A sequel to the Odd Sangaku problem by Michael Huang.

Geometry Level 3

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.


The answer is 82.76.

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1 solution

David Vreken
Jan 15, 2021

Label the diagram as follows, add segments H F HF and F K FK , and let x = D E = E F = F G = G D x = DE = EF = FG = GD .

By the Pythagorean Theorem on A B C \triangle ABC , A C = 2 7 2 + 3 6 2 = 45 AC = \sqrt{27^2 + 36^2} = 45 .

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 \triangle EBK \cong \triangle FGI , so E B = x EB = x .

Since A B C A D E \triangle ABC \sim \triangle ADE , A E = E D A C B C = x 45 36 = 5 4 x AE = ED \cdot \cfrac{AC}{BC} = x \cdot \cfrac{45}{36} = \cfrac{5}{4}x and A D = E D A B B C = x 27 36 = 3 4 x AD = ED \cdot \cfrac{AB}{BC} = x \cdot \cfrac{27}{36} = \cfrac{3}{4}x

Then A B = A E + E B = 5 4 x + x = 27 AB = AE + EB = \cfrac{5}{4}x + x = 27 , which solves to x = 12 x = 12 .

That means the radius of the orange circle is r o = 1 2 12 = 6 r_o = \cfrac{1}{2} \cdot 12 = 6 .

It also means that A E = 5 4 x = 5 4 12 = 15 AE = \cfrac{5}{4}x = \cfrac{5}{4} \cdot 12 = 15 , and A D = 3 4 x = 3 4 12 = 9 AD = \cfrac{3}{4}x = \cfrac{3}{4} \cdot 12 = 9 , so the radius of the green circle is r g = 1 2 ( 12 + 9 15 ) = 3 r_g = \cfrac{1}{2}(12 + 9 - 15) = 3 .

From F G I A D E \triangle FGI \sim ADE , the purple incircle is F G A D = 12 9 = 4 3 \cfrac{FG}{AD} = \cfrac{12}{9} = \cfrac{4}{3} bigger than the green incircle, so the purple circle has a radius of r p = 4 3 r g = 4 3 3 = 4 r_p = \cfrac{4}{3}r_g = \cfrac{4}{3} \cdot 3 = 4 .

The red circle has the same radii as the purple circle, so r r = r p = 4 r_r = r_p = 4 .

Since E H F A B C \triangle EHF \sim \triangle ABC , E H = E F A B A C = 12 27 45 = 36 5 EH = EF \cdot \cfrac{AB}{AC} = 12 \cdot \cfrac{27}{45} = \cfrac{36}{5} .

Then H B = E B E H = 12 36 5 = 24 5 HB = EB - EH = 12 - \cfrac{36}{5} = \cfrac{24}{5} , so the radius of the blue circle is r b = 1 2 24 5 = 12 5 r_b = \cfrac{1}{2} \cdot \cfrac{24}{5} = \cfrac{12}{5} .

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 + 1 2 2 5 2 = 82.76 r_o^2 + r_g^2 + r_p^2 + r_r^2 + r_b^2 = 6^2 + 3^2 + 4^2 + 4^2 + \cfrac{12^2}{5^2} = \boxed{82.76} .

Brilliant. Note one thing though, in line 3, shouldn’t it be E B K F G I \triangle EBK \cong \triangle FGI ?

Jeff Giff - 4 months, 3 weeks ago

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Yes, thank you! I edited it.

David Vreken - 4 months, 3 weeks ago

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