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

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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}\\
 
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*}</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>
 
&=\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>
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== Solution 3 ==
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For convenience, let <math>a, b, c</math> denote the lengths of segments <math>BC, CA, AB,</math> respectively, and let <math>\alpha, \beta, \gamma</math> denote the measures of <math>\angle CAB, \angle ABC, \angle BCA,</math> respectively. Let <math>R</math> denote the circumradius of <math>\triangle ABC.</math>
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Clearly, <math>AO = R.</math> Since <math>AH^2 = 2\cdot AO^2,</math> we have <math>AH = \sqrt{2}R.</math> Thus, <math>AH\cdot AO = \sqrt{2}R^2.</math>
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Note that <math>AH = b\sin\gamma = c\sin\beta.</math> Then, since <math>\angle PHA = \beta</math> and <math>\angle QHA = \gamma,</math> we have:
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<cmath>AP = AH\sin\beta = c\sin^2\beta</cmath>
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<cmath>AQ = AH\sin\gamma = b\sin^2\gamma</cmath>
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The Extended Law of Sines states that:
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<cmath>\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.</cmath>
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Since <math>AP = \frac{b^2 c}{4R^2}</math> and <math>AQ = \frac{bc^2}{4R^2},</math>
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<math></math>AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.<math>
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We see that </math>AP\cdot AQ = AH\cdot AO.<math>
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Rearranging </math>AP\cdot AQ = AH\cdot AO,<math> we get: </math>\frac{AP}{AH} = \frac{AO}{AQ}.<math> We also have </math>\angle PAH = \angle OAQ = \frac{\pi}{2} - \beta,<math> so </math>\triangle PAH\sim\triangle OAQ<math> by SAS similarity. Thus, </math>\angle AOQ = \angle APH,<math> so </math>\angle AOQ<math> is a right angle.
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Rearranging </math>AP\cdot AQ = AH\cdot AO,<math> we get: </math>\frac{AP}{AO} = \frac{AO}{AH}.<math> We also have </math>\angle PAO = \angle HAQ = \frac{\pi}{2} - \gamma,<math> so </math>\triangle PAO\sim\triangle HAQ<math> by SAS similarity. Thus, </math>\angle AOP = \angle AQH,<math> so </math>\angle AOP<math> is a right angle.
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Since </math>\angle AOP<math> and </math>\angle AOQ<math> are both right angles, we get </math>\angle POQ = \pi,<math> so we conclude that </math>P, O, Q<math> are collinear, so we are done. We also obtain the extra fact that </math>AO\perp PQ.$
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{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:07, 27 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 1

It is well-known that $AH\cdot 2AO=AB\cdot AC$ (just use similar triangles or standard area formulas). Then by Power of a Point, \[AP\cdot AB=AH^2=AQ\cdot AC\] Consider the transformation $X\mapsto \Psi(X)$ which dilates $X$ from $A$ by a factor of $\dfrac{AB}{AQ}=\dfrac{AC}{AP}$ and reflects about the $A$-angle bisector. Then $\Psi(O)$ clearly lies on $AH$, and its distance from $A$ is \[AO\cdot\frac{AB}{AQ}=AO\cdot\frac{AB}{\frac{AH^2}{AC}}=AO\cdot\frac{AB\cdot AC}{AH^2}=\frac{AO\cdot AH\cdot 2AO}{AH^2}=\frac{2AO^2}{AH}=AH\] so $\Psi(O)=H$, hence we conclude that $O,P,Q$ are collinear, as desired.

Solution 2

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. $\blacksquare$


Solution 3

For convenience, let $a, b, c$ denote the lengths of segments $BC, CA, AB,$ respectively, and let $\alpha, \beta, \gamma$ denote the measures of $\angle CAB, \angle ABC, \angle BCA,$ respectively. Let $R$ denote the circumradius of $\triangle ABC.$

Clearly, $AO = R.$ Since $AH^2 = 2\cdot AO^2,$ we have $AH = \sqrt{2}R.$ Thus, $AH\cdot AO = \sqrt{2}R^2.$

Note that $AH = b\sin\gamma = c\sin\beta.$ Then, since $\angle PHA = \beta$ and $\angle QHA = \gamma,$ we have: \[AP = AH\sin\beta = c\sin^2\beta\] \[AQ = AH\sin\gamma = b\sin^2\gamma\] The Extended Law of Sines states that: \[\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.\]

Since $AP = \frac{b^2 c}{4R^2}$ and $AQ = \frac{bc^2}{4R^2},$ $$ (Error compiling LaTeX. Unknown error_msg)AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.$We see that$AP\cdot AQ = AH\cdot AO.$Rearranging$AP\cdot AQ = AH\cdot AO,$we get:$\frac{AP}{AH} = \frac{AO}{AQ}.$We also have$\angle PAH = \angle OAQ = \frac{\pi}{2} - \beta,$so$\triangle PAH\sim\triangle OAQ$by SAS similarity. Thus,$\angle AOQ = \angle APH,$so$\angle AOQ$is a right angle.

Rearranging$ (Error compiling LaTeX. Unknown error_msg)AP\cdot AQ = AH\cdot AO,$we get:$\frac{AP}{AO} = \frac{AO}{AH}.$We also have$\angle PAO = \angle HAQ = \frac{\pi}{2} - \gamma,$so$\triangle PAO\sim\triangle HAQ$by SAS similarity. Thus,$\angle AOP = \angle AQH,$so$\angle AOP$is a right angle.

Since$ (Error compiling LaTeX. Unknown error_msg)\angle AOP$and$\angle AOQ$are both right angles, we get$\angle POQ = \pi,$so we conclude that$P, O, Q$are collinear, so we are done. We also obtain the extra fact that$AO\perp PQ.$


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2016 USAJMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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