Given angle xOy, two points A, B and line segment DE. Construct ΔABC such that ∠ACB = ∠xOy and BC = DE.
Solution:
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.
Since circle C1 is determined, the problem now revolves around constructing circle C2.
Let K be an arbitrary point on C2, so ∠AKB = ∠ACB = ∠xOy and ∠AKO + ∠KBAB + ∠KAB = 2π => ∠ΑΚΒ + ∠ΚΒΑ < 2π. Then we can construct C2 and triangle ABC as follows:
1) Take arbitrary point I on the straight line (d) passing through point B, so that ∠IBA + ∠xOy < 2π.
2) Construct angle JIB, such that it is equal to angle xOy and J ∈ AB
3) Construct a straight line passing through point A and parallel to JI, which is cutting line (d) at K. We will have ∠AKB = ∠xOy = ∠JIB .
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 !
Proof: From above it is easy to prove that ∠AKB = ∠ACB = ∠xOy = ∠JIB and BC = DE .
Justification: 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:
1) r2 < ( DE / 2 ) => the problem has no solution;
2) r2 = DE/2 => the problem has one and only solution; and
3) r2 > (DE/2) => the problem has 2 distinct solutions.
1) r2 < ( DE / 2 ) => the problem has no solution;
2) r2 = DE/2 => the problem has one and only solution; and
3) r2 > (DE/2) => the problem has 2 distinct solutions.
30.1.2014
