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Difference between revisions of "2021 WSMO Team Round Problems/Problem 4"

(Created page with "==Problem== Consider a triangle <math>A_1B_1C_1</math> satisfying <math>A_1B_1=3,B_1C_1=3\sqrt{3},A_1C_1=6</math>. For all successive triangles <math>A_nB_nC_n</math>, we have...")
 
 
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==Problem==
 
==Problem==
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Consider a triangle <math>A_1B_1C_1</math> satisfying <math>A_1B_1=3,B_1C_1=3\sqrt{3},A_1C_1=6</math>. For all successive triangles <math>A_nB_nC_n</math>, we have <math>A_nB_nC_n\sim B_{n-1}A_{n-1}C_{n-1}</math> and <math>A_n=B_{n-1},C_n=C_{n-1}</math>, where <math>A_nB_nC_n</math> is outside of <math>A_{n-1}B_{n-1}C_{n-1}</math>. Find the value of<cmath>\left(\sum_{i=1}^{\infty}[A_iB_iC_i]\right)^2,</cmath>where <math>[A_iB_iC_i]</math> is the area of <math>A_iB_iC_i</math>.
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Consider a triangle <math>A_1B_1C_1</math> satisfying <math>A_1B_1=3,B_1C_1=3\sqrt{3},A_1C_1=6</math>. For all successive triangles <math>A_nB_nC_n</math>, we have <math>A_nB_nC_n\sim B_{n-1}A_{n-1}C_{n-1}</math> and <math>A_n=B_{n-1},C_n=C_{n-1}</math>, where <math>A_nB_nC_n</math> is outside of <math>A_{n-1}B_{n-1}C_{n-1}</math>. Find the value of <cmath>\left(\sum_{i=1}^{\infty}[A_iB_iC_i]\right)^2,</cmath> where <math>[A_iB_iC_i]</math> is the area of <math>A_iB_iC_i</math>.
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''Proposed by pinkpig''
  
 
==Solution==
 
==Solution==
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Note that <math>A_1B_1C_1</math> is a right triangle with right angle at <math>B_1</math>, so its area is 
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<cmath>[A_1B_1C_1] = \frac{1}{2}\cdot A_1B_1\cdot B_1C_1 = \frac{9\sqrt{3}}{2}.</cmath>
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For all <math>i</math>, we have 
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<cmath>\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}.</cmath>
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Thus, the ratio of areas is 
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<cmath>\frac{[A_iB_iC_i]}{[A_{i-1}B_{i-1}C_{i-1}]} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}.</cmath>
 +
 +
This forms a geometric series: 
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<cmath>\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}.</cmath>
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The final answer is 
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<cmath>(18\sqrt{3})^2 = 324\cdot3 = \boxed{972}.</cmath>
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~pinkpig

Latest revision as of 13:03, 9 September 2025

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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