2023 WSMO Speed Round Problems/Problem 6

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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2023 WSMO Speed Round Problems/Problem 6

Problem

Solution