2021 WSMO Team Round Problems/Problem 4

Problem

Consider a triangle $A_1B_1C_1$ satisfying $A_1B_1=3,B_1C_1=3\sqrt{3},A_1C_1=6$ . For all successive triangles $A_nB_nC_n$ , we have $A_nB_nC_n\sim B_{n-1}A_{n-1}C_{n-1}$ and $A_n=B_{n-1},C_n=C_{n-1}$ , where $A_nB_nC_n$ is outside of $A_{n-1}B_{n-1}C_{n-1}$ . Find the value of $\left(\sum_{i=1}^{\infty}[A_iB_iC_i]\right)^2,$ where $[A_iB_iC_i]$ is the area of $A_iB_iC_i$ .

Proposed by pinkpig

Solution

Note that $A_1B_1C_1$ is a right triangle with right angle at $B_1$ , so its area is $[A_1B_1C_1] = \frac{1}{2}\cdot A_1B_1\cdot B_1C_1 = \frac{9\sqrt{3}}{2}.$

For all $i$ , we have $\frac{A_iC_i}{A_{i-1}C_{i-1}} = \frac{B_{i-1}C_{i-1}}{A_{i-1}C_{i-1}} = \frac{B_1C_1}{A_1C_1} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}.$

Thus, the ratio of areas is $\frac{[A_iB_iC_i]}{[A_{i-1}B_{i-1}C_{i-1}]} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}.$

This forms a geometric series: $\sum_{i=1}^\infty [A_iB_iC_i] = [A_1B_1C_1] \sum_{i=0}^\infty \left(\frac{3}{4}\right)^i = \frac{9\sqrt{3}}{2} \cdot \frac{1}{1 - \frac{3}{4}} = \frac{9\sqrt{3}}{2} \cdot 4 = 18\sqrt{3}.$

The final answer is $(18\sqrt{3})^2 = 324\cdot3 = \boxed{972}.$ ~pinkpig

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2021 WSMO Team Round Problems/Problem 4

Problem

Solution