Triangle, Circle, and Hexagon

Geometry Level 4

In A B C \triangle ABC , A B = 3 AB=3 , B C = 4 BC=4 , and C A = 5 CA=5 . A circle cuts A B AB at C 1 C_1 and C 2 C_2 , cuts B C BC at A 1 A_1 and A 2 A_2 , and cuts C A CA at B 1 B_1 and B 2 B_2 . If A 1 A 2 = B 1 B 2 = C 1 C 2 = x A_1A_2=B_1B_2=C_1C_2=x , and that the area of the hexagon formed by the vertices A 1 , A 2 , B 1 , B 2 , C 1 A_1, A_2, B_1, B_2, C_1 and C 2 C_2 is 4 4 , find x x .


The answer is 1.639079157.

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1 solution

Maria Kozlowska
Jun 13, 2018

Our circle has the triangle incentre as its centre. Radius of the incircle is 1. Points of contact of incircle with triangle sides are midpoints of segments A 1 A 2 , B 1 B 2 , C 1 C 2 A_1 A_2, B_1 B_2, C_1 C_2 . This makes segment lengths A C 1 = A B 2 = ( 3 1 ) x 2 = 2 x 2 A C_1 = A B_2 = (3-1)-\dfrac{x}{2} = 2-\dfrac{x}{2} , B A 1 = B C 2 = 1 x 2 B A_1 = B C_2 = 1-\dfrac{x}{2} , C A 2 = C B 1 = 3 x 2 C A_2=C B_1 = 3-\dfrac{x}{2} .

A C 1 B 2 = 1 2 ( 2 x 2 ) 2 sin ( A ) = 1 2 ( 2 x 2 ) 2 4 5 \triangle A C_1 B_2 = \dfrac{1}{2}(2-\dfrac{x}{2})^2 \sin(A) = \dfrac{1}{2} (2-\dfrac{x}{2})^2 \dfrac{4}{5} B A 1 C 2 = 1 2 ( 1 x 2 ) 2 sin ( B ) = 1 2 ( 1 x 2 ) 2 \triangle B A_1 C_2 = \dfrac{1}{2}(1-\dfrac{x}{2})^2 \sin(B) = \dfrac{1}{2}(1-\dfrac{x}{2})^2 C A 2 B 1 = 1 2 ( 3 x 2 ) 2 sin ( C ) = 1 2 ( 3 x 2 ) 2 4 5 \triangle C A_2 B_1 = \dfrac{1}{2}(3-\dfrac{x}{2})^2 \sin(C) =\dfrac{1}{2} (3-\dfrac{x}{2})^2 \dfrac{4}{5}

A C 1 B 2 + B A 1 C 2 + C A 2 B 1 = A B C 4 = 2 \triangle A C_1 B_2+\triangle B A_1 C_2+\triangle C A_2 B_1 =\triangle ABC-4=2

This gives value of x = 11 37 3 1.64 x=\dfrac{11-\sqrt{37}}{3} \approx 1.64

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