuPw5DNbXawJNeinBL-7L8TExtj8t74Ghm2tOLj86j6q/s1600/IMG_7693.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Tools for drawing an inscribed and a circumscribed circles</td></tr>
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Notes:<br />
<br />
1) The picture is taken by Geometry Drawer for iPhone. It is the same way for drawing by Geometry Drawer for iPad.<br />
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2) After the above task, you can use circles tools to change style and colour of a drawn circles<br />
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<div style="background-color: white; color: #333333; font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 20px;">
11.6.2014</div>
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<a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8" style="color: #888888; text-decoration: none;">Plane Geometry Drawer for iPhone</a></div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-89426079458525294732014-03-16T00:36:00.001-07:002014-03-16T00:36:17.049-07:00How to draw perpendicular and parallel lines with points toolset<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhFlOGG5F3lyvG_4OryQhrEK6Jm4Gc-0OHTG4SVmbE0R55vw8jLC77QSHgtXjQk-wpMG9gCvLF20qulsheiSWvGBzwCwbXpsiTjd_DTuxgnFrsdu-EGP-ItW0c_47SdkXMLo1OXqEBWMaNu/s1600/IMG_7219.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhFlOGG5F3lyvG_4OryQhrEK6Jm4Gc-0OHTG4SVmbE0R55vw8jLC77QSHgtXjQk-wpMG9gCvLF20qulsheiSWvGBzwCwbXpsiTjd_DTuxgnFrsdu-EGP-ItW0c_47SdkXMLo1OXqEBWMaNu/s1600/IMG_7219.jpg" height="250" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">picture 1</td></tr>
</tbody></table>
Assume that we have a triangle ABC and we want to draw the height AH and parallelogram ABCD (draw the red lines on the picture 1).<br />
<br />
The following steps show how to do that.<br />
<br />
<b>Step 1: Draw line AH <span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13.333333015441895px; line-height: 12.319999694824219px;">⊥ BC</span></b><br />
<br />
Select point A. Drag the "add line" tool of the point toolset and drag it to line BC<br />
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The small green circles appeared to highlight "meaning points", one of them is a projected point of point A on BC (call it H).<br />
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Move the red pointer to H and drop. Point A will be connected with H. (Select AH and use color tool to change color of AH to red)<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj1zuHGCPCr9XzZD4qOWI3OEvkWEdHa4CtF_FOF8PjdCtdPuAmA4HeJkgnIXbd4IB4WrmoPl32Am13ovWdgKKzM9zn7FMDeYEu265VjaUI6vag191VHU2C8eN0hzgMftXUMUkoc2NJ3JC9E/s1600/IMG_7220.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj1zuHGCPCr9XzZD4qOWI3OEvkWEdHa4CtF_FOF8PjdCtdPuAmA4HeJkgnIXbd4IB4WrmoPl32Am13ovWdgKKzM9zn7FMDeYEu265VjaUI6vag191VHU2C8eN0hzgMftXUMUkoc2NJ3JC9E/s1600/IMG_7220.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">illustration of step 1</td></tr>
</tbody></table>
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<b>Step 2</b>: <b>Draw AD // BC</b><br />
<br />
Select point A, choose "parallel line" tool of the point toolset (//) and drag it into the area above of BC. Some points will be highlighted as shown in picture 3. Move the pointer to the point 2 and drop it there (see picture 3). Name the new point as D.<br />
<br />
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCW8lWKTRv4tTDCAQ4LbJtU_lX_axHy7h9Pv2xSg8-9yRCIm7P5n5EWkaZRPyfig_ju2LimCm_ejkhHwxEBzWFJW1ZWvhKo9Mq3hExVHJgcfNhWtO2GPcUzA9I8B3b9N9tUmOReXcUt6Ns/s1600/IMG_7245.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCW8lWKTRv4tTDCAQ4LbJtU_lX_axHy7h9Pv2xSg8-9yRCIm7P5n5EWkaZRPyfig_ju2LimCm_ejkhHwxEBzWFJW1ZWvhKo9Mq3hExVHJgcfNhWtO2GPcUzA9I8B3b9N9tUmOReXcUt6Ns/s1600/IMG_7245.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">picture 3</td></tr>
</tbody></table>
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Connect point C with point D and colour lines AD, CD by red. In the result we have a drawing which is shown in picture 1.<br />
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Note: There a variety ways to draw the construction. The above is method by using points toolset, we can also use line toolset, circles toolset or combinations of them. <br />
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<div style="background-color: white; color: #333333; font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 20px;">
16.3.2014</div>
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<a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8" style="color: #888888; text-decoration: none;">Plane Geometry Drawer for iPhone</a></div>
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<br />KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-56603612503559735302014-03-10T23:50:00.003-07:002014-03-11T00:06:38.714-07:00How to start drawing a geometry construction<b>Imagine your construction</b><br />
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Before starting drawing, it is important to think about the construction: which is the main object? (a triangle? a square? a trapezoid? or a circle?...) and where to start?... Try to imagine your construction and choose the object to start.<br />
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<b>Add the first object(s)</b><br />
<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgtVlsG-463VYs654PLBKymTaqmRl1TbLKlpVqG1twG3GomHGaKhT0T4hfXPzRp9qDUTawk5slENXGrG-Mqb5Iz0-roMP9-oJo3Df9yP1BrmcGOMiKT18H2fhuAQMIqrZmX_i4Ynt29Heuy/s1600/IMG_7197.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgtVlsG-463VYs654PLBKymTaqmRl1TbLKlpVqG1twG3GomHGaKhT0T4hfXPzRp9qDUTawk5slENXGrG-Mqb5Iz0-roMP9-oJo3Df9yP1BrmcGOMiKT18H2fhuAQMIqrZmX_i4Ynt29Heuy/s1600/IMG_7197.jpg" height="152" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>P1: Basic objects tools</i></td></tr>
</tbody></table>
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The drawing screen starts with the basic objects tools. Choose your object, drag into the drawing area and drop it! The most common object to start are triangle, line and circle.<br />
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After adding the first object, you can add the second,...<br />
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<span style="color: red;">Note</span>: <span style="color: #134f5c;"><i>basic objects tools add an object independently of other objects on your constructions. If you want to add an object with some relation with existing object (for example: draw line from a selected point to other point), choose points tools, line tools,... in accordance with your selected object.</i></span><br />
<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; text-align: right;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjE2XexkjoJXPmmEdXitPywC-mOpDUcC1grCrM2FxrYCvSJXL3tHKj0sXwBtYsLtK4KK007wdYk4AtO9YPTnupBLbynOaR3YzUgCWGssX5biwLbtE2YByc40xVzbwF6uKqupQ9r9U5umR48/s1600/IMG_7200.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjE2XexkjoJXPmmEdXitPywC-mOpDUcC1grCrM2FxrYCvSJXL3tHKj0sXwBtYsLtK4KK007wdYk4AtO9YPTnupBLbynOaR3YzUgCWGssX5biwLbtE2YByc40xVzbwF6uKqupQ9r9U5umR48/s1600/IMG_7200.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>P2: connecting points</i></td></tr>
</tbody></table>
<br />
<b>Draw line connecting a selected point with other point</b><br />
<br />
The picture P2 shows a simple construction with line segment BC and point A. Touch on point A to select it. Now you can use line drawing tool (the first one on the left of toolset) of points toolset to draw line from point A to any other point.<br />
<br />
Touch and hold the line tool, drag it into drawing area (do not drop it) and you can see 5 green circles appeared with the pink one connecting to A by a dotted red line (P3).<br />
<br />
1) Green circles are points A, B, C, projected point of A on BC and the midpoint of BC.<br />
<br />
2) The pink circle is pointer of current position, drag and drop it on the point you want to connect with point A.<br />
<br />
<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; text-align: left;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6oOhpdTbO3qqZ2qdzVUqtG4O1KsDPzqZWiYqEUxDg0rkPtQhBY-5BLVBlgsGY2aXCmGWA2L7cYOuQ0708h96Mukw37SDO7_bVEDbU-WFhIEuFT1_S9-O6wze5AAJKo0W9XtU61xb6Kc4O/s1600/IMG_7201.jpg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6oOhpdTbO3qqZ2qdzVUqtG4O1KsDPzqZWiYqEUxDg0rkPtQhBY-5BLVBlgsGY2aXCmGWA2L7cYOuQ0708h96Mukw37SDO7_bVEDbU-WFhIEuFT1_S9-O6wze5AAJKo0W9XtU61xb6Kc4O/s1600/IMG_7201.jpg" height="150" width="200" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">P3</td></tr>
</tbody></table>
<br />
<br />
<br />
For example, we do the following:<br />
<br />
- connect A with B (draw line AB)<br />
- connect A with C (draw line AC)<br />
- connect A with midpoint of BC (draw the median AM)<br />
- name the midpoint of BC as M<br />
<br />
Finally we have a triangle ABC with the median AM.<br />
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<span style="color: red;">Note</span>: <i><span style="color: #134f5c;">when you drag nearly to the highlighted points, the app with automatically adjust the pointer into it. This helps you draw easier and more exactly. </span></i><br />
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<div style="background-color: white; color: #333333; font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 20px;">
11.3.2014</div>
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<a href="https://www.facebook.com/pages/Geometry-Drawer/1449696861914126?ref=hl" style="color: #33aaff; font-size: 13px; text-decoration: none;">Geometry Drawer on Facebook</a></div>
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<a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8" style="color: #888888; text-decoration: none;">Plane Geometry Drawer for iPhone</a></div>
</div>
KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-50936920387347633862014-03-08T22:20:00.005-08:002014-03-08T22:20:51.371-08:00Tools News: Plane Geometry Drawer v1.3 is available on the AppStrore<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMck6cTAgRwBP7djB8uRRmEnbY7OhuiiUfy0Y2I6jLv-HOk_2eX4mMvq09oOalP_Na28xy2tMoisvJIIOU6u0Sif7QRbdtmUXm-Hqk10n2cFws4A5DBgQJjpz3ytSiv1tAaQwSAgGHxjrF/s1600/IMG_7180.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMck6cTAgRwBP7djB8uRRmEnbY7OhuiiUfy0Y2I6jLv-HOk_2eX4mMvq09oOalP_Na28xy2tMoisvJIIOU6u0Sif7QRbdtmUXm-Hqk10n2cFws4A5DBgQJjpz3ytSiv1tAaQwSAgGHxjrF/s1600/IMG_7180.jpg" height="320" width="180" /></a></div>
The new version (1.3) of "Plane Geometry Drawer" has the following new features/improvements:<br />
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<b><i><br /></i></b>
<b><i>1) Tools for set colour for lines and circles</i></b><br />
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When any line/circle is selected, the small circle will appear at the top of the drawing area (on the left). Touch on it to select colour for the selected line/circle.<br />
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<b><i>2) Improvement of the objects selection tools</i></b><br />
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Now if you touch on the "pink" circle, all options are listed and you could select the objects type by one touch. One more change: for selecting a line, choose the line come from/to a selected point.<br />
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<br /><a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8"><span style="color: blue;">FREE Download!</span></a><br /><br /><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgUFHx_QfLwYJ7ql5Pz_gvXUxPy274DDSs6uyoE82gXxWrpxDXhWIoUKeEorK5jSTijIBXs7mzowuQ-56lWJq8slHha11hALDUgSqNkmwtScf3ay2FM1JRBIxtKVWOmwiytc0pgP5a19SNv/s1600/IMG_7183.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgUFHx_QfLwYJ7ql5Pz_gvXUxPy274DDSs6uyoE82gXxWrpxDXhWIoUKeEorK5jSTijIBXs7mzowuQ-56lWJq8slHha11hALDUgSqNkmwtScf3ay2FM1JRBIxtKVWOmwiytc0pgP5a19SNv/s1600/IMG_7183.jpg" height="320" width="180" /></a><b><i>3) New keyboard with more geometry symbols</i></b><br />
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The new geometry keyboard add more geometry and math symbols and some of greek characters.<br />
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<b><i>4) Naming points</i></b><br />
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Automatic increase the next points name for complex names like A', B', (O), (I),... For example if you named a point as A', you can name the next point(s) as B', C' by dragging the naming tool for the point(s).<br />
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<a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8" style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13.333333015441895px; line-height: 12.319999694824219px; text-decoration: none;"><span style="color: blue;">FREE Download!</span></a><br style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13.333333015441895px; line-height: 12.319999694824219px;" /><br style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13.333333015441895px; line-height: 12.319999694824219px;" /><br />
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09.3.2014</div>
<div style="background-color: white; color: #333333; font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 20px;">
<div class="_wk" style="font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13px; line-height: 18px;">
<a href="https://www.facebook.com/pages/Geometry-Drawer/1449696861914126?ref=hl" style="color: #888888; font-size: 13px; text-decoration: none;">Geometry Drawer on Facebook</a></div>
<div class="_wk" style="font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13px; line-height: 18px;">
<a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8" style="color: #888888; text-decoration: none;">Plane Geometry Drawer for iPhone</a></div>
</div>
<div class="separator" style="background-color: white; clear: both; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13.333333015441895px; line-height: 12.319999694824219px; text-align: center;">
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-79936614825223568902014-03-07T08:45:00.002-08:002014-03-07T08:48:14.994-08:00Practice: how to draw rectangle and angle<i><b><span style="color: #274e13;">This article shows how to draw the rectangle ABCD, AD = 2AB and point E is on side BC and angle BAE = 15<span style="background-color: white; font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13.333333015441895px; line-height: 18px;">°</span></span></b></i><br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi835P6P9YlemSmXU6ZeSvH2zf1RESCIUh0wxbn6sPQ9bY8pMyAMm9EYW34jnIINhmhGauKKKDhmeZwqQ7gFHV-PuXlOrxypuBpLWAUyXXfw-3ojwdboGeorwnasatuy7dSsXWv8p5v5H-f/s1600/IMG_7174.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi835P6P9YlemSmXU6ZeSvH2zf1RESCIUh0wxbn6sPQ9bY8pMyAMm9EYW34jnIINhmhGauKKKDhmeZwqQ7gFHV-PuXlOrxypuBpLWAUyXXfw-3ojwdboGeorwnasatuy7dSsXWv8p5v5H-f/s1600/IMG_7174.jpg" height="248" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">picture 1</td></tr>
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<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjnJQF4x2b-bAVrcSF0B73o962fUgywMB4_1axNs8BO8shyXv6_Rud94V1WqgV9YXDw1V8CFfFt3RjiE5DCt2agBbxeKw6Q65OFTodB_Jj0oMgXlwl3tDqGP0gysO66A1rKPRSuNsm702Ek/s1600/IMG_7175.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjnJQF4x2b-bAVrcSF0B73o962fUgywMB4_1axNs8BO8shyXv6_Rud94V1WqgV9YXDw1V8CFfFt3RjiE5DCt2agBbxeKw6Q65OFTodB_Jj0oMgXlwl3tDqGP0gysO66A1rKPRSuNsm702Ek/s1600/IMG_7175.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">picture 2</td></tr>
</tbody></table>
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There are different ways to draw a geometry constuction shown in picture 1. The following steps are one of constructing methods:<br />
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1) From the basic tools select the "add line" tool and add a line, let name it AD<br />
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2) Select point A by touching on it, the points tools will be showed<span style="background-color: white; color: #37404e; line-height: 18px;"><span style="font-family: Times, Times New Roman, serif;"><br /></span></span>
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<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: left; text-align: left;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdVscjvn1dp7Sd5rmk0InbLOLlKEzfhL-bR4L89TQ5D_wCJJXcjXaoe2Pi3ADR5EhPWvEkeQDDHiysGTEMxxKwqxrMIAEHz0Ttkr6sWESn30imvTr6Ym6gMGxCuupMTwzP9Wrd_R4mAB9A/s1600/IMG_7177.jpg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdVscjvn1dp7Sd5rmk0InbLOLlKEzfhL-bR4L89TQ5D_wCJJXcjXaoe2Pi3ADR5EhPWvEkeQDDHiysGTEMxxKwqxrMIAEHz0Ttkr6sWESn30imvTr6Ym6gMGxCuupMTwzP9Wrd_R4mAB9A/s1600/IMG_7177.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">picture 3</td></tr>
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<br /><br />3) Use draw perpendicular line from A, draw AB ⊥ AD and AB = 1/2 AD (picture 2)<br /><br /><br />
4) Select point B, and draw BC // AD, BC = AD, draw line from C to D. Now we have a rectangle ABCD with AD = 2AB.<br />
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5) Select point A and from A draw arbitrary line AE, and drag the protractor to point A (picture 3), use protractor to rotate AE such as angle EAD = 75°.<br />
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6) Use change length tool to move point E to line BC. In the end we have the construction as required.<span style="background-color: white; color: #37404e; line-height: 18px;"><span style="font-family: Times, Times New Roman, serif;"><br /></span></span>
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<div style="background-color: white; color: #333333; font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 20px;">
7.3.2014</div>
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<a href="https://www.facebook.com/pages/Geometry-Drawer/1449696861914126?ref=hl" style="color: #33aaff; font-size: 13px;">Geometry Drawer on Facebook</a></div>
<div class="_wk" style="font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13px; line-height: 18px;">
<a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8" style="color: #888888; text-decoration: none;">Plane Geometry Drawer for iPhone</a></div>
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<span style="background-color: white; color: #37404e; line-height: 18px;"><span style="font-family: Times, Times New Roman, serif;"><br /></span></span>
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<span style="background-color: white; color: #37404e; font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13.333333015441895px; line-height: 18px;"><br /></span>KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-79905398946397284622014-03-07T04:43:00.002-08:002014-03-07T04:55:55.154-08:00Drawing with selected line<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghzio8UmMaXvn3jVXwb5PiNI0d8DZEvXD63A8F7CI8Ks5NIMb_KipSzsRCT81ld7K9hXLubUKN9O_SM7mVvG8Kjl4X55EtCUG3emis_A4pM_t_n2K3jI4xEk9DqGJDszC9D2plBuMZBkJP/s1600/IMG_7171.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEghzio8UmMaXvn3jVXwb5PiNI0d8DZEvXD63A8F7CI8Ks5NIMb_KipSzsRCT81ld7K9hXLubUKN9O_SM7mVvG8Kjl4X55EtCUG3emis_A4pM_t_n2K3jI4xEk9DqGJDszC9D2plBuMZBkJP/s1600/IMG_7171.jpg" height="320" width="180" /></a></div>
With the selected (straight) line segment (or "line" in short) the "Plane Geometry Drawer" app supports variety set of users actions:<br />
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1) Add a new point on selected line AB (highlighted)<br />
2) Change line style solid <-> dotted<br />
3) Draw a new line perpendicular to the selected line<br />
4) Draw a new line parallel to the selected line<br />
5) Change the length of the line<br />
6) Draw extensions of the line<br />
7) Draw a circle tangent to the line<br />
8) Draw a cicle with selected line as diameter<br />
9) Draw a triangle with selected line as one side and with given angle<br />
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With the PRO version you can also mark the selected line with different mark signs (see the picture). <br />
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<span style="color: red;">New!!!</span> <span style="color: #cc0000;">With the new, coming soon version 1.3, the tools will allow users to choose color for lines & circles. </span><br />
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<div style="background-color: white; color: #333333; font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 20px;">
06.3.2014</div>
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<a href="https://www.facebook.com/pages/Geometry-Drawer/1449696861914126?ref=hl" style="color: #888888; font-size: 13px; text-decoration: none;">Geometry Drawer on Facebook</a></div>
<div class="_wk" style="font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13px; line-height: 18px;">
<a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8" style="color: #888888; text-decoration: none;">Plane Geometry Drawer for iPhone</a></div>
</div>
KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-3538159519560911572014-03-07T04:23:00.004-08:002014-03-07T04:26:35.083-08:00Drawing with selected point objects<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiD9qyXQszZIb75JmM0uXfe7C541tnoo2FpqUh5PwzXpxTNaHbSf6pCOQDnkAKuSzcyjpNsZ14teMMAGb4ptselT9OPhrO13qjKOy1jlOoEgLKnSnLKfLzjPSefFoKXFp0w-g_kpazUDPjU/s1600/IMG_7170.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiD9qyXQszZIb75JmM0uXfe7C541tnoo2FpqUh5PwzXpxTNaHbSf6pCOQDnkAKuSzcyjpNsZ14teMMAGb4ptselT9OPhrO13qjKOy1jlOoEgLKnSnLKfLzjPSefFoKXFp0w-g_kpazUDPjU/s1600/IMG_7170.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Draw line from A perpendicular to CD</td></tr>
</tbody></table>
When you tap on any point of construction, the point will be highlighted by the small green circle around it and point tools will be appeared on the bottom of screen.<br />
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There many useful tools for drawing with selected points (A). To select a point touch on it or use selection tools (point option):<br />
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1) Draw line form point A to other points or new point<br />
2) Draw straight line from point A perpendicular to arbitrary line. A picture shows how to do that.<br />
3) Draw straight line from point A parallel to arbitrary line<br />
4) Draw an arc of center A, with required length on any direction<br />
5) Draw a circle with a diameter from A to any point<br />
6) Draw a circle passing through A and tangent line<br />
7) Draw a circle of center A and passing through any other point<br />
8) Draw a circle passing through A and two other points<br />
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Simply drag the tool and drop it where you needed!<br />
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06.3.2014</div>
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<div class="_wk" style="font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13px; line-height: 18px;">
<a href="https://www.facebook.com/pages/Geometry-Drawer/1449696861914126?ref=hl" style="color: #888888; font-size: 13px; text-decoration: none;">Geometry Drawer on Facebook</a></div>
<div class="_wk" style="font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13px; line-height: 18px;">
<a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8" style="color: #888888; text-decoration: none;">Plane Geometry Drawer for iPhone</a></div>
</div>
KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-6390453224092378042014-03-07T04:01:00.006-08:002014-03-07T04:24:08.818-08:00Users Guides: How to draw basic geometry objects<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLW6MhcQm3Q4PNRE2x_j4wu-0duq1iX0lTeWVYG_8srF8dC5_lgYO0Qg9d_zj1PQv9BZTwuuEAHz3NQsj8EwGXU4DjXCJeS093XJ3q55JjdATGfSYAuRZB3HRclR70sKPHRSmU86JduLWM/s1600/IMG_7169.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLW6MhcQm3Q4PNRE2x_j4wu-0duq1iX0lTeWVYG_8srF8dC5_lgYO0Qg9d_zj1PQv9BZTwuuEAHz3NQsj8EwGXU4DjXCJeS093XJ3q55JjdATGfSYAuRZB3HRclR70sKPHRSmU86JduLWM/s1600/IMG_7169.jpg" height="320" width="180" /></a><br />
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Drawing basic objects (point, line, circle, angle, triangle, square, ...) is easy task: tap on the respective tool icon, drag it to the drawing area and drop it where you want. But before that you need to show basic tools by tapping on any point on the drawing area (not on existing objects).<br />
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<span style="color: red;">Usually, use the basic tools to add an "basic object" or an "main object" or the first object of your construction. </span><br />
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The tools adds objects independently, the new object doesn't have any relations with previous objects.<br />
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To draw objects with relations with existing objects, we use other toolsets: points tools, line tools,....<br />
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06.3.2014</div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-19844006479884272172014-02-19T22:15:00.001-08:002014-02-22T23:13:54.033-08:00Plane Geometry Drawer v1.2 is now available on the App Store<b>Tools News: <a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8">FREE Download</a></b><br />
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The new version of the drawing tools is now available on the Apple Appstore is running on devices with iOS 7.0 or later. The version provides new useful features:<br />
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<b><i>New tools to select drawing objects on a iPhone/iPod Touch screens</i></b>: Tools for selecting an object to work with. Now you can choose the needed objects which are closed to others. Usually this is an difficult action:<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_9CufaeinzYzfmwANd0_qe3Og6ssEK84_Aumqquzybg4j9EsUY-2i8cjvxkHnpDWkmqEaZ1MhkEfNJtTZzngZZW5U_7oV9A9QwSEEzWCnOen0j0l6KdyiF56EudMKJuLK5Sm39vBMk4VJ/s1600/IMG_7027.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_9CufaeinzYzfmwANd0_qe3Og6ssEK84_Aumqquzybg4j9EsUY-2i8cjvxkHnpDWkmqEaZ1MhkEfNJtTZzngZZW5U_7oV9A9QwSEEzWCnOen0j0l6KdyiF56EudMKJuLK5Sm39vBMk4VJ/s1600/IMG_7027.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">It is not easy to select objects like: angle AHP or segment BH</td></tr>
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In the following screen you can see a small pink circle on the top-left of the drawing area. It will appear when you touch on any point of the drawing picture. There are selection modes for different types of objects: angles, triangles, circles, lines and points. The initial mode is for selecting a line, <i>touch on the pink circle to change the working mode</i>.<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDQ9y0UcZGzMjIxBt-JJ2_u9eZ1NgjQzkZ32uoi3UVztMeReph2XXJ-tBGXxaYNxMQQRJ25CL2MQ1glBKWBQu6orjHbq-0P_oolF_qQ_Xu5WoQiJ6Mr-wITtkAUTrH3LQcfNC3gxxA4Idn/s1600/IMG_7025.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDQ9y0UcZGzMjIxBt-JJ2_u9eZ1NgjQzkZ32uoi3UVztMeReph2XXJ-tBGXxaYNxMQQRJ25CL2MQ1glBKWBQu6orjHbq-0P_oolF_qQ_Xu5WoQiJ6Mr-wITtkAUTrH3LQcfNC3gxxA4Idn/s1600/IMG_7025.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Plane Geometry Drawer for iPhone v1.2: Object selection tools (pink circle)</td></tr>
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To use selection tools, please follow the steps below:<br />
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1) Touch on random point of the drawing construction, pink circle will appeared<br />
2) Pan on the drawing area from any point to move the circle to the required object:<br />
- In point mode: move close to the point to be selected<br />
- In line mode: move close to the midpoint of the line<br />
- In angle mode: move to the midpoint of the line formed by two points on two rays of the angle<br />
- In triangle mode: Selected one vertex of triangle, then move the pink circle to the midpoint of the opposite side of the needed triangle<br />
- In circle mode: Move the pink circle to any point near the circle<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAUSP1FUO8hDCdW9s8aqk-VxmFr0z-M0PnmIQj9sf4YmBRBeFPLdAvsbrgdrwa9dEYLDLMQxBi_t7xG-lo3S7nvfgeJsSdfI0RkhEQYBiGbHTfzuKqYUfjOjsJ-szEp7WBL0v-qVmWnG4c/s1600/IMG_7042.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAUSP1FUO8hDCdW9s8aqk-VxmFr0z-M0PnmIQj9sf4YmBRBeFPLdAvsbrgdrwa9dEYLDLMQxBi_t7xG-lo3S7nvfgeJsSdfI0RkhEQYBiGbHTfzuKqYUfjOjsJ-szEp7WBL0v-qVmWnG4c/s1600/IMG_7042.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Plane Geometry Drawer for iPhone v1.2: Selected a line segment OH</td></tr>
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<b><i>Toolset for marking right angles or equal line segments/angles</i></b>: This feature is available for PRO version only. It helps users to mark equal/not equal angles or line segments. The toolset will appear on the right side of drawing area when any angle or line is selected! Touch on the button to mark the selected objects.<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYd9o9gGBihVY6KUBebpL_urqHl3RSpWffQGVDkoMaTE7WxOVGwfQ3iRGvoKPlbHarP_R5b7wAuflPwYcHtXIgbNOUjFqzk7RZw5SBt9RZtfd6FTA4GMtlHI2RrAAkNFuRc7bS1FwgrHAE/s1600/IMG_7037.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYd9o9gGBihVY6KUBebpL_urqHl3RSpWffQGVDkoMaTE7WxOVGwfQ3iRGvoKPlbHarP_R5b7wAuflPwYcHtXIgbNOUjFqzk7RZw5SBt9RZtfd6FTA4GMtlHI2RrAAkNFuRc7bS1FwgrHAE/s1600/IMG_7037.jpg" height="320" width="306" /></a></div>
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<b><i>Other improvements</i></b>: some geometric symbols are added to the geometry text editor keyboard like: <span style="background-color: white; color: #37404e; font-family: 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 13.333333015441895px; line-height: 18px;">° (degree) or π (Pi) or some tools are added to existing toolsets.</span><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjk7qzHhTm-X9un50OAPbOORFMhyphenhyphenCh975O4iJmfIn4uenQYnBGDNmKNJYfJIECzIGMNiGsM3U_Qao7dwvEvL5ImNHJpP0bWRyBDQB3GkMD02D9xGuHHN-xDBHtUv4QDdXFpW43XNvwlmHtp/s1600/IMG_7044.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjk7qzHhTm-X9un50OAPbOORFMhyphenhyphenCh975O4iJmfIn4uenQYnBGDNmKNJYfJIECzIGMNiGsM3U_Qao7dwvEvL5ImNHJpP0bWRyBDQB3GkMD02D9xGuHHN-xDBHtUv4QDdXFpW43XNvwlmHtp/s1600/IMG_7044.jpg" height="320" width="180" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">Plane Geometry Drawer for iPhone v1.2: New additional symbols (geometry text editor tools)</td></tr>
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20.2.2014</div>
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<br />KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-91438006590388973612014-02-10T07:23:00.002-08:002014-02-10T07:23:50.000-08:00Requirements of geometric drawing: The science<div class="p1">
Science in the drawing geometry? This topic is somewhat a bit confusing, so to make things simpler we will consider a example.</div>
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<i>Example: Given ΔABC, the height AH, median AM (H and M on BC).</i></div>
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Very simple, isn’t it? Then how will we draw such a figure? We can draw as below:</div>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjn86ZHcjk71NvFFRwYDyLFnvCXetMb_FBnZ2Dkx4hI0XMh1-YPDQOQf_mb74i4zInaatLuAZ0qsQZ4c_o5i_usVx9hNss6oxsPF3luSdhM8aoNTch9sndUxn5tzXsxvGmSJh4rRaIXKx-0/s1600/IMG_6954.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjn86ZHcjk71NvFFRwYDyLFnvCXetMb_FBnZ2Dkx4hI0XMh1-YPDQOQf_mb74i4zInaatLuAZ0qsQZ4c_o5i_usVx9hNss6oxsPF3luSdhM8aoNTch9sndUxn5tzXsxvGmSJh4rRaIXKx-0/s1600/IMG_6954.jpg" height="260" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="text-align: start;"><span style="font-size: x-small;"><i>Figure 1: Triangle ABC, the height AH, the median AM</i></span></span></td></tr>
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It is absolutely accurate according to the original data. The problem here is that points M and H are too close together, and that will lead to difficulty to distinguish between them and between two lines AH and AM, making simple drawing become unrecognizable.</div>
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What if after that, we need to identify the circumcenter and center of gravity of ΔABC? In this case, we can say Figure 1 is "not scientific". In general, it would be a lot better if drawing like Figure 2 below:</div>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgR3RZ8oZgf2pEb4Uo-zeg6jpV9eXPpaw_YhYdDRnk3jZxSZTWt9abtBhrpE3oZeuuCb-_psENAfa1JYI4rLSpMWFeAtgFKM4iAajVOaxaLBPllNjOu9njTvMrS6kjiLlfX_8l0FzDn0Q9w/s1600/IMG_6955.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgR3RZ8oZgf2pEb4Uo-zeg6jpV9eXPpaw_YhYdDRnk3jZxSZTWt9abtBhrpE3oZeuuCb-_psENAfa1JYI4rLSpMWFeAtgFKM4iAajVOaxaLBPllNjOu9njTvMrS6kjiLlfX_8l0FzDn0Q9w/s1600/IMG_6955.jpg" height="236" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i style="font-size: small; text-align: start;">Figure 2: Triangle ABC, the height AH, the median AM</i></td></tr>
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Similarly, it can be noticed that Figure 3 or Figure 4 below are not scientific (except for cases that are required to draw as such):</div>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiLrqFUCxw7R6eloA7JJRlteWsCmnzxH8PuiBn0BTWvPBsQzc6UEtW3efIKeOBlt4Tq6wdAXdq2QV_2WCi0YaUiRwWG7gQOWrD2N6jn6sIsvCumIODzPbbsXzJjeS7RRE7A9HyddcbkI_3/s1600/IMG_6956.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiLrqFUCxw7R6eloA7JJRlteWsCmnzxH8PuiBn0BTWvPBsQzc6UEtW3efIKeOBlt4Tq6wdAXdq2QV_2WCi0YaUiRwWG7gQOWrD2N6jn6sIsvCumIODzPbbsXzJjeS7RRE7A9HyddcbkI_3/s1600/IMG_6956.jpg" height="266" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i style="font-size: small; text-align: start;">Figure 3: Triangle ABC, the height AH, the median AM</i></td></tr>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKaXxho1pnKhtMl20XMMDhT3AkpqQd7w-XlyBh0ZP5wo88g68UEGzUbIZKig7_bLNn-VG-QtvfJLL6qtjoQsGQrhouIMP5bPwdnf74qHsAsrtAHJFvCo_-rsOFn2dohEmFAURmL36bIxjP/s1600/IMG_6957.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKaXxho1pnKhtMl20XMMDhT3AkpqQd7w-XlyBh0ZP5wo88g68UEGzUbIZKig7_bLNn-VG-QtvfJLL6qtjoQsGQrhouIMP5bPwdnf74qHsAsrtAHJFvCo_-rsOFn2dohEmFAURmL36bIxjP/s1600/IMG_6957.jpg" height="142" width="320" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i style="font-size: small; text-align: start;">Figure 4: Triangle ABC, the height AH, the median AM</i></td></tr>
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So to draw scientifically, what do we have to note? The following suggestions may be helpful and should be applied when you start drawing (if not bound by other facts):</div>
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1) When drawing angles or triangles, try not to draw the vertices with special values, such as 30°, 60°, 45°, 90°, 120°, 180°, ...</div>
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2) The edges of the triangles, parallelograms or rectangle should not be drawn approximately equal. For example, if you draw a rectangle ABCD, the lengths of AB and BC should not be close in value and also one should not double the other!</div>
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3) If you draw a trapezoid or a triangle, you should not draw an isosceles one. Normally, the two trapezoidal sides should be approximately 1/3 different in lengths. For example, the trapezoid ABCD, AB // CD, you should draw the side BC = about 2/3 or 4/3 of the DA.</div>
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4) Use the solid or interrupted lines logically. For example, if two sides of a triangle are in solid lines, the 3rd edge should be also be solid. Or when you have multiple solid lines converge at one point, then an additional line should be an interrupted line...</div>
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There may be many other points to note, but the above points have helped a great deal. Most importantly, the figure should be easy to read, easy to understand and highlight the given facts as well as the relationships between them.</div>
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<i>Related articles:</i></div>
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<i><br /></i></div>
<div>
<a href="http://planegeometrydrawer.blogspot.com/2014/02/geometric-drawing-requirements-accuracy.html">Geometric drawing requirements: accuracy</a></div>
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<a href="http://planegeometrydrawer.blogspot.com/2014/02/plane-geometry-requirements-for-drawing.html">Requirements for geometry drawing: overview</a></div>
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9.2.2014</div>
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<a href="https://itunes.apple.com/us/app/plane-geometry-drawer/id777703891?ls=1&mt=8" style="color: #888888; text-decoration: none;">Plane Geometry Drawer for iPhone</a></div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-90535891309689136192014-02-06T07:56:00.002-08:002014-02-06T07:56:31.375-08:00Geometric drawing requirements: accuracy<div class="p1">
<i><b>Related article(s)</b></i>: <a href="http://planegeometrydrawer.blogspot.com/2014/02/plane-geometry-requirements-for-drawing.html">Plane geometry: The requirements for the drawing - an overview</a></div>
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In the previous article about an overview of the requirements for the plane geometry drawing mentioned the requirements for accuracy, scientific and aesthetics. This article will further analyse the first requirement: accuracy! Accuracy in drawing is especially important and reflected in the following key points:</div>
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1) Angles with a given value as of 90 °, 60 °, 45 °, 30 °, ... need to be accurately constructed. Or an angle bisector must divide that angle into two congruent parts,... Let’s look at the diagram Pic 1: </div>
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Suppose ΔABC = ΔJIK and 2 straight lines AD and IL are the bisector of the angle BAC and JIK respectively. Did you think that both figures were correct?</div>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0VK80L7Cbl0eI-dc01RMqBlUkc5PsZZ5dp15BbtezyrezXZGvI128pmDXqzVBQ-fLotGvrorbIWDj82bTiHZybUHj4wRjHUkEXy6CF-qGYvpBKFuFHVJPfMNh2gVCByuSA9D2wbdrv6xI/s1600/IMG_6799.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0VK80L7Cbl0eI-dc01RMqBlUkc5PsZZ5dp15BbtezyrezXZGvI128pmDXqzVBQ-fLotGvrorbIWDj82bTiHZybUHj4wRjHUkEXy6CF-qGYvpBKFuFHVJPfMNh2gVCByuSA9D2wbdrv6xI/s1600/IMG_6799.jpg" height="320" width="236" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="text-align: start;"><span style="font-size: x-small;"><i>Pic 1: AD is the bisector of angle BAC and IL is the angle bisector of JIK</i></span></span></td></tr>
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Next, draw two circles C1, C2 of center D and L that are tangent to the rays of <span class="s1">∠</span>BAC and <span class="s1">∠</span>JIK respectively! Pic 2 shows what would happen:</div>
<div class="p1">
<br /></div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7gjNKy19t5u4TS542uDxS4Ou5BrEzUhccgsUAa1C5Pk1RUdj-ezlynF1phrrm6UYWqKx0iDnDoS0pT_WBVk-zddnGXfusxAJhsQbPmA2eAIC35W-Lm_ii6DQfu5iaDurvHZdqBPH7M1Uf/s1600/IMG_6800.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7gjNKy19t5u4TS542uDxS4Ou5BrEzUhccgsUAa1C5Pk1RUdj-ezlynF1phrrm6UYWqKx0iDnDoS0pT_WBVk-zddnGXfusxAJhsQbPmA2eAIC35W-Lm_ii6DQfu5iaDurvHZdqBPH7M1Uf/s1600/IMG_6800.jpg" height="400" width="255" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="text-align: start;"><span style="font-size: x-small;"><i>Pic 2: Draw a circle of center D, tangent to AB and a circle of center L, tangent to IJ</i></span></span></td></tr>
</tbody></table>
<div class="p1">
<br /></div>
<div class="p1">
C1 touches both rays AB and AC while C2 only touches IJ but not IK! The problem is that bisector IL is drawn incorrectly (specifically in this case <span class="s1">∠</span>JIL < <span class="s1">∠</span>LIK). Clearly, although looking at pic 1, both figures seem to be good, the 2nd picture in fact is inaccurate and will not allow us to continue to solve the problem as figure Pic 2.</div>
<div class="p2">
<br /></div>
<div class="p1">
2) Accuracy is also reflected in the correlation in length of the line segments. For example, given the triangle ΔABC with AC > AB, in the drawing, the length of AC should be greater than AB. If given two straight lines of equal length, then their lengths must be equal in the drawing. Or if you draw two perpendicular lines, the angle created by two lines in the figure must be exactly 90°. For example, take a look at the figure below:</div>
<div class="p2">
<br /></div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiv44EUe_rBuXZkutJ86HeOzCC8eHw2FoHFqzUzyucWHRshNUSuxUlr9cGSdrdfgHdheWjITg1Or7lyDA9AN3MENqtzxao27-2VDsKzaAUgrye-hu-59CvVu1E-5CWNy9Ax9cwhIIsFOg-e/s1600/IMG_6802.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiv44EUe_rBuXZkutJ86HeOzCC8eHw2FoHFqzUzyucWHRshNUSuxUlr9cGSdrdfgHdheWjITg1Or7lyDA9AN3MENqtzxao27-2VDsKzaAUgrye-hu-59CvVu1E-5CWNy9Ax9cwhIIsFOg-e/s1600/IMG_6802.jpg" height="400" width="280" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="text-align: start;"><span style="font-size: x-small;"><i>Pic 3: Given ΔABC, angle A = 90°</i></span></span></td></tr>
</tbody></table>
<div class="p2">
<br /></div>
<div class="p1">
Which triangle is correctly drawn? Are both or either correct? Given angle A is a right angle, hence if we draw a circle of diameter BC, then A will be one of the points on such circle. We have:</div>
<div class="p2">
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFyEuuZDRTiwC1JwJwLAH_9tVAeK5jGEfLCcS7EgMkGkGQ6Mfg3lusl_0vEStyVTI-kc1Vt-tt4ZWoHEFhpClQkj98mWkIBNc50x8rT2yEqjSUdGahFQMfE4Kul1fsbT2dwP0kjjXy2fjA/s1600/IMG_6801.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFyEuuZDRTiwC1JwJwLAH_9tVAeK5jGEfLCcS7EgMkGkGQ6Mfg3lusl_0vEStyVTI-kc1Vt-tt4ZWoHEFhpClQkj98mWkIBNc50x8rT2yEqjSUdGahFQMfE4Kul1fsbT2dwP0kjjXy2fjA/s1600/IMG_6801.jpg" height="400" width="230" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="text-align: start;"><span style="font-size: x-small;"><i>Pic 4: for ΔABC, angle A = 90°, draw a circle of diameter BC</i></span></span></td></tr>
</tbody></table>
<div class="p2">
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<br />
<div class="p1">
Figure Pic 4 showed that apparently the below triangle in Figure Pic 3 is not correct. And if we continue to draw, the figure would become unreasonable. Those above are just two of many examples showing that if we did not draw accurately, we could not continue to study or solve geometric problems. The accuracy in drawing is absolutely important and is the first requirement!</div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-23842834000258351812014-02-06T01:48:00.001-08:002014-02-06T01:54:02.577-08:00Napoleon Problem: Magic of the circles<b>Napoleon Bonaparte </b>(1769 - 1821) well known as a genius general, a political leader, an Emperor of the French (1804 - 1814). Napoleon also was known to be an excellent amateur mathematician! The following problem to find a centre of a given circle widely known as "Napoleon Problem":<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcVuI6O3Ns0ZOhZ5jshpql2KhfcKPu-YSaxC026o-bcgwgAN2-2XAU6h7BMjTnrDdZ_OzZGkK-9lcj7srnUrAxfSj_giQXWkqDga-XWdpQ86dC2Iyksl1eI2mFf2rzOjDWU3ZlZahBunJj/s1600/1613778_353535708118905_1419547272_n.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhcVuI6O3Ns0ZOhZ5jshpql2KhfcKPu-YSaxC026o-bcgwgAN2-2XAU6h7BMjTnrDdZ_OzZGkK-9lcj7srnUrAxfSj_giQXWkqDga-XWdpQ86dC2Iyksl1eI2mFf2rzOjDWU3ZlZahBunJj/s1600/1613778_353535708118905_1419547272_n.jpg" height="320" width="307" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>pic 1: Find a center of a given circle, using the pair of compasses alone (no straight edge)</i></td></tr>
</tbody></table>
The problem looks simple to be understood, but the solution is surprised magic of 7 circles! The following figure describes how to detect a centre of a given circle:<br />
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgR5vsawUbkA1OmUIiRYRDBv9hMgas2AZVk90rEh_l8YNkwrekBROYgthlX-71gknydISEfyjs-M_krxKZNeN4jynAkEVuvoNaL41vpYjO3XqTdssraxKcHM5hm23G2AUdG9Fhyphenhyphen6lqgQznm/s1600/IMG_6911.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgR5vsawUbkA1OmUIiRYRDBv9hMgas2AZVk90rEh_l8YNkwrekBROYgthlX-71gknydISEfyjs-M_krxKZNeN4jynAkEVuvoNaL41vpYjO3XqTdssraxKcHM5hm23G2AUdG9Fhyphenhyphen6lqgQznm/s1600/IMG_6911.jpg" height="320" width="308" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>pic 2: How to detect a centre of a given circle</i></td></tr>
</tbody></table>
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<div class="separator" style="clear: both; text-align: center;">
</div>
<b>Constructing steps:</b><br />
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1) Select an arbitrary point A on the circle and draw a circle C2 of centre A, which intersects with a given circle (C1) at points B and C.<br />
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2) Draw circles C3, C4 of centres B, C and radius r = BA, which intersect each other at point D (D ≢ A), and draw a circle C5 of centre D and radius r = DA.<br />
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3) Circle C 5 intersects with circle C2 at points E, F. Draw circles C6, C7 of centres E, F and radius r = EA, the two circles intersect each other at A and an other point - this point is a centre of given circle C1!<br />
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It is a magic, isn't it? The problem is solved by drawing of 6 new cicles by compass alone, and there is not any lines or straight segments on the figue!<br />
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<b>Proof:</b><br />
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To prove that the above solution works, we will prove the following lemma:<br />
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<div class="p1">
<b><i>Lemma</i></b>: Lets (C) be a circle of centre O radius a, AE be a diameter of circle (C), B, C be points on (C) and BC is perpendicular to OA. Point D is on AE and BD = AD, b is length of BC. So we will have AD = b*b/r (see picture)</div>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgIsspMEuZmPSLmTlD_GhuUamF6WFXewUUJjP7Oo-I_Ms4PSbuuNWBi96T4_eosAaeXSsjLSTYUlBLx4Qqn3qa2zZhSJd186_Q2DNpKuh7JlUHQ9O8UYi0ucZhj4Z2sOZvqHnBew2OmyEgi/s1600/IMG_6918.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgIsspMEuZmPSLmTlD_GhuUamF6WFXewUUJjP7Oo-I_Ms4PSbuuNWBi96T4_eosAaeXSsjLSTYUlBLx4Qqn3qa2zZhSJd186_Q2DNpKuh7JlUHQ9O8UYi0ucZhj4Z2sOZvqHnBew2OmyEgi/s1600/IMG_6918.jpg" height="320" width="284" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>pic 3: Lemma</i></td></tr>
</tbody></table>
<i><b>Lemma's proof</b></i>: Triangle EBA has a right angle B, and EA ⊥ BC => ∆EBA ∾ ∆BHA => EA/BA = BA/AH => AH = BA * BA / EA = (b x b) / (2 x a) => AD = 2AH = b^2 / a.<br />
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In the above construction of centre (pic 2), the configuration of lemma appears twice:<br />
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1) Let O be a centre of a circle C1, radius OA = r, see a circle C1, points A, B, C, D and O. Application of lemma give us: R = BA = CA = BD = CD => AD = R x R / r<br />
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2) See a circle C5 of centre D, radius R, points E, F, A lie on C5, and point O' = C6 ∩ C7 => O' ∈ DA and EO' = EA = FO' = FA. Application of lemma gives us: AO' = R * R / AD = R * R / (R * R / r) = r => O' ≣ O (<i>or O' is a centre of a circle C1</i>).<br />
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Reference: wikipedia.org<br />
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<br />KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-88408045845110842302014-02-04T09:01:00.002-08:002014-02-06T07:57:39.490-08:00Plane geometry: The requirements for drawing - an overview<div class="p1">
A good illustration has a great help for learning, problem solving and research. For solving the plane geometric problems, completing constructing already helps to solve half of the problem. The main benefits of a good drawing include:</div>
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1) Describing the requirements: good drawing accurately describe the original elements and the mathematical relations between them , highlighting the mathematical logics to help find solutions to the problems.</div>
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2) Supporting creativity: easier to visualise, analyse, relate these elements to advance with new objects/figures.</div>
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3) Creating compelling: Mathematic always has the hidden beauty of wisdom. A good picture help make a solution or an article more appealing, more attractive to readers. This is very important, whether you sit for an examination or make a scientific report.</div>
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So what requirements must a good drawing must? Here are some basic requirements:</div>
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1) The accuracy (<a href="http://planegeometrydrawer.blogspot.com/2014/02/geometric-drawing-requirements-accuracy.html">read more details ...</a>)</div>
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2) The scientific</div>
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3) The aesthetics</div>
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<span style="font-family: Courier New, Courier, monospace;"><b><i>Example</i></b>: Given ΔABC, centroid G and straight line d orbiting around G, cutting edge AB at M and edge AC at N. Prove that regardless of thethe specific location of the line d, we always have:</span></div>
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<span style="font-family: Courier New, Courier, monospace;"><br /></span></div>
<div class="p1">
<span style="font-family: Courier New, Courier, monospace;">1) AB/AM + AC/AN = 3</span></div>
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<span style="font-family: Courier New, Courier, monospace;">2) BM/AM + CN/AN = 1</span></div>
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<span style="font-family: Courier New, Courier, monospace;"><br /></span></div>
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<span style="font-family: Courier New, Courier, monospace;">Illustration for the above problem </span></div>
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<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgFCBPv5ton7FCY8O-vRVVu514Yo7ehzOxbLlgxi7Vq8egW3faej7O9foHt9KLN0t5imrziKSEuTx1Gbe4OKqPVqlduIhZhHseNvcfE0yLp6AtRI0blxV26qrTCja7okWwgZJUiQktyrBn1/s1600/IMG_6884.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgFCBPv5ton7FCY8O-vRVVu514Yo7ehzOxbLlgxi7Vq8egW3faej7O9foHt9KLN0t5imrziKSEuTx1Gbe4OKqPVqlduIhZhHseNvcfE0yLp6AtRI0blxV26qrTCja7okWwgZJUiQktyrBn1/s1600/IMG_6884.jpg" height="298" width="320" /></a></div>
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<span style="font-family: Courier New, Courier, monospace;">Look at the illustration! It clearly describes a way to find the solution for the problem, isn't it?</span></div>
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<div class="p1">
The next article will in turn to a closer analyse more in details of the above requirements as well as guiding principles to help practice basic drawing skills. </div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-50396738562896178202014-02-03T10:43:00.003-08:002014-02-03T10:43:59.240-08:00Problem & Solution (3): Draw a square ABCD with given vertext A, point M on BC, point N on CD <div class="p1">
<b>Problem</b>: <i>Construct a square ABCD given point A and two points M <span class="s1">∈</span> BC, N <span class="s1">∈</span> CD</i></div>
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<b>Solution:</b></div>
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<div class="p1">
Suppose we have constructed the square ABCD as required. Connect the points A, M and N. Observe figure 1, we can draw the following comments :</div>
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<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisDNcUWcWyAu7BaEqoBsG5-RgEObTarM8bmGg2I9ZNsVoTCvcbgrHfAAVrPUFbgzom3KhWTJK7pq3egeLBp7YKJG8nIJf2Sn_zedqYdxklY-vPj1Dmk_rE5XAYESaSY0QjTQKw_xGARc1P/s1600/IMG_6773.jpg" imageanchor="1" style="clear: left; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisDNcUWcWyAu7BaEqoBsG5-RgEObTarM8bmGg2I9ZNsVoTCvcbgrHfAAVrPUFbgzom3KhWTJK7pq3egeLBp7YKJG8nIJf2Sn_zedqYdxklY-vPj1Dmk_rE5XAYESaSY0QjTQKw_xGARc1P/s1600/IMG_6773.jpg" height="320" width="260" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>figure 1</i></td></tr>
</tbody></table>
<div class="p1">
1) <span class="s1">∠</span>BCD = π / 2 = > <span class="s1">∠</span>MCN = π/2, where M, N are known. Thus vertex C is on the circle C1 of diameter MN and is on the opposite side of the vertex A through MN.</div>
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<div class="p1">
2) Through A, construct a straight line perpendicular to AN and cuts the straight line CB at P. Consider the triangles ABP and DNA: AB = AD, ∠ADN = <span class="s1">∠</span>ABP = π/2, <span class="s1">∠</span>DAN = <span class="s1">∠</span>BAP = > ΔADN = ΔABP => AP = AN => point P is determined. So, C is an intersection of PM with a circle C1 (C should in the opposit side compared to point A through line MN )</div>
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3) <span class="s1">∠</span>MAN ≤ BAD = π/2</div>
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From the above analysis we can determine the square ABCD and point C as follows:</div>
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<br /></div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgZlkDAqjylsCRf8a8rdTetvD1LXVGzjdsfPrJ82yyJJRzhINreQoou-TAE95Deh5MrOQ3RoHqCiG7hzAbOgBd7WRQMYT9D_K_3JrvrGeuEfVhM-8ygXbmyMsAZD11EPMPfBShZcI50xaJn/s1600/IMG_6778.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgZlkDAqjylsCRf8a8rdTetvD1LXVGzjdsfPrJ82yyJJRzhINreQoou-TAE95Deh5MrOQ3RoHqCiG7hzAbOgBd7WRQMYT9D_K_3JrvrGeuEfVhM-8ygXbmyMsAZD11EPMPfBShZcI50xaJn/s1600/IMG_6778.jpg" height="320" width="268" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>figure 2</i></td></tr>
</tbody></table>
<div class="p1">
<br /></div>
<div class="p1">
<i>Step 1</i>: Construct the line segment MN, construct the circle C1 of diameter MN ( easily done with basic constructing skills )</div>
<div class="p2">
<br /></div>
<div class="p1">
<i>Step 2</i>: Construct the line segment NA and a straight line perpendicular to AN at A , get point P at the same plane side with point M ( through the straight lineAN ) so that AP = AN .</div>
<div class="p2">
<br /></div>
<div class="p1">
<i>Step 3</i>: Construct the straight line (d1) passing through P and M, which cuts circle C1 at C (a different point from M). Also construct the straight line (d2) through C and N. Through A, construct the straight lines perpendicular to (d1), (d2) at B and D respectively.</div>
<div class="p2">
<br /></div>
<div class="p1">
<i>ABCD is the square we need to construct (see figure 2)</i></div>
<div class="p1">
<br /></div>
<div class="p1">
Proof: According to how we construct, the four vertices A,B,C, D are all square angles = > ABCD is a rectangle ( * ) . Considering the triangles ABP and DNA : AN = AP ( as above ) , ADN = <span class="s1">∠</span>ABP = π/2; <span class="s1">∠</span>DAN = BAP => ∆ADN = ∆ABP => AD = AB (**). From ( * ) and ( ** ) we have ABCD is a square .</div>
<div class="p2">
<br /></div>
<div class="p1">
Justification :</div>
<div class="p2">
<br /></div>
<div class="p1">
a) The problem will have no solution if <span class="s1">∠</span>MAN > π/2, or</div>
<div class="p1">
b) or point C lies on the same plane side with point A through line MN (including the case such that C ≡ M. This case occurs when AP <span class="s1">≧</span> AP' with P' = AP <span class="s1">⋂</span> NM <=> ∠ANM = ∠ANP' <span class="s1">≦</span> ∠ANP = π/4 or P' = AP <span class="s1">⋂</span> NM <=> ∠ANM = ∠ANP' <span class="s1">≦</span> ∠ANP = π/4 <=> ∠ANM <span class="s1">≦</span> π/4 or <span class="s1">∠</span>AMN <span class="s1">≦</span> π/4 (see figure 3)</div>
<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhP3AJJJSfB0SBkRPckCqc6Qz7l5x63UJLEWozAzRNw2LjYA7AVTDrSLvRHvZOr2a40Egf1FuRLr6UhRkAyXn8ce7L02U6xr5IxgdnfCY-1IxDCQkCZcDxflqK_ZscVX2aWic0V8OjmqAvI/s1600/IMG_6780.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhP3AJJJSfB0SBkRPckCqc6Qz7l5x63UJLEWozAzRNw2LjYA7AVTDrSLvRHvZOr2a40Egf1FuRLr6UhRkAyXn8ce7L02U6xr5IxgdnfCY-1IxDCQkCZcDxflqK_ZscVX2aWic0V8OjmqAvI/s1600/IMG_6780.jpg" height="320" width="250" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><i>figure 3</i></td></tr>
</tbody></table>
<div class="p1">
<br /></div>
<br />
<div class="p1">
c) In the remaining cases the problem has a unique solution.</div>
<div class="p1">
<br /></div>
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4.2.2014</div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-30723628035415147282014-02-03T03:20:00.001-08:002014-02-04T07:54:10.635-08:00Problem & solution (1): Draw a circle which passes through a given point P and tangents on rays of given angle<b>Problem</b>: Let xAy be a given angle and P be a given point inside of ∠xAy. Construct a circle which passes through point P and tangents on rays Ax, Ay.<br />
<br />
<b>Solution</b>:<br />
<br />
<br />
<div class="p1">
Suppose that a circle C<span style="font-size: xx-small;">1</span>(O, OP) has been constructed (see a picture). We have following remarks:</div>
<div class="p2">
<br /></div>
<div class="p1">
1) Center O of drawn circle lies on a bisector Az of an angle xAy.</div>
<div class="p2">
<br /></div>
<div class="p1">
2) Let O<span style="font-size: xx-small;">1</span> be an arbitrary point on Az and C2(O<span style="font-size: xx-small;">1</span>, O<span style="font-size: xx-small;">1</span>J) be a circle of center O<span style="font-size: xx-small;">1</span> which tangents on Ay, then C<span style="font-size: xx-small;">2 </span>also tangents on Ax.</div>
<div class="p2">
<br /></div>
<div class="p1">
3) Let I, J be tangent points of C<span style="font-size: xx-small;">1</span>, C<span style="font-size: xx-small;">2</span> on Ay, then OI and O<span style="font-size: xx-small;">1</span>J are perpendicular to Ay => OI // O<span style="font-size: xx-small;">1</span>J => AO<span style="font-size: xx-small;">1</span> / AO = O<span style="font-size: xx-small;">1</span>J / OI (a)</div>
<div class="p2">
<br /></div>
<div class="p1">
4) Draw a straight line which passes O<span style="font-size: xx-small;">1</span> and is parrallel to OP. This line intersects AP at point Q. We have O<span style="font-size: xx-small;">1</span>Q / OP = AO<span style="font-size: xx-small;">1</span> / AP, on other side OP = OI => O<span style="font-size: xx-small;">1</span>Q / OI = AO<span style="font-size: xx-small;">1</span> / AO (b)</div>
<div class="p2">
<br /></div>
<div class="p1">
5) From (a) and (b) we have O<span style="font-size: xx-small;">1</span>Q = O<span style="font-size: xx-small;">1</span>J => Q is on the circle C<span style="font-size: xx-small;">2</span></div>
<div class="p2">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfSKARsKJpaNMdi1ThR89jaiyVOvaBE-P0Nugcy6NsMCqOPOJ4MnjyuT2OYRc6dEtHW_1P9_AHRDWYVPMMWQM7OLdO7BCM6G64L6Xo4QS4s6AmNBsKB-NwHClxeZxR6TQVnt8mnrT1-DML/s1600/IMG_6870.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfSKARsKJpaNMdi1ThR89jaiyVOvaBE-P0Nugcy6NsMCqOPOJ4MnjyuT2OYRc6dEtHW_1P9_AHRDWYVPMMWQM7OLdO7BCM6G64L6Xo4QS4s6AmNBsKB-NwHClxeZxR6TQVnt8mnrT1-DML/s1600/IMG_6870.jpg" height="342" width="400" /></a></div>
<div class="p2">
<br /></div>
<div class="p2">
<br /></div>
<div class="p1">
By the above, we can construct the circle C<span style="font-size: xx-small;">1</span> as bellowing steps:</div>
<div class="p2">
<br /></div>
<div class="p1">
1) Draw a bisector Az of an angle xAy, take an arbitrary point O<span style="font-size: xx-small;">1</span> on Az and draw a circle C<span style="font-size: xx-small;">2</span> of center O<span style="font-size: xx-small;">1</span> which tangents on Ay.</div>
<div class="p2">
<br /></div>
<div class="p1">
2) Let Q be a intersection point of C<span style="font-size: xx-small;">2</span> and straight line passing through A and P. From P draw a straight line which is parallel to O<span style="font-size: xx-small;">1</span>Q. Let O be an intersection of this new line with Az.</div>
<div class="p2">
<br /></div>
<div class="p1">
A circle of center O, radius OP is tangent on Ax and Ay and passes through point P!</div>
<div class="p2">
<br /></div>
<div class="p1">
<b>Proof</b>: Draw OI ⊥ Ay, I ∈ Ay and O<span style="font-size: xx-small;">1</span>J ⊥ Ay, J ∈ Ay => ∆ AOI ∾ ∆AO<span style="font-size: xx-small;">1</span>J and ∆ AOP ∾ ∆AO<span style="font-size: xx-small;">1</span>Q => OP / AO = O<span style="font-size: xx-small;">1</span>Q / AO1 = O<span style="font-size: xx-small;">1</span>J / AO<span style="font-size: xx-small;">1</span> = OJ / AO => OP = OJ = P ∈ C<span style="font-size: xx-small;">2</span>. In other side C<span style="font-size: xx-small;">1</span> tangents on Ay => C<span style="font-size: xx-small;">1</span> tangents on Ax = C<span style="font-size: xx-small;">1</span> is a required circle!</div>
<div class="p2">
<br /></div>
<br />
<div class="p1">
<b>Justification</b>: the problem has two distinct solutions if P is not on Ax, Ay. In other cases we have one unique solution!</div>
<div class="p1">
<br /></div>
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3.2.2014</div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-77259133866330072912014-01-30T07:34:00.004-08:002014-02-02T01:59:59.562-08:00Given angle xOy, two points A, B and line segment DE. Construct ΔABC such that ∠ACB = ∠xOy and BC = DE<div class="p1">
<br /></div>
<div class="p1">
<i>Given angle xOy, two points A, B and line segment DE. Construct ΔABC such that <span class="s1">∠</span>ACB = ∠xOy and BC = DE.</i></div>
<div class="p2">
<br /></div>
<div class="p1">
<b>Solution:</b></div>
<div class="p1">
<br /></div>
<div class="p1">
Suppose that we have constructed triangle ΔABC. Vertices A and B are known, hence we only need to determine vertex C. According to the question, ∠ACB = ∠xOy and BC = DE. Therefore point C is the intersection of a circle of centre B, radius r1 = DE (let call it C1) and the arc C2 which consists of a group of points C ' such that AC'B equals angle xOy.</div>
<div class="p2">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg0vbJSgCfWSjYIEmn_7UoxGsPZF-5QkK7Pvka33CpVTdExM1u8cduHh1SQC-vG6lK5djZ1nmeEtjz0lj7g_UpBB5yswfPm945TInFDOK1glR2CvNToUdQaXehZV3148mduNkmh79BdQyo8/s1600/IMG_6752.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg0vbJSgCfWSjYIEmn_7UoxGsPZF-5QkK7Pvka33CpVTdExM1u8cduHh1SQC-vG6lK5djZ1nmeEtjz0lj7g_UpBB5yswfPm945TInFDOK1glR2CvNToUdQaXehZV3148mduNkmh79BdQyo8/s1600/IMG_6752.jpg" height="400" width="270" /></a></div>
<div class="p2">
<br /></div>
<div class="p2">
<br /></div>
<div class="p1">
Since circle C1 is determined, the problem now revolves around constructing circle C2.</div>
<div class="p2">
<br /></div>
<div class="p1">
Let K be an arbitrary point on C2,<span class="s1"> so </span>∠AKB = ∠ACB = ∠xOy and ∠AKO + ∠KBAB + ∠KAB = 2π => ∠ΑΚΒ + ∠ΚΒΑ < 2π. Then we can construct C2 and triangle ABC as follows:</div>
<div class="p2">
<br /></div>
<div class="p1">
1) Take arbitrary point I on the straight line (d) passing through point B, so that <span class="s1">∠</span>IBA + <span class="s1">∠</span>xOy < 2π.</div>
<div class="p1">
2) Construct angle JIB, such that it is equal to angle xOy and J ∈ AB</div>
<div class="p1">
3) Construct a straight line passing through point A and parallel to JI, which is cutting line (d) at K. We will have <span class="s1">∠</span>AKB = <span class="s1">∠</span>xOy = <span class="s1">∠</span>JIB .</div>
<div class="p1">
4) Construct the circumscribed circle C2 of triangle ΔAKB, this circle cuts circles C1 at point C. Triangle ABC is a triangle we need to construct !</div>
<div class="p2">
<br /></div>
<div class="p1">
<b>Proof:</b> From above it is easy to prove that <span class="s1">∠</span>AKB = <span class="s1">∠</span>ACB = <span class="s1">∠</span>xOy = <span class="s1">∠</span>JIB and BC = DE .</div>
<div class="p2">
<br /></div>
<br />
<div class="p1">
<b>Justification:</b> Let V be the centre of circle C2, r2 is the radius of C2 , we have r2 = (AB/2 ) / sin(xOy) and the following circumstances:<br />
<br />
1) r2 < ( DE / 2 ) => the problem has no solution;<br />
2) r2 = DE/2 => the problem has one and only solution; and<br />
3) r2 > (DE/2) => the problem has 2 distinct solutions.</div>
<div class="p1">
<br /></div>
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30.1.2014</div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-78380231082442403672014-01-27T08:12:00.002-08:002014-02-02T01:59:31.944-08:00Plane geometry: constructing techniques (2)<div class="p1">
<b>(6) Constructing a straight line parallel to a given straight line (d) and passing through a given point A</b></div>
<div class="p2">
<br /></div>
<div class="p1">
Construct AH such that AH is perpendicular to (d). Construct a straight line (e) that passes through A and is perpendicular to the straight line AH , (e) is the straight line passing through A and (e) // (d) </div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXysOXAyZmljUOCF5t3AG_JhBMeQ8C-KzCiiU6xFVPw_99tEsE1LencKOVStXvXXLcwGMW2Iu2OrL-7CTJcTTWx8OSaVI7BOSc28SfG1QKOxxHhpmrE5z9KeCnbaixzSSrADHIUWdJ4vjJ/s1600/IMG_6713.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXysOXAyZmljUOCF5t3AG_JhBMeQ8C-KzCiiU6xFVPw_99tEsE1LencKOVStXvXXLcwGMW2Iu2OrL-7CTJcTTWx8OSaVI7BOSc28SfG1QKOxxHhpmrE5z9KeCnbaixzSSrADHIUWdJ4vjJ/s1600/IMG_6713.jpg" height="228" width="320" /></a></div>
<div class="separator" style="clear: both; text-align: center;">
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<div class="p2">
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<div class="p1">
<b>(7) Constructing the circumscribed circle (or circumcircle) of triangle ABC</b></div>
<div class="p2">
<br /></div>
<div class="p1">
Draw the perpendicular bisectors of BC and AC , the two lines intersect at O. The circle of center O and radius OA will pass through the points A, B and C (in other words, the circle of centre O, radius OA is the circumcircle of ΔABC).</div>
<div class="p2">
<br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgKOfxYl0oJUCkb46g6FBPc9iAN07AUGjJaGvTAeXnzH1-bAZ3eL0XPWRw6jhrs5Lm5B4kHrYCfznTx2jh9Qcjp438BUpMtW1XgPypo8ydKweVvoS-v2G2N_f3sZhKKwY04RPPQAtOlt7O3/s1600/IMG_6697.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgKOfxYl0oJUCkb46g6FBPc9iAN07AUGjJaGvTAeXnzH1-bAZ3eL0XPWRw6jhrs5Lm5B4kHrYCfznTx2jh9Qcjp438BUpMtW1XgPypo8ydKweVvoS-v2G2N_f3sZhKKwY04RPPQAtOlt7O3/s1600/IMG_6697.jpg" height="320" width="314" /></a></div>
<div class="p2">
</div>
<div class="p2">
<br /></div>
<div class="p1">
<b><br /></b>
<b>(8) Construction incircle (or inscribed circle) of triangle ΔABC</b></div>
<div class="p2">
<br /></div>
<div class="p1">
Draw the bisectors of angle BAC and angle ABC . This two lines intersect at I. Draw the straight line IJ such that IJ <span class="s1">⊥</span> BC ( point J is on BC). The circle of centre I, radius IJ is the incircle of triangle ΔABC (the circle contained in the triangle and <a href="http://en.wikipedia.org/wiki/Tangent">tangent</a> to the three sides)</div>
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<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiiNOjf1qLC5AiqD6eo1DUc0t21XndTIq1XF11PVJjH532nN4ZvmM8ax2acBCDKXA3vZXB7R6Z_AO_YZyrB5cC7HqofEKHsjm29EYyXr2si7Lpmg4AL08llVFy3GCxlUFSeuTwdadieyTvR/s1600/IMG_6714.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiiNOjf1qLC5AiqD6eo1DUc0t21XndTIq1XF11PVJjH532nN4ZvmM8ax2acBCDKXA3vZXB7R6Z_AO_YZyrB5cC7HqofEKHsjm29EYyXr2si7Lpmg4AL08llVFy3GCxlUFSeuTwdadieyTvR/s1600/IMG_6714.jpg" height="254" width="320" /></a></div>
<div class="p1">
<br /></div>
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<br /></div>
<div class="p1">
<div class="p1">
<b>(9) Constructing a straight line (d) passing through point A of a given line segment AB and forming with AB an angle which is equal with a given angle xOy.</b></div>
<div class="p2">
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwHXWFsreESMVxljP2AC2YFG_xVLk4gy5NYYqZaismBQFlsSm1eT4sXflBA2lCVBowVKSc-tstCivxL44lJ9lUCZ-Q3qUa9qI2Z7QBYUuzaht02Rw6wB9vu-mVy2SVi1Y4Flf2G1xlG8qx/s1600/IMG_6702.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjwHXWFsreESMVxljP2AC2YFG_xVLk4gy5NYYqZaismBQFlsSm1eT4sXflBA2lCVBowVKSc-tstCivxL44lJ9lUCZ-Q3qUa9qI2Z7QBYUuzaht02Rw6wB9vu-mVy2SVi1Y4Flf2G1xlG8qx/s1600/IMG_6702.jpg" height="320" width="226" /></a></div>
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Draw a circle of center O, radius r cutting the sides Ox, Oy respectively in P and Q. Draw a circle of center A, radius r, cutting AB at point R (AR = OP = OQ = r). Draw a circle of centre R and radius PQ which cuts the circle of centre A at point S. The straight line AS is the line we need to construct.</div>
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<b>(10) Constructing a parallelogram ABCD with a given side BC, length AB = a and <span class="s1">∠</span>ABC is equal to a given <span class="s1">∠</span>xOy </b></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi-xMUe60QpgfXkpkxNU5l3HKnDJXR9FKOw_p0sP0QK4S9kQlCb6Rhc4g7gAkNbv_akk8mVmbeo3CYZt5uCHL9YmTki-DL6XJh_DMIPmmGmXNVyUNGu2q9pHuBXSi0RIj6AydX5Yt86lGFd/s1600/IMG_6711.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi-xMUe60QpgfXkpkxNU5l3HKnDJXR9FKOw_p0sP0QK4S9kQlCb6Rhc4g7gAkNbv_akk8mVmbeo3CYZt5uCHL9YmTki-DL6XJh_DMIPmmGmXNVyUNGu2q9pHuBXSi0RIj6AydX5Yt86lGFd/s1600/IMG_6711.jpg" height="320" width="302" /></a></div>
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Solution: Draw the straight line passing through B and forming with side BC an angle that is equal to angle xOy. This line cuts the circle of center B, radius a at point A. Draw the circle of center A, radius r = BC and the circle of center C and radius a, two circles intersect at point D (as opposed to point B). ABCD is a parallelogram we need to construct.</div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-49541249341789005662014-01-27T02:59:00.002-08:002014-02-02T01:58:59.577-08:00Plane geometry: constructing techniques (1)<div class="p1">
<b><i>"The journey of a thousand miles begins with a single step" (Lao Tzu)</i></b></div>
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The constructing techniques (drawing) by basic rulers and compasses is fairly simple. Only when you have mastered the use of this technique, you can then construct more complex geometric figures accurately, scientifically and clearly.</div>
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<b>(1) Constructing a perpendicular bisector of a given line segment AB</b></div>
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Perpendicular bisector of the line segment is a line perpendicular to the given line segment at its midpoint . To draw the perpendicular bisector of the line segment AB, we just draw two circles of centre A and B respectively and of equal radius R ( R > AB / 2 ) . Two circles intersect at two points I and K. The line passing through these 2 points I, K is the perpendicular bisector of AB .</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_n9hy1UXESiu066ew_uZ0OLWrQpjKhmSqHVnU3RbcKtonCeFhgdi5UKn3eIfR73Ja6DKwwh6f3CsghqT-ncVapxv746hllXEY8Gd-isf5kA2pF6p_d6ODu7Ztfe4fE6ABByZ0QpkSS5zN/s1600/IMG_6694.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_n9hy1UXESiu066ew_uZ0OLWrQpjKhmSqHVnU3RbcKtonCeFhgdi5UKn3eIfR73Ja6DKwwh6f3CsghqT-ncVapxv746hllXEY8Gd-isf5kA2pF6p_d6ODu7Ztfe4fE6ABByZ0QpkSS5zN/s1600/IMG_6694.jpg" height="296" width="320" /></a></div>
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<b>(2) Constructing a line through point A and perpendicular to a given line (d)</b></div>
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To draw the line through point A and perpendicular to line (d), draw a circle of center A. This circle cuts the straight line (d) at two points B and C. Draw two circles of centre B and C with radius R = AB . Two circles intersect at points A and D. The straight line AD is the line we need to construct ( AD is perpendicular to (d) ) .</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAW1ux512nlZDs4OeKG98bfZQbyte9wOXAiBxijd2q51cByl6CcTihjaQNvq1imtxm06jK9WBqih7RvHLJ3OaJYSgT9UDxFBifPYCazxNZnGEiXrCMJg47UKaKJ8916cK36eb8UKBj3g04/s1600/IMG_6693.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiAW1ux512nlZDs4OeKG98bfZQbyte9wOXAiBxijd2q51cByl6CcTihjaQNvq1imtxm06jK9WBqih7RvHLJ3OaJYSgT9UDxFBifPYCazxNZnGEiXrCMJg47UKaKJ8916cK36eb8UKBj3g04/s1600/IMG_6693.jpg" height="320" width="294" /></a></div>
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<b>(3) Constructing bisector of a given angle A</b></div>
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Method: Draw a circle of center A and of any radius R ( R1 > 0 ). This circle cuts the sides Ax, Ay of the given angle at two points I and J. Draw two circles of center I and J, with radius R2 such as the two circles can intersect. Let K be the intersection of two circles of center I, J as above. The straight line passing through the two points A and K is the bisector of angle A.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhz_QMri3sDa_JcAPsX3a_t9uzfW8XxlAcUdaKYycuCapc7AvU2hoomJ5qjwuoPUzQCIh65J7XH86KTbBQE3gXLIHBolyt__crtohquy5ox5IpZZxJ09VOvpYvkRLZS0X6TQEHQRoO_DyQr/s1600/IMG_6696.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhz_QMri3sDa_JcAPsX3a_t9uzfW8XxlAcUdaKYycuCapc7AvU2hoomJ5qjwuoPUzQCIh65J7XH86KTbBQE3gXLIHBolyt__crtohquy5ox5IpZZxJ09VOvpYvkRLZS0X6TQEHQRoO_DyQr/s1600/IMG_6696.jpg" height="250" width="320" /></a></div>
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<b>(4) Constructing a tangent to the circle of center O from a given external point A</b></div>
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Draw a circle of diameter AO which cuts the circle of center point O at two points P and Q. The two straight lines AP and AQ are two tangents to the circle from point A.</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJOX5PphvUL5AT6T55YuV_FJW7RtRKcjoWf-sn8R6RFvNacKfRZ4_5T5C2rjds9KMCOdsOcX36vW-Yj_bRH6QDnkimDOEKa0W_LmLtFI7FZPq80nlGOjjCOatYgxdhEBPlT9lqmooZpWhh/s1600/IMG_6660.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJOX5PphvUL5AT6T55YuV_FJW7RtRKcjoWf-sn8R6RFvNacKfRZ4_5T5C2rjds9KMCOdsOcX36vW-Yj_bRH6QDnkimDOEKa0W_LmLtFI7FZPq80nlGOjjCOatYgxdhEBPlT9lqmooZpWhh/s1600/IMG_6660.jpg" height="234" width="320" /></a></div>
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<b>(5) Divide a line equally in three</b></div>
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PROBLEM: From point O outside the line segment AB , construct two straight lines OM , ON ( M , N are the two points on line segment AB ) that divide AB into three equal line segments .</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvaBLTXF7HYgQLNE9dX7fYBmFd2yVk7kNO_chgVFfR31Xx9vNiqO2jO5nPmOLKXrZcdCi0Lh7afVLHVIUMepwghsXxXikmZLRWZV3ihd05qW_ieBsYFsmRGALbjvZq9mf4E7Db1BjslUFq/s1600/IMG_6663.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvaBLTXF7HYgQLNE9dX7fYBmFd2yVk7kNO_chgVFfR31Xx9vNiqO2jO5nPmOLKXrZcdCi0Lh7afVLHVIUMepwghsXxXikmZLRWZV3ihd05qW_ieBsYFsmRGALbjvZq9mf4E7Db1BjslUFq/s1600/IMG_6663.jpg" height="278" width="320" /></a></div>
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Solution: Draw a circle of center O and of radius AO that cuts the straight line AO at P (different from point A). Then, Draw a circle of center P and of radius AO that cuts the straight line AO at another point Q (Q is different from P). Draw the straight line QB. Through O , P, draw the straight lines that are parallel to QB , cutting AB at M , N respectively. The straight lines OM, ON are the ones we need to construct.</div>
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27.1.2014</div>
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KTKhttp://www.blogger.com/profile/15932834689844272783noreply@blogger.com0tag:blogger.com,1999:blog-5803417181121820838.post-56001709384841863682014-01-26T06:42:00.000-08:002014-02-02T01:58:31.877-08:00Plane Geometry: Mathematical Imagination<div class="p1">
Mathematics is an exact science of imagination and beauty! You don’t have to be numerate students or mathematicians to be interested in this. A simple example is drawing a trapezoid ABCD (AB // CD):</div>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi24BcwT0vwbLIt9m1GqgBg3AMDhbC6Sr6q9QafQDOksn_6zsW_q8eU0vkSmBiptuXAnbQBioSx47ZcjmtbiQ2ZC9verTXWN2QO2iLmZZJkLFkvEJljyBFe0LcBHTz0kPywh2DAPgnPjbj0/s1600/IMG_6677.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi24BcwT0vwbLIt9m1GqgBg3AMDhbC6Sr6q9QafQDOksn_6zsW_q8eU0vkSmBiptuXAnbQBioSx47ZcjmtbiQ2ZC9verTXWN2QO2iLmZZJkLFkvEJljyBFe0LcBHTz0kPywh2DAPgnPjbj0/s1600/IMG_6677.jpg" height="250" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: small; text-align: start;"><i>Pic 1: Drawings on paper/board</i></span></td></tr>
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Look at pic 1, what do you see here? A fairly simple drawing - a trapezoid only. However while studying maths, you will have at least the following imaginations in your mind:</div>
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1 ) The two diagonals AC and BD and point O - an intersection of them</div>
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2 ) The heights AH , BI , CJ and DK</div>
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3 ) The midpoints M, N, R and S of AD, BC, AB and CD respectively</div>
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4 ) The mid-segment MN</div>
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5 ) The intersection point P of the two sides AD and BC</div>
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<table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody>
<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiT5dcaETBTQTEbfTi347cZm9NGXuACpE9Ug2VUEQ-_9TuSzoLrGo2dMoa7mJ9QANrBVTmUHd3tHAiGpBLgcXQ0Z6dvkwmGJ3JHGsslieBnV4KDzSI1-BCda8POTYr0t-5eYt6KJ_9QZwLM/s1600/IMG_6683.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto; text-align: center;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiT5dcaETBTQTEbfTi347cZm9NGXuACpE9Ug2VUEQ-_9TuSzoLrGo2dMoa7mJ9QANrBVTmUHd3tHAiGpBLgcXQ0Z6dvkwmGJ3JHGsslieBnV4KDzSI1-BCda8POTYr0t-5eYt6KJ_9QZwLM/s1600/IMG_6683.jpg" height="365" width="400" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;"><span style="font-size: small; text-align: start;"><i>Pic 2: The drawing in imagination when looking at the Pic 1</i></span></td></tr>
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If you add a characteristic, such as that trapezoid ABCD is isosceles, then in your visualization there will be the circumcircle (or circumscribed circle) of such trapezoid and its circumcenter! Not to mention the logics & relationship between those factors:</div>
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a) The area of a trapezoid = ( AB + CD ) x AH / 2</div>
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b ) MN = ( AB + CD ) / 2</div>
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c ) AH = BI = CJ = DK</div>
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d ) Do you think the points P , R , O , S are on a straight line ?</div>
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Pretty impressive isn’t it? This explains why mathematicians need imagination ! Mathematics , problem solving in high school with normal standards (not specialized in mathematics nor with the goal to be a mathematician in the future), if given proper care, can help develop and enrich imaginations naturally!</div>
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