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Difference between revisions of "2021 WSMO Team Round Problems/Problem 4"
 
Line 6: Line 6:
 
==Solution==
 
==Solution==
 
Note that <math>A_1B_1C_1</math> is a right triangle with right angle at <math>B_1</math>, so its area is   
 
Note that <math>A_1B_1C_1</math> is a right triangle with right angle at <math>B_1</math>, so its area is   
−
<cmath>[A_1B_1C_1] = \frac{1}{2}\cdot A_1B_1\cdot B_1C_1 = \frac{9\sqrt{3}}{2}</cmath>
+
<cmath>[A_1B_1C_1] = \frac{1}{2}\cdot A_1B_1\cdot B_1C_1 = \frac{9\sqrt{3}}{2}.</cmath>
  
 
For all <math>i</math>, we have   
 
For all <math>i</math>, we have   
−
<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>
+
<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>
  
 
Thus, the ratio of areas is   
 
Thus, the ratio of areas is   
−
<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>
+
<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:   
 
This forms a geometric series:   
−
<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>
+
<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>
  
 
The final answer is   
 
The final answer is   
−
<cmath>(18\sqrt{3})^2 = 324\cdot3 = \boxed{972}</cmath>
+
<cmath>(18\sqrt{3})^2 = 324\cdot3 = \boxed{972}.</cmath>
 
~pinkpig
 
~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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