Plane Geometry Drawer v1.2 is now available on the App Store

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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:

New tools to select drawing objects on a iPhone/iPod Touch screens: 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:

It is not easy to select objects like: angle AHP or segment BH

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, touch on the pink circle to change the working mode.

Plane Geometry Drawer for iPhone v1.2: Object selection tools (pink circle)

To use selection tools, please follow the steps below:

1) Touch on random point of the drawing construction, pink circle will appeared
2) Pan on the drawing area from any point to move the circle to the required object:
    - In point mode: move close to the point to be selected
    - In line mode: move close to the midpoint of the line
    - In angle mode: move to the midpoint of the line formed by two points on two rays of the angle
    - In triangle mode: Selected one vertex of triangle, then move the pink circle to the midpoint of the opposite side of the needed triangle
    - In circle mode: Move the pink circle to any point near the circle

Plane Geometry Drawer for iPhone v1.2: Selected a line segment OH

Toolset for marking right angles or equal line segments/angles: 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.

Other improvements: some geometric symbols are added to the geometry text editor keyboard like: ° (degree) or π (Pi) or some tools are added to existing toolsets.

Plane Geometry Drawer for iPhone v1.2: New additional symbols (geometry text editor tools)

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20.2.2014

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Requirements of geometric drawing: The science

Science in the drawing geometry? This topic is somewhat a bit confusing, so to make things simpler we will consider a example.

Example: Given ΔABC, the height AH, median AM (H and M on BC).

Very simple, isn’t it? Then how will we draw such a figure? We can draw as below:

Figure 1: Triangle ABC, the height AH, the median AM

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.

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:

Figure 2: Triangle ABC, the height AH, the median AM

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):

Figure 3: Triangle ABC, the height AH, the median AM

Figure 4: Triangle ABC, the height AH, the median AM

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):

1) When drawing angles or triangles, try not to draw the vertices with special values, such as 30°, 60°, 45°, 90°, 120°, 180°, ...

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!

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.

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...

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.

Related articles:

Geometric drawing requirements: accuracy

Requirements for geometry drawing: overview

9.2.2014

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Geometric drawing requirements: accuracy

Related article(s): Plane geometry: The requirements for the drawing - an overview

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:

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: 

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?

Pic 1: AD is the bisector of angle BAC and IL is the angle bisector of JIK

Next, draw two circles C1, C2 of center D and L that are tangent to the rays of ∠BAC and ∠JIK respectively! Pic 2 shows what would happen:

Pic 2: Draw a circle of center D, tangent to AB and a circle of center L, tangent to IJ

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 ∠JIL < ∠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.

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:

Pic 3: Given ΔABC, angle A = 90°

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:

Pic 4: for ΔABC, angle A = 90°, draw a circle of diameter BC

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!

6.2.2014

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Napoleon Problem: Magic of the circles

Napoleon Bonaparte (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":

pic 1: Find a center of a given circle, using the pair of compasses alone (no straight edge)

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:

pic 2: How to detect a centre of a given circle

Constructing steps:

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.

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.

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!

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!

Proof:

To prove that the above solution works, we will prove the following lemma:

Lemma: 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)

pic 3: Lemma

Lemma's proof: 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.

In the above construction of centre (pic 2), the configuration of lemma appears twice:

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

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 (or O' is a centre of a circle C1).

Reference: wikipedia.org

6.2.2014

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Plane geometry: The requirements for drawing - an overview

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:

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.

2) Supporting creativity: easier to visualise, analyse, relate these elements to advance with new objects/figures.

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.

So what requirements must a good drawing must? Here are some basic requirements:

1) The accuracy (read more details ...)

2) The scientific

3) The aesthetics

Example: 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:

1) AB/AM + AC/AN = 3

2) BM/AM + CN/AN = 1

Illustration for the above problem 

Look at the illustration! It clearly describes a way to find the solution for the problem, isn't it?

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. 

5.2.2014

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Problem & Solution (3): Draw a square ABCD with given vertext A, point M on BC, point N on CD

Problem: Construct a square ABCD given point A and two points M ∈ BC, N ∈ CD

Solution:

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 :

figure 1

1) ∠BCD = π / 2 = > ∠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.

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  = ∠ABP = π/2, ∠DAN = ∠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 )

3) ∠MAN ≤ BAD = π/2

From the above analysis we can determine the square ABCD and point C as follows:

figure 2

Step 1: Construct the line segment MN, construct the circle C1 of diameter MN ( easily done with basic constructing skills )

Step 2: 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 .

Step 3: 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.

ABCD is the square we need to construct (see figure 2)

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 = ∠ABP = π/2; ∠DAN = BAP =>  ∆ADN = ∆ABP => AD = AB (**). From ( * ) and ( ** ) we have ABCD is a square .

Justification :

a) The problem will have no solution if ∠MAN > π/2, or

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 ≧ AP' with  P' = AP ⋂ NM <=> ∠ANM = ∠ANP' ≦ ∠ANP = π/4 or P' = AP ⋂ NM <=> ∠ANM = ∠ANP' ≦ ∠ANP = π/4 <=> ∠ANM ≦ π/4 or ∠AMN ≦ π/4  (see figure 3)

figure 3

c) In the remaining cases the problem has a unique solution.

4.2.2014

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Problem & solution (1): Draw a circle which passes through a given point P and tangents on rays of given angle

Problem: 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.

Solution:

Suppose that a circle C1(O, OP) has been constructed (see a picture). We have following remarks:

1) Center O of drawn circle lies on a bisector Az of an angle xAy.

2) Let O1 be an arbitrary point on Az and C2(O1, O1J) be a circle of center O1 which tangents on Ay, then C2 also tangents on Ax.

3) Let I, J be tangent points of C1, C2 on Ay, then OI and O1J are perpendicular to Ay => OI // O1J => AO1 / AO = O1J / OI (a)

4) Draw a straight line which passes O1 and is parrallel to OP. This line intersects AP at point Q. We have O1Q / OP = AO1 / AP, on other side OP = OI => O1Q / OI = AO1 / AO (b)

5) From (a) and (b) we have O1Q = O1J => Q is on the circle C2

By the above, we can construct the circle C1 as bellowing steps:

1) Draw a bisector Az of an angle xAy, take an arbitrary point O1 on Az and draw a circle C2 of center O1 which tangents on Ay.

2) Let Q be a intersection point of C2 and straight line passing through A and P. From P draw a straight line which is parallel to O1Q. Let O be an intersection of this new line with Az.

A circle of center O, radius OP is tangent on Ax and Ay and passes through point P!

Proof: Draw OI ⊥ Ay, I ∈ Ay and O1J ⊥ Ay, J ∈ Ay => ∆ AOI ∾ ∆AO1J and ∆ AOP ∾ ∆AO1Q => OP / AO = O1Q / AO1 = O1J / AO1 = OJ / AO => OP = OJ = P ∈ C2. In other side C1 tangents on Ay => C1 tangents on Ax = C1 is a required circle!

Justification: the problem has two distinct solutions if P is not on Ax, Ay. In other cases we have one unique solution!

3.2.2014

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