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