Pretty easy for a JMO5 in my opinion (and certainly easier than JMO1); it took me about 20 minutes to solve during DiffEq today.

Anyone else thought this was much easier than JMO/1? It is well-known that $AH\cdot 2AO=AB\cdot AC$ (just use similar triangles or $\sqrt{bc}$ inversion). 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.

Length bashing with trig to show PO+OQ=PQ!

coordinate bashed gg

Complex bash also worked on this.

Barycentric bash also worked on this.

Coordinate bashed with $A=(0,a)$, $B=(b,0)$, $C=(c,0)$, $H=(0,0)$. It didn't turn out that bad, though I did get some messy expressions.

I bashed the area of triangle $\triangle PQO$, which worked out pretty decently I guess.

Trig should show that O'A =R, where O' is intersection of PQ and reflection of AH over angle bisector. Is this sufficient?

definitely easier than jmo 1 first wlog R=1, then bc = 2sqrt2, then law of sines and law of cosines on AOP and AOQ to get that angles AOP and AOQ are right.

Since the reflection of AH over the angle bisector is AO, this is sufficient, though you should add what I just said to the solution. oops sniped

mathwizard888 wrote: Coordinate bashed with $A=(0,a)$, $B=(b,0)$, $C=(c,0)$, $H=(0,0)$. It didn't turn out that bad, though I did get some messy expressions. yes. mine was 6 pages , two exceptions when b and c are negatives or o and p are vertical

Did anyone else prove that $P$ and $Q$ were on the diameter opposite of $A$ (immediately implying the solution)?

bestwillcui1 wrote: mathwizard888 wrote: Coordinate bashed with $A=(0,a)$, $B=(b,0)$, $C=(c,0)$, $H=(0,0)$. It didn't turn out that bad, though I did get some messy expressions. yes. mine was 6 pages , two exceptions when b and c are negatives or o and p are vertical Wait dealing with OP vertical is easy, and you know exactly one of b and c is negative because ABC is acute.

FlakeLCR wrote: Did anyone else prove that $P$ and $Q$ were on the diameter opposite of $A$ (immediately implying the solution)? I actually saw that it looked true at like 1:30 (PT) but I was too lazy to do it b/c I already had a bashy solution. It seriously looked like it though (I constructed it)... I saw that it looked like it after I rotated $AH$ onto the circle and I thought that $P$ and $Q$ seriously resembled the intersection b/w the line connecting the two rotations of $AH$ on the circle and $AB$ and $AC$ respectively.

I showed that $BPQC$ was cyclic and that the extension of $AO$ is perpendicular to $PQ$. Letting this intersection be $O'$, I showed $O=O'$ by length bashing using angle equalities from the cyclic quads and right triangles.

I coordinate bashed with $A=(0,0) B=(1,0) C=(b, c)$ and found expressions for $O, P, Q,H$. I wrote out the collinearity stuff and the $2AO^2=AH^2$, and said they were equivalent. How many points would I get? Hopefully >0?

Hey look I got to do this problem because my proctor gave me the wrong test. Unfortunately I realized this was USAJMO immediately after solving it Since $AO\perp PQ$, $O$ lies on $PQ$ iff it corresponds to $H$ in similar $AQP,ABC$. Let $AD=k\cdot AH$, so the ratio of similarity is $k\cos A$, and $AH=2kR\cos A$, $AO=R$, so the problem holds iff $AO=\cos A\cdot AH \Leftrightarrow R = 2k^2R\cos^2 A \Leftrightarrow AH^2=2AO^2$.

http://www.mit.edu/~evanchen/handouts/bary/bary-full.pdf thank

Basicaly by the 90º angles we have that $(BPH),(CQH)$ are tangent at $H$ and the tangent line is $AH$ so by PoP $AP \cdot AB=AQ \cdot AC=AH^2=2AO^2=AO \cdot AA'$ where $A'$ is the $A$-antipode in $(ABC)$, now by inversion with center $A$ and radius $AH$ we have that $P \to B$, $C \to Q$, $O \to A'$ so $BPOA',CQOA'$ are cyclic but note that $(ABC) \to PQ$ so $O$ lies in $PQ$, thus we are done

The desired condition can be written as $[AOP] + [AOQ ] = [APQ]$, or \begin{align*} R(AP \cos C + AQ \cos B) &= AP \cdot AQ \cdot \sin A \\ \sin^2 B\sin C \cos C + \sin^2 C \sin B \cos B &= 2\sin A \sin^3 B \sin^3 C \\ \sin^2 B \sin^2 C = \frac 12 \end{align*}as $AP = AH \sin B = 2R \sin^2 B \sin C$. But $AH^2 = 2AO^2$ is obviously equivalent to this as well, so we are done.

We can equivalently prove $[AOP] + [AOQ ] = [APQ],$ and using the sine area formula gives $$R \cdot AQ \cdot \sin \angle BAO + R\cdot AP \cdot \sin \angle CAO = AQ \cdot AP \cdot \sin \angle A.$$Since $O$ and $H$ are isogonal conjugates, we note that $\angle BAO = \angle HAC = 90^\circ - \angle C$ and similarly $\angle CAO = \angle BAH =90^\circ - \angle B.$ Since $\sin (90^\circ-x)=\cos x,$ we simplify our equation to $$R \cdot AQ \cdot \cos \angle C + R\cdot AP \cdot \cos \angle B = AQ \cdot AP \cdot \sin \angle A.$$In right triangle $AHP,$ observe that $\angle AHP=90^\circ - \angle HAP = \angle C,$ so $\sin \angle C = \tfrac{AP}{AH},$ but right triangle $AHC$ gives $\sin \angle C =\tfrac{AH}{b},$ hence equating gives $AP=\tfrac{AH^2}{b}=\tfrac{2R^2}{b}$ by the condition. Symmetrically, $AQ=\tfrac{2R^2}{c}.$ Substituting back in, $$R \cdot \frac{2R^2}{c} \cdot \cos \angle C + R\cdot \frac{2R^2}{b} \cdot \cos \angle B = \frac{2R^2}{c} \cdot \frac{2R^2}{b} \cdot \sin \angle A$$which directly simplifies to $$b \cos\angle C+c\cos \angle B=2R \sin \angle A=a,$$where the last step follows by the Extended Law of Sines. Now, expressing the cosines of the angles in terms of side lengths using the Law of Cosines and substituting, it remains to show $$a=b \cdot \frac{a^2+b^2-c^2}{2ab}+c \cdot \frac{a^2+c^2-b^2}{2ac},$$which is trivial upon simplification. Remarks: I needed a hint for the area re-interpretation, and the rest was blind trigging

Invert around $A$ with radius $AH$. Notice that line $PQ$ swaps with $(ABC)$ since $AP \cdot AB=AH^2$ by similar triangles. Let the inverse of $O$ be $O^*$. From the condition, $AO\cdot AO^*=AH^2=2AO^2,$ so $O^*$ is the reflection of $A$ wrt $O,$ the antipode of $A$. However, $O^*$ clearly lies on $(ABC),$ which means that $O$ lies on $PQ$ as desired.

trigbash ftw Letting $R$ be the circumradius, the given length condition is $AH = \sqrt{2}R$. Let $\angle HAB = x$ and $\angle HAC= y$ so that $\angle ABC = 90^\circ- x$ and $\angle ACB = 90^\circ - y$. Additionally, note that $\angle OAP = 90^\circ - \angle ACB = y$ and similarly $\angle OAQ = x$. Now, to prove that $P, O, Q$ are collinear, it suffices to show that $[APQ] = [AOP] + [AOQ]$. First, we will compute $[APQ]$. Note that $AP = \sqrt{2}R\cos x$ and $AQ = \sqrt{2}R\cos y$, so \[ [APQ] = \frac{1}{2}(\sqrt{2}R\cos x)(\sqrt{2}R\cos y)\sin (x + y) = R^2\cos x\cos y\sin (x + y).\]Now, noting that $AO = R$, we can also compute \[ [AOP] = \frac{1}{2}AP\cdot AO\sin \angle OAP = \frac{\sqrt{2}}{2}R^2\cos x\sin y,\]and similarly $[AOQ] = \tfrac{\sqrt{2}}{2}R^2\cos y\sin x$. It follows that $[AOP] + [AOQ] = \frac{\sqrt{2}}{2}R^2\sin (x + y)$, so to show this equals $[APQ]$, it suffices to show $\cos x\cos y = \tfrac{\sqrt{2}}{2}$. Note that \[ [ABC] = \frac{1}{2}AH\cdot BC= 2R^2\sin A\sin B\sin C,\]\[ \frac{1}{2}\sqrt{2}R\cdot 2R\sin A = 2R^2\sin A\sin B\sin C\implies \sin B\sin C = \frac{\sqrt{2}}{2}.\]However, $\sin B\sin C = \sin (90^\circ - x)\sin (90^\circ - y) = \cos x\cos y$, so we have $\cos x \cos y = \tfrac{\sqrt{2}}{2}$ and we are done.

Let $O_{1}$ be the intersection of line $AO$ and $PQ$. Note that $\angle{APQ} \cong \angle{AHQ} \cong \angle{ACB}$ because $APHQ$ is cyclic. Thus, $\triangle{APQ} \sim \triangle{ABC}$. Also note that $\angle{PAO_{1}} = 90-\angle{C}$ because of circumcircle properties. Thus, $\angle{PO_{1}A} = \angle{90}$. Thus, $AO_{1}$ is the altitude. \[\frac{AQ}{AB} = \frac{AO_{1}}{AH}\] Note, \[AQ = AH\sin{C}\]and \[\frac{AB}{\sin{C}} = 2AO\]Thus, \begin{align*} AO_{1} &= \frac{AQ\cdot AH}{AB} \\ &= \frac{AH^2\sin{C}}{AB} \\ &= \frac{2AO^2}{2AO} \\ &= AO \end{align*}Thus, $O = O_{1}$ proving that $P$, $O$ and $Q$ are collinear.

Let A' be the diametrically opposite point of A. Then $$AO*AD=AH^2=AQ*AC=AP*AB,$$which readily implies the cyclic quads $$BPOA',CQOA'\implies POA'+QOA'=180-PBA'+180-QCA'=180,$$as desired. $\blacksquare$

Set circumcircle of $\triangle ABC$ as the unit circle and perform a Complex Numbers Coordinate bash. It works. Full proof here: https://infinityintegral.substack.com/p/usajmo-2016-contest-review

Let $AO = R$. We have \begin{align*} OP^2 &= \left(AP-\frac{AB}{2} \right)^2 + \left(R^2 - \left(\frac{AB}{2} \right)^2 \right) \\ &= AP^2-AB \cdot AP +\frac{AB^2}{4} +R^2 - \frac{AB^2}{4} \\ & = AP^2- AH^2 +R^2 \\ &= AP^2-R^2. \end{align*} Hence, $\angle AOP = 90^\circ$. We can also find that $\angle AOQ = 90^\circ$ in a similar fashion so we are done. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.731310895342761, xmax = 21.47317966094688, ymin = -2.2122120152209988, ymax = 13.666927038291828; /* image dimensions */ /* draw figures */ draw((5,12)--(0,0), linewidth(1)); draw((0,0)--(15.053205510543378,0.11333175599989247), linewidth(1)); draw((15.053205510543378,0.11333175599989247)--(5,12), linewidth(1)); draw((0.7665722796619195,1.839773471188607)--(10.935438145969488,4.982080785085927), linewidth(1)); draw((5,12)--(5.0900564335560805,0.0383217403993734), linewidth(1)); draw((5.0900564335560805,0.0383217403993734)--(0.7665722796619195,1.839773471188607), linewidth(1)); draw((5.0900564335560805,0.0383217403993734)--(10.935438145969488,4.982080785085927), linewidth(1)); draw((5,12)--(7.497534055385944,3.917694143589191), linewidth(1)); /* dots and labels */ dot((5,12),dotstyle); label("$A$", (5.085595268282164,12.200257790738304), NE * labelscalefactor); dot((0,0),dotstyle); label("$B$", (0.17936423663279954,0.09312555076678242), NE * labelscalefactor); dot((15.053205510543378,0.11333175599989247),dotstyle); label("$C$", (15.137168511515654,0.31045909057106436), NE * labelscalefactor); dot((7.497534055385944,3.917694143589191),linewidth(4pt) + dotstyle); label("$O$", (7.569155194139466,4.115132364308087), NE * labelscalefactor); dot((5.0900564335560805,0.0383217403993734),linewidth(4pt) + dotstyle); label("$H$", (5.163817628151685,-0.39312555076678242), NE * labelscalefactor); dot((0.7665722796619195,1.839773471188607),linewidth(4pt) + dotstyle); label("$P$", (0.4420322453606324,1.9922398277657725), NE * labelscalefactor); dot((10.935438145969488,4.982080785085927),linewidth(4pt) + dotstyle); label("$Q$", (11.010939028398404,5.1406898125140055), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Anyone try inversion about A with radius AH

We proceed by complex with $O$ as the origin, and $A$, $B$, and $C$ lying on the unit circle. By complex foot, we have \[h=\frac{1}{2}\left(a+b+c-\frac{bc}{a}\right).\]Hence, the given condition reduces to \begin{align*} |a-h|^2=2|a|^2&=2 \\ (a-h)\overline{(a-h)}&=2 \\ \frac{1}{2}\left(a-b-c+\frac{bc}{a}\right)\overline{\frac{1}{2}\left(a-b-c+\frac{bc}{a}\right)}&=2 \\ \left(a-b-c+\frac{bc}{a}\right)\left(\frac{1}{a}-\frac{1}{b}-\frac{1}{c}+\frac{a}{bc}\right)&=8 \\ \frac{a^2}{bc}+\frac{bc}{a^2}-\frac{2a}{b}-\frac{2b}{a}-\frac{2a}{c}-\frac{2c}{a}+\frac{b}{c}+\frac{c}{b}&=4. \end{align*}Then, by complex foot again, we have \[p=\frac{1}{2}\left(a+b+h-ab\overline{h}\right),\]which expands to \begin{align*} p&=\frac{1}{4}\left(2a+2b+\left(a+b+c-\frac{bc}{a}\right)-ab\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{a}{bc}\right)\right) \\ &=\frac{1}{4}\left(2a+2b+c-\frac{bc}{a}-\frac{ab}{c}+\frac{a^2}{c}\right). \end{align*}Analogously, we have \[q=\frac{1}{4}\left(2a+2c+b-\frac{bc}{a}-\frac{ac}{b}+\frac{a^2}{b}\right).\]By the complex collinearity criterion, it suffices to show that \[\frac{0-p}{0-q}=\frac{p}{q} \in \mathbb{R}.\]Thus, we calculate \begin{align*} \frac{p}{q}&=\frac{2a+2b+c-\frac{bc}{a}-\frac{ab}{c}+\frac{a^2}{c}}{2a+2c+b-\frac{bc}{a}-\frac{ac}{b}+\frac{a^2}{b}} \\ &=\frac{2+\frac{2b}{a}+\frac{c}{a}-\frac{bc}{a^2}-\frac{b}{c}+\frac{a}{c}}{2+\frac{2c}{a}+\frac{b}{a}-\frac{bc}{a^2}-\frac{c}{b}+\frac{a}{b}}. \end{align*}Now, let $x$ and $y$ be the numerator and denominator of this fraction, respectively, so that $\frac{p}{q}=\frac{x}{y}$. Note that \[x+\overline{x}=4+\left(-\frac{a^2}{bc}-\frac{bc}{a^2}+\frac{2a}{b}+\frac{2b}{a}+\frac{2a}{c}+\frac{2c}{a}-\frac{b}{c}-\frac{c}{b}\right)=4-4=0\]by substituting directly into the result we derived from the given condition. Similarly, we get $y+\overline{y}=0$. Hence, \[\frac{x}{y}=\frac{-\overline{x}}{-\overline{y}}=\frac{\overline{x}}{\overline{y}} \implies \frac{x}{y} \in \mathbb{R},\]which clearly finishes.

Note that $APHQ$ is cyclic since $\angle APH + \angle HQA = 90^\circ + 90^\circ = 180^\circ$, and inscribed angles and AA Similarity easily shows $\triangle APQ \sim ACB$ (inversely similar.) Now $$\angle AOP = 180^\circ - \angle PAO - \angle APO = 180^\circ - \angle CAH - \angle AHQ = 180^\circ - 90^\circ = 90^\circ.$$Therefore, $AO \perp PQ$, so it is the altitude of $\triangle APQ$ with respect to $A$. Then $$\dfrac{AO}{AH} = \dfrac{AQ}{AB} \implies \dfrac{c}{2b \sin(C)^2} = \dfrac{b \sin(C)^2}{c} \implies \dfrac{c}{2b \sin(C)^2} = \dfrac{1}{\sqrt{2}}$$which yields $AH = AO \sqrt{2} \implies AH^2 = 2 AO^2$ as desired.

Easy for a JMO P5 Geometry. Toss the figure on the coordinate plane. Set $A \equiv (1,b), B \equiv(0,0), C \equiv (a,0)$ and $H \equiv (1,0)$. Now to find coordinates of $P,Q$ we can use: Lemma: Let the foot of perpendicular from a point $(x_{0},y_{0})$ to a line $ax+by+c=0$ be $(h,k)$. Then, $$ \boxed{\dfrac{h-x_{0}}{a}=\dfrac{k-y_{0}}{b}=-\left(\dfrac{ax_{0}+by_{0}+c}{a^{2}+b^{2}}\right)}.$$Thus we shall get, $\boxed{P \equiv \left(\dfrac{1}{b^{2}+1},\dfrac{b}{b^{2}+1}\right)}$ and $\boxed{Q \equiv \left(\dfrac{a^{2}-2a+ab^{2}+1}{a^{2}-2a+b^{2}+1},\dfrac{a^{2}b-2ab+b}{a^{2}-2a+b^{2}+1}\right)}$. Now we can find the coordinates of $O$ by intersection of perpendicular bisectors of sides. Thus, $$\boxed{O \equiv \left(\dfrac{a}{2},\dfrac{b^{2}-a+1}{2b}\right)}$$Now clearly we have $AH^{2}=b^{2}$ and $AO^{2}=\left(\dfrac{a}{2}-1\right)^{2}+\left(\dfrac{b^{2}-a+1}{2b}-b\right)^{2}$. Thus, $$ AH^{2}=2AO^{2}$$$$ \implies b^{2}=2\left(\left(\dfrac{a}{2}-1\right)^{2}+\left(\dfrac{b^{2}-a+1}{2b}-b\right)^{2}\right)$$$$ \implies 2b^{4}=b^{4}+2b^{2}+1+a^{2}-2a+a^{2}b^{2}-2ab^{2}$$$$ \implies b^{4}-2b^{2}-1-a^{2}+2a-a^{2}b^{2}+2ab^{2}=0$$$$ \implies \boxed{b^{2}(b^{2}-a^{2}+2a-2)=(a-1)^{2}}$$Now we need to find the equation of line $PQ$ and satisfy the coordinates of $O$ in it to get the condition for $O,P$ and $Q$ to be collinear. Let $m_{PQ}$ be the slope of line $PQ$. Then, $$ m_{PQ}=\left(\dfrac{\dfrac{b}{b^{2}+1}-\dfrac{a^{2}b-2ab+b}{a^{2}-2a+b^{2}+1}}{\dfrac{1}{b^{2}+1}-\dfrac{a^{2}-2a+ab^{2}+1}{a^{2}-2a+b^{2}+1}}\right) $$$$ \implies m_{PQ}=\dfrac{b^{3}(-2a+a^{2})}{a^{2}b^{2}-ab^{2}+ab^{4}}$$$$ \implies m_{PQ}=\dfrac{b(a-2)}{a+b^{2}-1}$$Now we can find the equation of line $PQ$ as: $$\boxed{y=\left(\dfrac{b(a-2)}{a+b^{2}-1}\right)x+\left(\dfrac{b}{a+b^{2}-1}\right)}$$Now we can just plug the coordinates of $O$ into it to get the desired condition. Thus for $O,P,Q$ to be collinear we must have, $$\dfrac{b^{2}-a+1}{2b}=\dfrac{\dfrac{ab(a-2)}{2}+b}{a+b^{2}-1}$$$$ \implies 2b^{2}+b^{2}(a^{2}-2a)=b^{4}-(a-1)^{2} $$$$ \implies \boxed{b^{2}(b^{2}-a^{2}+2a-2)=(a-1)^{2}}.$$This condition is exactly equivalent to that of $AH^{2}=2AO^{2}$ and hence, we are done. $\blacksquare$

The condition we want to prove is equivalent to $\angle AQO = \angle PQA$ or $\angle AQO = \angle PHA$ by a cyclic quadrilateral. Because $AH$ and $AO$ are isogonal in $A$, this reduces to proving that $\triangle APH \sim \triangle AOQ$ or eqiuvalently $\triangle AOQ \sim AHB$ ergo $$\frac{AH^2}{AO^2} = \frac{AB^2}{AQ^2} $$$$2 = \frac{AB^2}{AQ^2}$$So from our condition we just need to prove $2 AQ^2 = AB^2$ and we are done. Now we resort to trig. We have that $AO \sin{\angle C} = \frac{AB}{2}$ and $AH \sin{\angle C} = AQ$ by basic right triangles. Plugging this into our given condition we get $$2 \cdot \left( \frac{AB}{2 \sin{\angle C}} \right)^2 = \left( \frac{AQ}{\sin{\angle C}} \right)^2$$which reduces to $$AB^2 = 2AQ^2$$, done.