In the right triangle ABC having side lengths:
AB = 27 and BC = 36
The square and the rectangle are positioned inside this triangle, such that the purple and red circles have the same radii.
Input the sum of the squares of the radii of all five circles as your answer.
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Label the diagram as follows, add segments H F and F K , and let x = D E = E F = F G = G D .
By the Pythagorean Theorem on △ A B C , A C = 2 7 2 + 3 6 2 = 4 5 .
All the triangles are similar to each other by AA similarity, and since the purple and red circles have the same radii, △ E B K ≅ △ F G I , so E B = x .
Since △ A B C ∼ △ A D E , A E = E D ⋅ B C A C = x ⋅ 3 6 4 5 = 4 5 x and A D = E D ⋅ B C A B = x ⋅ 3 6 2 7 = 4 3 x
Then A B = A E + E B = 4 5 x + x = 2 7 , which solves to x = 1 2 .
That means the radius of the orange circle is r o = 2 1 ⋅ 1 2 = 6 .
It also means that A E = 4 5 x = 4 5 ⋅ 1 2 = 1 5 , and A D = 4 3 x = 4 3 ⋅ 1 2 = 9 , so the radius of the green circle is r g = 2 1 ( 1 2 + 9 − 1 5 ) = 3 .
From △ F G I ∼ A D E , the purple incircle is A D F G = 9 1 2 = 3 4 bigger than the green incircle, so the purple circle has a radius of r p = 3 4 r g = 3 4 ⋅ 3 = 4 .
The red circle has the same radii as the purple circle, so r r = r p = 4 .
Since △ E H F ∼ △ A B C , E H = E F ⋅ A C A B = 1 2 ⋅ 4 5 2 7 = 5 3 6 .
Then H B = E B − E H = 1 2 − 5 3 6 = 5 2 4 , so the radius of the blue circle is r b = 2 1 ⋅ 5 2 4 = 5 1 2 .
Therefore, the sum of the squares of the five radii is r o 2 + r g 2 + r p 2 + r r 2 + r b 2 = 6 2 + 3 2 + 4 2 + 4 2 + 5 2 1 2 2 = 8 2 . 7 6 .