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, 2023

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let $ABC$ be an equilateral triangle of side length $15$ . Let $A_b$ and $B_a$ be points on side $AB$ , $A_c$ and $C_a$ be points on side $AC$ , and $B_c$ and $C_b$ be points on side $BC$ such that $\triangle AA_bA_c$ , $\triangle BB_cB_a$ , and $\triangle CC_aC_b$ are equilateral triangles with side lengths $3$ , $4$ , and $5$ , respectively. Compute the radius of the circle tangent to segments $\overline{A_bA_c}$ , $\overline{B_aB_c}$ , and $\overline{C_aC_b}$ .

Solution

Solution:

Let $\triangle XYZ$ be the triangle formed by lines $A_bA_c$ , $B_aB_c$ , and $C_aC_b$ . Then, the desired circle is the incircle of $\triangle XYZ$ , which is equilateral. We have
$\begin{aligned} YZ & = YA_c + A_cA_b + A_bZ \\ & = A_cC_a + A_cA_b + A_bB_a \\ & = (15 - 3 - 5) + 3 + (15 - 3 - 4) \\ & = 18, \end{aligned}$
and so the inradius is $\frac{1}{2\sqrt{3}} \cdot 18 = 3\sqrt{3}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.