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

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== Solution 1==
 
== Solution 1==
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<asy>
 +
size(8cm);
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pair O=(0,0);
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pair A=dir(110);
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pair B=dir(-29);
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pair C=dir(209);
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pair H=foot(A,B,C);
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pair P=foot(H,A,B);
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pair Q=foot(H,A,C);
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draw(A--B--C--A--H--P);
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draw(circle(O,1));
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draw(Q--H);
 +
 +
dot("$A$", A, dir(A));
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dot("$B$", B, dir(B));
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dot("$C$", C, dir(C));
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dot("$H$", H, S);
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dot("$P$", P, NE);
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dot("$Q$", Q, NW);
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dot("$O$", O, S);
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</asy>
 +
 
It is well-known that <math>AH\cdot 2AO=AB\cdot AC</math> (just use similar triangles or standard area formulas). Then by Power of a Point,
 
It is well-known that <math>AH\cdot 2AO=AB\cdot AC</math> (just use similar triangles or standard area formulas). Then by Power of a Point,
 
<cmath>AP\cdot AB=AH^2=AQ\cdot AC</cmath> Consider the transformation <math>X\mapsto \Psi(X)</math> which dilates <math>X</math> from <math>A</math> by a factor of <math>\dfrac{AB}{AQ}=\dfrac{AC}{AP}</math> and reflects about the <math>A</math>-angle bisector. Then <math>\Psi(O)</math> clearly lies on <math>AH</math>, and its distance from <math>A</math> is <cmath>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</cmath> so <math>\Psi(O)=H</math>, hence we conclude that <math>O,P,Q</math> are collinear, as desired.
 
<cmath>AP\cdot AB=AH^2=AQ\cdot AC</cmath> Consider the transformation <math>X\mapsto \Psi(X)</math> which dilates <math>X</math> from <math>A</math> by a factor of <math>\dfrac{AB}{AQ}=\dfrac{AC}{AP}</math> and reflects about the <math>A</math>-angle bisector. Then <math>\Psi(O)</math> clearly lies on <math>AH</math>, and its distance from <math>A</math> is <cmath>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</cmath> so <math>\Psi(O)=H</math>, hence we conclude that <math>O,P,Q</math> are collinear, as desired.
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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>
 +
 +
 +
== Solution 3 ==
 +
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>
 +
 +
Since the central angle <math>\angle AOB</math> subtends the same arc as the inscribed angle <math>\angle ACB</math> on the circumcircle of <math>\triangle ABC,</math> we have <math>\angle AOB = 2\gamma.</math> Note that <math>OA = OB,</math> so <math>\angle OAB = \angle OBA.</math> Thus, <math>\angle OAB = \frac{\pi}{2} - \gamma.</math> Similarly, one can show that <math>\angle OAC = \frac{\pi}{2} - \beta.</math> (One could probably cite this as well-known, but I have proved it here just in case.)
 +
 +
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>
 +
 +
Note that <math>AH = b\sin\gamma = c\sin\beta.</math> The Extended Law of Sines states that:
 +
<cmath>\frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R.</cmath>
 +
Therefore, <math>AH = \frac{bc}{2R} = \sqrt{2}R.</math> Thus, <math>bc = \sqrt{2}R^2.</math>
 +
 +
Since <math>\angle PHA = \beta</math> and <math>\angle QHA = \gamma,</math> we have:
 +
<cmath>AP = AH\sin\beta = c\sin^2\beta = \frac{b^2 c}{4R^2}</cmath>
 +
<cmath>AQ = AH\sin\gamma = b\sin^2\gamma = \frac{bc^2}{4R^2}</cmath>
 +
It follows that:
 +
<cmath>AP\cdot AQ = \frac{b^3 c^3}{16R^4} = \frac{16\sqrt{2}R^6}{16R^4} = \sqrt{2}R^2.</cmath>
 +
We see that <math>AP\cdot AQ = AH\cdot AO.</math>
 +
 +
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.
 +
 +
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.
 +
 +
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, and we are done. (We also obtain the extra interesting fact that <math>AO\perp PQ.</math>)
 +
 +
== Solution 4==
 +
Draw the altitude from <math>O</math> to <math>AB</math>, and let the foot of this altitude be <math>D</math>.
 +
 +
Then, by the Right Triangle Altitude Theorem on triangle <math>AHB</math>, we have: <math>AB\cdot AP=AH^{2}</math>.
 +
 +
Since <math>OD</math> is the perpendicular bisector of <math>AB</math>, <math>2\cdot AD = AB</math>.
 +
 +
Substituting this into our previous equation gives <math>2\cdot AD \cdot AP = AH^{2}</math>, which equals <math>2\cdot AO^{2}</math> by the problem condition.
 +
 +
Thus, <math>2\cdot AD\cdot AP = 2\cdot AO^{2} \implies AD\cdot AP = AO^{2}</math>.
 +
 +
Again, by the Right Triangle Altitude Theorem, angle <math>AOP</math> is right.
 +
 +
By dropping an altitude from <math>O</math> to <math>AC</math> and using the same method, we can find that angle <math>AOQ</math> is right. Since <math>\angle AOP=\angle AOQ=90</math>, <math>P</math>, <math>O</math>, <math>Q</math> are collinear and we are done.
 +
 +
~champion999
 +
 +
== Solution 5==
 +
We use complex numbers. Let lower case letters represent their respective upper case points, with <math>|a| = |b| = |c| = 1</math>. Spamming the foot from point to segment formula, we obtain <cmath>h = \dfrac{1}{2}\left(a+b+c-\dfrac{bc}{a}\right),</cmath> <cmath>p = \dfrac{1}{2}(h+a+b-ab\overline{h}) = \dfrac{1}{4}\left(2a+2b+c-\dfrac{bc}{a}-\dfrac{ab}{c}+\dfrac{a^2}{c}\right),</cmath> and <cmath>q = \dfrac{1}{2}(h+a+c-ac\overline{h}) = \dfrac{1}{4}\left(2a+b+2c-\dfrac{bc}{a}-\dfrac{ac}{b}+\dfrac{a^2}{b}\right).</cmath> We now simplify the given length condition: <cmath>\begin{align*} AH^2 &= 2AO^2 \\ \implies (h-a)\overline{(h-a)} &= 2 \\ \implies \dfrac{1}{2}\left(-a+b+c-\dfrac{bc}{a}\right) \cdot \dfrac{1}{2}\left(-\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{a}{bc}\right) &= 2 \\ \implies (ab+ac-bc-a^2)^2 &= 8a^2bc \\ \implies (a-b)^2(a-c)^2 &= 8a^2bc.\end{align*}</cmath> We would like to show that <math>P</math>, <math>O</math>, <math>Q</math> are collinear, or <cmath>\begin{vmatrix} p & \overline{p} & 1 \\ 0 & 0 & 1 \\ q & \overline{q} & 1 \end{vmatrix} = \overline{p}q - p\overline{q} = 0.</cmath> After some factoring (or expanding) that takes about 15 minutes, this eventually reduces to <cmath>\overline{p}q - p\overline{q} = 0 \iff (a-b)(a-c)(b-c)((a-b)^2(a-c)^2 - 8a^2bc) = 0,</cmath> which is true. <math>\square</math>
 +
 +
-MP8148
 +
 +
==Solution 6==
 +
Claim: <math>\triangle AOQ \sim \triangle AHB</math>
 +
 +
Proof: We compute the area of <math>\triangle ABC</math> using two methods. Let <math>R=AO</math> be the circumradius of <math>\triangle ABC</math>.
 +
 +
First, by extended law of sines, <math>BC=2R \sin \angle BAC</math>. We are also given that <math>AH= R \sqrt{2}</math>.
 +
<math>AH \perp BC</math>, so <cmath>[ABC]=\frac{2R \sin \angle BAC \cdot R \sqrt{2}}{2}=R^2 \sin \angle BAC \sqrt{2}.</cmath>
 +
 +
Second, we compute the area using <math>[ABC]=\frac{1}{2} \cdot AB \cdot AC \sin \angle BAC</math>.
 +
 +
Equating these two expressions for the area of <math>\triangle ABC</math> and reducing, we get <cmath>AB \cdot AC = 2 R^2 \sqrt{2}.</cmath>
 +
But <math>2R^2 =2 AO^2 = AH^2</math>, so <math>AB \cdot AC = AH^2 \sqrt{2}</math> and <math>\bf{AB =\frac{AH^2 \sqrt{2}}{AC}}</math>.
 +
 +
Since <math>\angle AQH=\angle AHC =90^\circ</math>, and both triangles share the angle <math>\angle HAC</math>, <math>\triangle AHC \sim \triangle AQH</math>. This tells us <math>\frac{AQ}{AH}=\frac{AH}{AC}</math> or <math>AQ =\frac{AH^2}{AC}</math>.
 +
 +
Substituting into the bolded equation, we get <math>AB=AQ \sqrt{2}</math>.
 +
 +
The original length condition can be written as <math>AH= AO \sqrt{2}</math>.
 +
 +
We also have <cmath>\angle OAQ= \angle OAC= 90- \angle ABC = \angle HAB.</cmath>
 +
 +
Therefore, by SAS similarity, <math>\triangle AOQ \sim \triangle AHB</math>.
 +
-----------------------------------------
 +
We can prove analogously that <math>\triangle AOP \sim \triangle AHC</math>.
 +
We now have <math>\angle AOQ=\angle AHB =90^\circ</math> and <math>\angle AOP=\angle AHC =90^\circ</math>.
 +
This implies that <math>\angle POQ=\angle POA +\angle QOA=180^\circ</math> which tells us <math>P,O,</math> and <math>Q</math> are collinear, as desired.<math>\square</math>
 +
 +
-vvluo
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:15, 14 June 2020

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

[asy] size(8cm); pair O=(0,0); pair A=dir(110); pair B=dir(-29); pair C=dir(209); pair H=foot(A,B,C); pair P=foot(H,A,B); pair Q=foot(H,A,C); draw(A--B--C--A--H--P); draw(circle(O,1)); draw(Q--H); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$H$", H, S); dot("$P$", P, NE); dot("$Q$", Q, NW); dot("$O$", O, S); [/asy]

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, and we are done. (We also obtain the extra interesting fact that $AO\perp PQ.$)

Solution 4

Draw the altitude from $O$ to $AB$, and let the foot of this altitude be $D$.

Then, by the Right Triangle Altitude Theorem on triangle $AHB$, we have: $AB\cdot AP=AH^{2}$.

Since $OD$ is the perpendicular bisector of $AB$, $2\cdot AD = AB$.

Substituting this into our previous equation gives $2\cdot AD \cdot AP = AH^{2}$, which equals $2\cdot AO^{2}$ by the problem condition.

Thus, $2\cdot AD\cdot AP = 2\cdot AO^{2} \implies AD\cdot AP = AO^{2}$.

Again, by the Right Triangle Altitude Theorem, angle $AOP$ is right.

By dropping an altitude from $O$ to $AC$ and using the same method, we can find that angle $AOQ$ is right. Since $\angle AOP=\angle AOQ=90$, $P$, $O$, $Q$ are collinear and we are done.

~champion999

Solution 5

We use complex numbers. Let lower case letters represent their respective upper case points, with $|a| = |b| = |c| = 1$. Spamming the foot from point to segment formula, we obtain \[h = \dfrac{1}{2}\left(a+b+c-\dfrac{bc}{a}\right),\] \[p = \dfrac{1}{2}(h+a+b-ab\overline{h}) = \dfrac{1}{4}\left(2a+2b+c-\dfrac{bc}{a}-\dfrac{ab}{c}+\dfrac{a^2}{c}\right),\] and \[q = \dfrac{1}{2}(h+a+c-ac\overline{h}) = \dfrac{1}{4}\left(2a+b+2c-\dfrac{bc}{a}-\dfrac{ac}{b}+\dfrac{a^2}{b}\right).\] We now simplify the given length condition: \begin{align*} AH^2 &= 2AO^2 \\ \implies (h-a)\overline{(h-a)} &= 2 \\ \implies \dfrac{1}{2}\left(-a+b+c-\dfrac{bc}{a}\right) \cdot \dfrac{1}{2}\left(-\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{a}{bc}\right) &= 2 \\ \implies (ab+ac-bc-a^2)^2 &= 8a^2bc \\ \implies (a-b)^2(a-c)^2 &= 8a^2bc.\end{align*} We would like to show that $P$, $O$, $Q$ are collinear, or \[\begin{vmatrix} p & \overline{p} & 1 \\ 0 & 0 & 1 \\ q & \overline{q} & 1 \end{vmatrix} = \overline{p}q - p\overline{q} = 0.\] After some factoring (or expanding) that takes about 15 minutes, this eventually reduces to \[\overline{p}q - p\overline{q} = 0 \iff (a-b)(a-c)(b-c)((a-b)^2(a-c)^2 - 8a^2bc) = 0,\] which is true. $\square$

-MP8148

Solution 6

Claim: $\triangle AOQ \sim \triangle AHB$

Proof: We compute the area of $\triangle ABC$ using two methods. Let $R=AO$ be the circumradius of $\triangle ABC$.

First, by extended law of sines, $BC=2R \sin \angle BAC$. We are also given that $AH= R \sqrt{2}$. $AH \perp BC$, so \[[ABC]=\frac{2R \sin \angle BAC \cdot R \sqrt{2}}{2}=R^2 \sin \angle BAC \sqrt{2}.\]

Second, we compute the area using $[ABC]=\frac{1}{2} \cdot AB \cdot AC \sin \angle BAC$.

Equating these two expressions for the area of $\triangle ABC$ and reducing, we get \[AB \cdot AC = 2 R^2 \sqrt{2}.\] But $2R^2 =2 AO^2 = AH^2$, so $AB \cdot AC = AH^2 \sqrt{2}$ and $\bf{AB =\frac{AH^2 \sqrt{2}}{AC}}$.

Since $\angle AQH=\angle AHC =90^\circ$, and both triangles share the angle $\angle HAC$, $\triangle AHC \sim \triangle AQH$. This tells us $\frac{AQ}{AH}=\frac{AH}{AC}$ or $AQ =\frac{AH^2}{AC}$.

Substituting into the bolded equation, we get $AB=AQ \sqrt{2}$.

The original length condition can be written as $AH= AO \sqrt{2}$.

We also have \[\angle OAQ= \angle OAC= 90- \angle ABC = \angle HAB.\]

Therefore, by SAS similarity, $\triangle AOQ \sim \triangle AHB$.

We can prove analogously that $\triangle AOP \sim \triangle AHC$. We now have $\angle AOQ=\angle AHB =90^\circ$ and $\angle AOP=\angle AHC =90^\circ$. This implies that $\angle POQ=\angle POA +\angle QOA=180^\circ$ which tells us $P,O,$ and $Q$ are collinear, as desired.$\square$

-vvluo

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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