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Difference between revisions of "2023 WSMO Speed Round Problems/Problem 6"

(Created page with "==Problem== Let <math>ABC</math> be an equilateral triangle of side length <math>6.</math> Points <math>A_1,A_2,A_3,B_1,B_2,B_3,C_1,C_2,C_3</math> are chosen such that <math>...")
 
 
(One intermediate revision by the same user not shown)
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
Let <math>[XYZ]</math> denote the area of triangle <math>XYZ</math>. We have
 +
<cmath>\begin{align*}
 +
[A_iB_jC_k] &= [ABC]-[AB_jC_k]-[BC_kA_i]-[CA_iB_j]\implies\\
 +
&= 6^2\cdot\frac{\sqrt{3}}{4}-\frac{1}{2}\sin(60^{\circ})(AB_j)(AC_k)-\frac{1}{2}\sin(60^{\circ})(BC_k)(BA_i)-\frac{1}{2}\sin(60^{\circ})(CA_i)(CB_j) \\
 +
&= 9\sqrt{3}-\frac{\sqrt{3}}{4}\left((AB_j)(AC_k)+(BC_k)(BA_i)+(CA_i)(CB_j)\right).\\
 +
\end{align*}</cmath>
 +
Let <math>AB_j=c</math>, <math>BC_k=a,</math> and <math>CA_i=b</math>. This directly implies that <math>B_jC=6-c</math>, <math>C_kA=6-a</math>, and <math>A_iB=6-b</math>. So, we have
 +
<cmath>\begin{align*}
 +
[A_iB_jC_k] &= 9\sqrt{3}-\frac{\sqrt{3}}{4}\left(c(6-a)+a(6-b)+b(6-c)\right)\\
 +
&=9\sqrt{3}-\frac{\sqrt{3}}{4}\left(6(a+b+c)-(ab+ac+bc)\right)\implies\\
 +
\mathbb{E}([A_iB_jC_k)) &= 9\sqrt{3}-\frac{\sqrt{3}}{4}\left(6\cdot(\mathbb{E}(a)+\mathbb{E}(b)+\mathbb{E}(c))-(\mathbb{E}(ab)+\mathbb{E}(ac)+\mathbb{E}(bc)\right).\\
 +
\end{align*}</cmath>
 +
We have <cmath>\mathbb{E}(a) = \mathbb{E}(b) = \mathbb{E}(c) = \frac{6}{2} = 3</cmath> and <cmath>\mathbb{E}(ab) = \mathbb{E}(ac) = \mathbb{E}(bc) = \left(\mathbb{E}(a)\right)^2 = 9.</cmath> Substituting,
 +
<cmath>\begin{align*}
 +
\mathbb{E}([A_iB_jC_k]) &= 9\sqrt{3}-\frac{\sqrt{3}}{4}\left(6\cdot\left(3+3+3\right)-\left(9+9+9\right)\right)\\
 +
&= 9\sqrt{3} - \frac{\sqrt{3}}{4}\left(54-27\right) = 9\sqrt{3}-\frac{27\sqrt{3}}{4}\\
 +
&=\frac{9\sqrt{3}}{4}\Rightarrow 9+3+4 = \boxed{16}.
 +
\end{align*}</cmath>
 +
 +
~pinkpig

Latest revision as of 11:29, 12 September 2025

Problem

Let $ABC$ be an equilateral triangle of side length $6.$ Points $A_1,A_2,A_3,B_1,B_2,B_3,C_1,C_2,C_3$ are chosen such that $A_1,A_2,A_3$ divide $BC$ into four equal segments, $B_1,B_2,B_3$ divide $AC$ into four equal segments, and $C_1,C_2,C_3$ divide $AB$ into four equal segments. If $i,j,k$ are chosen from the set ${1,2,3}$ independently and randomly, the expected area of $A_iB_jC_k$ is $\frac{a\sqrt{b}}{c},$ for squarefree $b$ and relatively prime positive integers $a$ and $c.$ Find $a+b+c.$

Solution

Let $[XYZ]$ denote the area of triangle $XYZ$. We have \begin{align*} [A_iB_jC_k] &= [ABC]-[AB_jC_k]-[BC_kA_i]-[CA_iB_j]\implies\\ &= 6^2\cdot\frac{\sqrt{3}}{4}-\frac{1}{2}\sin(60^{\circ})(AB_j)(AC_k)-\frac{1}{2}\sin(60^{\circ})(BC_k)(BA_i)-\frac{1}{2}\sin(60^{\circ})(CA_i)(CB_j) \\ &= 9\sqrt{3}-\frac{\sqrt{3}}{4}\left((AB_j)(AC_k)+(BC_k)(BA_i)+(CA_i)(CB_j)\right).\\ \end{align*} Let $AB_j=c$, $BC_k=a,$ and $CA_i=b$. This directly implies that $B_jC=6-c$, $C_kA=6-a$, and $A_iB=6-b$. So, we have \begin{align*} [A_iB_jC_k] &= 9\sqrt{3}-\frac{\sqrt{3}}{4}\left(c(6-a)+a(6-b)+b(6-c)\right)\\ &=9\sqrt{3}-\frac{\sqrt{3}}{4}\left(6(a+b+c)-(ab+ac+bc)\right)\implies\\ \mathbb{E}([A_iB_jC_k)) &= 9\sqrt{3}-\frac{\sqrt{3}}{4}\left(6\cdot(\mathbb{E}(a)+\mathbb{E}(b)+\mathbb{E}(c))-(\mathbb{E}(ab)+\mathbb{E}(ac)+\mathbb{E}(bc)\right).\\ \end{align*} We have \[\mathbb{E}(a) = \mathbb{E}(b) = \mathbb{E}(c) = \frac{6}{2} = 3\] and \[\mathbb{E}(ab) = \mathbb{E}(ac) = \mathbb{E}(bc) = \left(\mathbb{E}(a)\right)^2 = 9.\] Substituting, \begin{align*} \mathbb{E}([A_iB_jC_k]) &= 9\sqrt{3}-\frac{\sqrt{3}}{4}\left(6\cdot\left(3+3+3\right)-\left(9+9+9\right)\right)\\ &= 9\sqrt{3} - \frac{\sqrt{3}}{4}\left(54-27\right) = 9\sqrt{3}-\frac{27\sqrt{3}}{4}\\ &=\frac{9\sqrt{3}}{4}\Rightarrow 9+3+4 = \boxed{16}. \end{align*}

~pinkpig

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