2016 USAJMO Problems/Problem 5

Problem

Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC}$ , and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$ , respectively.

Given that $AH^2=2\cdot AO^2,$ prove that the points $O,P,$ and $Q$ are collinear.

Solution

We will use barycentric coordinates with respect to $\triangle ABC.$ The given condition is equivalent to $(\sin B\sin C)^2=\frac{1}{2}.$ Note that $O=(\sin(2A):\sin(2B):\sin(2C)), P=(\cos^2B,\sin^2B,0), Q=(\cos^2C,0,\sin^2C).$ Therefore, we must show that $\begin{vmatrix} \sin(2A) & \sin(2B) & \sin(2C) \\ \cos^2B & \sin^2B & 0 \\ \cos^2C & 0 & \sin^2C \\ \end{vmatrix}=0.$ Expanding, we must prove $\sin(2A)\sin^2B\sin^2C=\cos^2C\sin^2B\sin(2C)+\sin^2C\cos^2B\sin(2B)$ $\frac{\sin(2A)}{2}=\sin^2B(1-\sin^2C)\sin(2C)+\sin^2C(1-\sin^2B)\sin(2B)$ $\begin{align*} \frac{\sin(2A)+\sin(2B)+\sin(2C)}{2}&=\sin^2B\sin(2C)+\sin^2C\sin(2B)\\ &=2\sin B\sin C(\sin B\cos C+\cos B\sin C) \\ &=2\sin B\sin C\sin A.\end{align*}$

Let $x=e^{iA}, y=e^{iB}, z=e^{iC},$ such that $xyz=-1.$ The left side is equal to $\frac{x^2+y^2+z^2-\frac{1}{x^2}-\frac{1}{y^2}-\frac{1}{z^2}}{4i}.$ The right side is equal to $\begin{align*} 2\cdot \frac{x-\frac{1}{x}}{2i}\cdot \frac{y-\frac{1}{y}}{2i}\cdot \frac{z-\frac{1}{z}}{2i}&=\frac{xyz-\frac{1}{xyz}-\frac{xy}{z}-\frac{yz}{x}-\frac{xz}{y}+\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}}{-4i}\\ &=\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-x^2-y^2-z^2}{-4i},\end{align*}$ which is equivalent to the left hand side. Therefore, the determinant is $0,$ and $O,P,Q$ are collinear. $\blacksquare$

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2016 USAJMO Problems/Problem 5

Problem

Solution