Difference between revisions of "2016 USAJMO Problems/Problem 5"

(Created page with "== Problem == Let <math>\triangle ABC</math> be an acute triangle, with <math>O</math> as its circumcenter. Point <math>H</math> is the foot of the perpendicular from <math>A...")
 
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\cos^2B & \sin^2B & 0 \  
 
\cos^2B & \sin^2B & 0 \  
 
\cos^2C & 0 & \sin^2C \  
 
\cos^2C & 0 & \sin^2C \  
\end{vmatrix}=0.</cmath> Expanding, we must prove <cmath>\sin(2A)\sin^2B\sin^2C=\cos^2C\sin^2B\sin(2C)+\sin^2C\cos^2B\sin(2B)</cmath> <cmath>\frac{\sin(2A)}{2}=\sin^2B(1-\sin^2C)\sin(2C)+\sin^2C(1-\sin^2B)\sin(2B)</cmath> (to be finished)
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\end{vmatrix}=0.</cmath> Expanding, we must prove <cmath>\sin(2A)\sin^2B\sin^2C=\cos^2C\sin^2B\sin(2C)+\sin^2C\cos^2B\sin(2B)</cmath> <cmath>\frac{\sin(2A)}{2}=\sin^2B(1-\sin^2C)\sin(2C)+\sin^2C(1-\sin^2B)\sin(2B)</cmath>  
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<cmath>\begin{align*}
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\frac{\sin(2A)+\sin(2B)+\sin(2C)}{2}&=\sin^2B\sin(2C)+\sin^2C\sin(2B)\
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&=2\sin B\sin C(\sin B\cos C+\cos B\sin C) \
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&=2\sin B\sin C\sin A.\end{align*}</cmath>
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 +
Let <math>x=e^{iA}, y=e^{iB}, z=e^{iC},</math> such that <math>xyz=-1.</math> The left side is equal to <cmath>\frac{x^2+y^2+z^2-\frac{1}{x^2}-\frac{1}{y^2}-\frac{1}{z^2}}{4i}.</cmath> The right side is equal to
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<cmath>\begin{align*}
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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}\
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&=\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-x^2-y^2-z^2}{-4i},\end{align*}</cmath> which is equivalent to the left hand side. Therefore, the determinant is <math>0,</math> and <math>O,P,Q</math> are collinear. <math>\blacksquare</math>

Revision as of 15:09, 21 April 2016

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. $\blacksquare$