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2016 USAJMO Problems/Problem 5

Revision as of 03:16, 27 April 2016 by Sujaykazi (talk | contribs) (→‎Solution 3)

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 $\triangle ABC.$

Since the central angle $\angle AOB$ subtends the same arc as the inscribed angle $\angle ACB$ on the circumcircle of $\triangle ABC,$ we have $\angle AOB = 2\gamma.$ Note that $OA = OB,$ so $\angle OAB = \angle OBA.$ Thus, $\angle OAB = \frac{\pi}{2} - \gamma.$ Similarly, one can show that $\angle OAC = \frac{\pi}{2} - \beta.$ (One could probably cite this as well-known, but I have proved it here just in case.)

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 $AH = b\sin\gamma = c\sin\beta.$ The Extended Law of Sines states that: \[\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.\] Therefore, $AH = \frac{bc}{2R} = \sqrt{2}R.$ Thus, $bc = \sqrt{2}R^2.$

Since $\angle PHA = \beta$ and $\angle QHA = \gamma,$ we have: \[AP = AH\sin\beta = c\sin^2\beta = \frac{b^2 c}{4R^2}\] \[AQ = AH\sin\gamma = b\sin^2\gamma = \frac{bc^2}{4R^2}\] It follows that: \[AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.\] 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 $AP\cdot AQ = AH\cdot AO,$ 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 $\angle AOP$ 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 interesting 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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