N.T.TUAN wrote: Let $ABC$ be a triangle. Triangles $PAB$ and $QAC$ are constructed outside of triangle $ABC$ such that $AP = AB$ and $AQ = AC$ and $\angle{BAP}= \angle{CAQ}$. Segments $BQ$ and $CP$ meet at $R$. Let $O$ be the circumcenter of triangle $BCR$. Prove that $AO \perp PQ.$ Well, let's solve this one with complex numbers. We denote $\angle{BAP}= \angle{CAQ}=\phi$ then because the triangles $PAC$ and $QAB$ coincide after rotation with center $A$ and angle $\phi$, we have that the angle formed by the segments $CP$ and $QB$ is equal to $\phi$. Then we get that $\angle{BOC}=2\phi$ and $BO=CO$. Now we take $A$ as the origin of the complex plane. Then if $z=\cos{\phi}+i\sin{\phi}$ we have that $p=\frac{b}{z}$ and $q=cz$ and also $\frac{b-o}{c-o}=z^{2}$ thus $o=\frac{z^{2}c-b}{z^{2}-1}=\frac{q-p}{z-\frac{1}{z}}$. Hence $\frac{o}{\bar{o}}=-\frac{q-p}{\bar{q}-\bar{p}}$ which means that $OA$ is perpendicular to $PQ$, as desired.

Here is my synthetic solution There is no new philosophy in this problem. The idea is how to relate the circumcenter $O$ with $PQ$. Since we have so many pairs of equal angles here, probably it would help if we can take advantage of some hidden cyclic quadrilaterals and some spirality. Let $\measuredangle{PAB}= \measuredangle{CAQ}= \alpha$. It is not hard to see that $APBR$ and $AQCR$ are cyclic. Indeed, the congruence of two triangles $APC$ and $ABQ$ gives us $\measuredangle{ABR}= \measuredangle{APR}$, $\measuredangle{AQR}= \measuredangle{ACR}$. Let's draw the circles $(ARBP)$ and $(ARCQ)$. Denote $U, V$ respectively the intersections of $AO$ with these two circles. Notice that $\measuredangle{BOC}= 2 \alpha$. Hence, let 's bisect this angle by calling $T$ the midpoint of the arc $BC$ of the circumcircle of $BRC$. Then we have some spiralities here: $BPA$ and $BTO$, $CQA$ and $CTO$ $(*)$. Thus $BUTO$ is cyclic and $P, T, U$ are collinear. Similarly, $COVT$ is cyclic and $C, T, V$ are collinear. We need now to prove $\measuredangle{PUA}+\measuredangle{UPQ}= 90^{0}$. This is equivalent to proving $\measuredangle{TPQ}= \frac{\alpha}{2}$. Hence, it is sufficient for us to prove that actually $PTQ$ is isosceles at $T$ and $\measuredangle{PTQ}= 180^{0}-\alpha$. The latter is very easy by some angle-chasing. To prve $TP = TQ$, use $(*)$.

although it's a little longer than the proofs above, the lemma $AB\perp CD \Leftrightarrow AC^{2}+BD^{2}=AD^{2}+BC^{2}$ and some trigonometry solve the problem

What's surprise! the main idea of my first solution is similar with yours, treegoner. The idea is how to relate the circumcenter O with PQ This is my first solution Triangles $APC$ and $ABQ$ are congruent so $\angle APR=\angle ABR$. then $A,P,B,R$ are cyclic. So $\angle BRP=\angle PAB$. But $\angle PRB= \angle BOC/2$ so $\angle PAB=\angle BOC/2$ Let the perpendicular line of $BC,AP,AQ$ through $O,B,C$ meet $BC,AP,AQ$ at $D,E,F$. triangle $CFA$ and $BEA$ are similar so $AE/AF=AB/AC=AP/AQ$ so $EF//PQ$ $(1)$ We have triangle $CDO$ and $CFA$ are similar. Then triangle $CFD$ and $CAO$ are similar thus $FD/AO=CD/OC$ and $\angle FDC=\angle AOC$ $(2)$ Similarly $ED/AO=BD/BO$ and $\angle BDE=\angle BOA$ $(3)$ from (2) and (3) $DE=DF$ and $\angle EDF=180-\angle BOC$ then $\angle EFD=\angle BOC/2=\angle CAQ$ $(4)$ $FD$ meets $AO$ at $K$. we have $\angle CFK=\angle CAK$ so $A,K,C,F$ are cyclic. then $\angle CAQ=\angle CKF$ $(5)$ from (4) and (5) $EF//CK$ $(6)$ from (1) and (6) $PQ//CK$ but $\angle CKA=\angle CFA=90$ so $PQ\bot AO$ $q.e.d$

May be this soluiton is shorter Triangle $APC$ and $ABQ$ are congruent so $\angle APR=\angle ABR$. Then $A,P,B,R$ are on circle $(O1)$. Similarly $A,R,C,Q$ are on circle $(O2)$ $(O1), (O2)$ meet $AQ,AP$ at $E,F$ Let $(I)$ be the circumcircle of triangle $EAF$ $O1I,O2I$ meet $AE,AF$ at $M,N$. we have $\angle AEP=\angle ARC=\angle AFQ$ so triangles $AEP, AFQ$ are similar. $M,N$ are midpoint of $AE,AF$ so triangles $PAM$ and $QAN$ are similar so $\angle MPN=\angle MQN$. Then $P,M,N,Q$ are cyclic And with $I,M,A,N$ are cyclic we have $\angle AQP=\angle MNA=\angle MIA$. So $I,M,H,Q$ are cyclic. ($IA$ meets $PQ$ at $H$) Thus $IA\bot PQ$ But $PQ$ is the radical axis of $(I)$ and $(O)$ so $IO\bot PQ$. (because $PA*PF=PR*PC$ and $QA*QE=QB*QR$) This means $AO\bot PQ$ $q.e.d$

Wow, amazing second proof, Huyen Vu.

May be this metrical proof is longest ! I"ll show that $OP^{2}-OQ^{2}=c^{2}-b^{2}$, i.e. $AO\perp PQ$. Denote $2x=m(\widehat{BAP})$, the circumcircle $C(O,\rho )$ of $\triangle BRC$ and $S\in AR\cap BC$. Thus, $\{\begin{array}{c}AC=AQ\\\ AP=AB\\\ \widehat{CAP}\equiv\widehat{QAB}\end{array}\|$ $\implies$ $\triangle ACP\sim\triangle AQB$ $\implies$ $\{\begin{array}{c}CP=QB\\\ \widehat{APC}\equiv\widehat{ABQ}\\\ \widehat{ACP}\equiv\widehat{AQB}\end{array}\|$ $\implies$ $\{\begin{array}{c}\widehat{APR}\equiv\widehat{ABR}\\\ \widehat{ACR}\equiv\widehat{AQR}\end{array}\|$ $\implies$ the quadrilaterals $APBR$, $AQCR$ are cyclically with the radical axis $AR$. Observe that $\{\begin{array}{c}m(\widehat{BRS})=m(\widehat{BPA})=90^{\circ}-x\\\ m(\widehat{CRS})=m(\widehat{CQA})=90^{\circ}-x\\\ m(\widehat{BRP})=m(\widehat{BAP})=2x\end{array}\|$ $\implies$ the ray $[RS$ is the bisector of the angle $\widehat{BRC}$. Prove easily that $\{\begin{array}{c}m(\widehat{BOC})=4x\\\ a=2\rho\sin 2x\end{array}$. Thus, $\{\begin{array}{c}PA=c\ ,\ QA=b\\\ PB=2c\cdot\sin x\\\ QC=2b\cdot \sin x\end{array}$ and $\{\begin{array}{c}m(\widehat{CBO})=m(\widehat{BCO})=90^{\circ}-2x\\\ m(\widehat{PBO})=180^{\circ}+3x-B\\\ m(\widehat{QCO})=180^{\circ}+3x-C\end{array}$. Apply the generalized Pythagoras' theorem in the triangles $POB$ , $QOC$ to the sides $OP$, $OQ$ respectively : $\{\begin{array}{c}OP^{2}=BP^{2}+\rho^{2}-2\rho\cdot BP\cos \widehat{PBO}\\\ OQ^{2}=CQ^{2}+\rho^{2}-2\rho\cdot CQ\cos \widehat{QCO}\end{array}$ $\implies$ $OP^{2}-OQ^{2}=4(c^{2}-b^{2})\sin^{2}x+4c\rho\sin x\cos (B-3x)-4b\rho \sin x\cos (C-3x)=$ $4(c^{2}-b^{2})\sin^{2}x+4c\rho \sin x(\cos B\cos 3x+\sin B\sin 3x)$ $-4b\rho \sin x(\cos C\cos 3x+\sin C\sin 3x)=$ $4(c^{2}-b^{2})\sin^{2}x+4\rho\sin x\cos 3x$ $(c\cos B-b\cos C)+4\rho \sin x\sin 3x(c\sin B-b\sin C)=$ $4(c^{2}-b^{2})\sin^{2}x+4\rho\sin x\cos 3x(\frac{a^{2}+c^{2}-b^{2}}{2a}-\frac{a^{2}+b^{2}-c^{2}}{2a})=$ $4(c^{2}-b^{2})\sin^{2}x+\frac{4\rho (c^{2}-b^{2})}{a}\sin x\cos 3x=$ $(c^{2}-b^{2})(4\sin^{2}x+\frac{2\sin x\cos 3x}{\sin 2x})=c^{2}-b^{2}$. I used the simple relations $c\sin B=b\sin C$ and $4\sin^{2}x+\frac{2\sin x\cos 3x}{\sin 2x}=1$.

This problem was in Brazilian Training Lists! I solved this using Spiral Similarity, some Trigonometry and the lemma mentioned by campos!

I found a really nice proof:

Let the perpendicular bisectors of $AB$ and $AC$ intersect $AP$ and $AQ$ at $O_1$ and $O_2$, respectively. Since quadrilaterals $ABPO_1$ and $ACQO_2$ are similar, it follows that $AO_1/AP=AO_2/AQ$ and hence that $PQ$ and $O_1 O_2$ are parallel. Now let $\mathcal{O}_1$ and $\mathcal{O}_2$ be the circles centered at $O_1$ and $O_2$ with radii $O_1 A$ and $O_2 A$, respectively. Let the second intersection of $\mathcal{O}_1$ and $\mathcal{O}_2$ be $K$. By the inscribed angle theorem, it follows that $\angle{(KB,KA)}=\angle{(KA,KC)}=90^\circ - \angle{(AP, AB)}$ which implies that $\angle{CK, BK}=2\angle{(AP, AB)}$. Now note that since $A$ is the center of spiral similarity taking $PB$ to $CQ$, it follows that $APBR$ and $AQCR$ are cyclic which implies that $\angle{(CR, BR)}=\angle{(AP, AB)}$ and hence that $\angle{(CO, BO)}=2\angle{(AP, AB)}=\angle{(CK, BK)}$. Therefore $BOCK$ is cyclic and since $AK$ bisects $\angle{BKC}$ and $OB=OC$, it follows that $A$, $K$ and $O$ are collinear. Hence $AO$ is the radical axis of $\mathcal{O}_1$ and $\mathcal{O}_2$, which implies that $AO$ is perpendicular to $O_1 O_2$ and therefore perpendicular to $PQ$ since $PQ$ is parallel to $O_1 O_2$.

Hello, I have a very funny solution to this problem.

Yes ! But the formulation of the problem screams for a complex bash

Dear Mathlinkers, you can also see : http://perso.orange.fr/jl.ayme vol. 16 Deux triangles adjacents... p. 62 which indicate p. 60-61 . Sincerely Jean-Louis

My solution: Let $ O' $ be the circumcenter of $ \triangle ACQ $ . Let $ M, N $ be the midpoint of $ CQ, CR $, respectively . Easy to see $ R \in (O') $ . Since $ O', M, N, C $ are concyclic , so we get $ \angle AO'O=\angle QCP $ . ... $ (1) $ Since $ \angle RO'O=\angle BQC, \angle O'OR=\angle CBQ $ , so we get $ \triangle ORO' \sim \triangle BCQ $ , hence $ \frac{O'A}{CQ}=\frac{O'R}{CQ}=\frac{O'O}{QB}=\frac{O'O}{CP} $ . ... $ (2) $ From $ (1) $ and $ (2) $ we get $ \triangle AOO' \sim \triangle QPC $ , so from $ OO' \perp PC $ and $ AO' \perp QC \Longrightarrow AO \perp QP $ . Q.E.D

My solution: Lemma: $AO \perp PQ$ if only if $AP^2-AQ^2=OP^2-OQ^2$ (It's well known) It is easy too see that: $\triangle APC\cong \triangle ABQ$ $\Longrightarrow $ $BQ=CP=1$ and $RP=d,RQ=c$ Let $AP=AB=a,CA=CQ=b$ and $AR=x$ $\Longrightarrow $ By power of point in $\odot (RBC)$ we get : $AP^2-AQ^2=OP^2-OQ^2$ if only if $a^2-b^2=d-c$ since $OP^2-OQ^2=d-c$ (By power of point in $\odot (RBC)$ By Stewart's theorem in $\triangle ACP$ we get: $a^2(1-d)+b^2d=x^2+d(1-d)...(1)$ By Stewart's theorem in $\triangle ABQ$ we get: $b^2(1-c)+a^2c=x^2+c(1-c)...(2)$ By $(1)-(2)$ we get:$a^2-b^2=d-c$ which it is true $\Longrightarrow $ $a^2-b^2=d-c$ $\Longrightarrow $ $AO \perp PQ$...

Needing help for a complex number solution, I'm trying to learn that technique at the moment and working through an article by Robin Park. And well, there is a step in the calculation that I can't seem to be able to comprehend. Noticing those rotations, we can find $p = a+e^{-i \theta}(b-a)$, $q=a+e^{i \theta}(c-a)$ and $o= \tfrac{b-ce^{2i \theta}}{1-e^{2i \theta}}$, where $\theta = \angle{BAP}$. It remains to verify $\tfrac{a-o}{\overline{a}-\overline{o}}=\tfrac{p-q}{\overline{p}-\overline{q}}$. To ease the amount of calculation we also set $A$ to be the origin. Also, we let $z = e^{i \theta}$.Then we get \[ \frac{o}{\overline{o}} = \frac{p-q}{\overline{p}-\overline{q}} \iff \frac{\frac{b-cz^2}{1-z^2}}{\frac{\frac{1}{b}-\frac{1}{cz^2}}{1-\frac{1}{z^2}}} = \frac{\frac{b}{z}-cz}{\frac{z}{b}-\frac{1}{cz}} \iff \dots \]That above is how the author of the article presented the solution; I didn't formulate it like him. Now to my problem: how did he get to that equivalent form? Especially, I assume he got $\overline{p}-\overline{q}=\tfrac{z}{b}-\tfrac{1}{cz}$. Why is that true? Also, admittingly, I'm still very weak at manipulating complex numbers as I rarely use those.

Here's a solution similar to Robin Park's. Whoops, I'm bad at this complex bashing stuff. We use complex numbers. Let the complex number corresponding to a point be its lowercase letter. Set $a$ as the origin, $c=1$, and $\angle PAB = \angle QAC = \theta$. By standard formulas, we have $p=e^{-i \theta}b$ and $q=e^{i \theta}c$. Also, I claim $APBR$ and $AQCR$ are both cyclic. Observe $\triangle PAB \sim \triangle CAQ$, with spiral center at $A$, so the standard Yufei spiral similarity configuration tells us $R \in \odot(ABP)$ and $R \in \odot(ACQ)$. Now angle chase to get $\angle BOC = 2 \theta$. By the rotation formula, you can get $o = \frac{b-ce^{i \theta}}{1-e^{i 2 \theta}}$. Now we want $\frac{o-a}{p-q}$ as pure imaginary, that is \[ \frac{(b-e^{i 2\theta})}{(1-e^{i 2\theta})(e^{-i \theta}b-e^{i \theta})} = \frac{-(\overline{b}-\frac{1}{e^{i 2\theta}})}{(1-\frac{1}{e^{i 2\theta}})(e^{i \theta} \overline{b}-\frac{1}{e^{i \theta}})} \]After some simplifying, one can see that this is clearly true, so $OA \perp PQ$ and we are done.

Finally got around to actually doing this. Looks like my solution bears resemblance to Huyền Vũ's in post 5, though the last part is a bit strange because I didn't know how to construct the argument elegantly. Construct point $T$ outside $\triangle ABC$ such that $\triangle TBC\sim\triangle ABP\sim\triangle ACQ$. By considering the rotation sending $\triangle APC$ to $\triangle ABQ$ (which are congruent by SAS) we see that the angle between lines $PC$ and $QB$ is equal to $\angle BAP = \angle BTC$. Thus $B$, $C$, $T$, and $R$ lie on a common circle, so $O$ is the circumcenter of $\triangle TBC$. We will subsequently dispose of point $R$. Denote by $D_1$ and $D_2$ the projections of $B$ onto $AP$ and $C$ onto $AQ$ respectively, and let $M$ be the midpoint of $\overline{BC}$. I claim that $D_1M = D_2M$. To prove this, remark that by simple angle chasing $\angle BOC = \angle ABD_1$, so $\triangle AD_1B$ is directly similar to $\triangle CMB$. By the duality of spiral similarity, $\triangle BD_1M\sim\triangle BAO$ with similarity ratio $r = \tfrac{BM}{BO}$. Analogously, $\triangle CD_2M\sim\triangle CAO$ with the same similarity ratio $r = \tfrac{CM}{CO}$. Since $AO$ is taken to $D_1M$ and $D_2M$ under these spiral similarities, we must have $D_1M = D_2M$. Now let $\psi$ denote the measure of angle $\angle ABD_1 = \angle ACD_2$, and remark that \[\angle D_1MD_2 = 180^\circ - \angle BMD_1 - \angle CMD_2 = 180^\circ - \angle BOC = 2\psi.\]It follows that, upon letting the perpendicular from $M$ to $D_1D_2$ intersect $AC$ at $X$, we have $\angle XMD_1 = \psi$, so quadrilateral $CD_1XM$ is cyclic. Thus the angle between $MX$ and $AC$ is equal to $\angle MD_1C = \angle OAC$, which implies $AO\perp D_1D_2$. But $D_1D_2\parallel PQ$ by similarity, and so we obtain the desired perpendicularity. $\blacksquare$

Here is a synthetic solution constructing only one point. I would appreciate if someone simplifies my final directed angle chasing. Let $\theta = \angle BAP$. By Spiral Similarity, $ \triangle APC\stackrel{+}{\sim} \triangle ABQ$ so $\angle APR = \angle ABR\implies \angle BRP = \theta$. Thus $\angle BOC = 2\theta$. Now for the key step, construct point $X$ outside $ \triangle BOC$ such that $\triangle OXB\sim \triangle ABC$. Then $\angle XOC = \angle BAC + 2\theta = \angle APQ$ thus, $$\frac{OX}{OC} = \frac{OX}{OB} = \frac{AB}{AC} = \frac{AP}{AQ}\implies \triangle OXC\stackrel{-}{\sim} \triangle APQ.$$Furthermore, $\angle XBC = \angle C + (90^{\circ} - \theta) = \angle OCA$ so $$\frac{BC}{BX} = \frac{AC}{OB} = \frac{AC}{OC}\implies \triangle BCX\stackrel{-}{\sim} \triangle CAO.$$Hence, using directed angle, we find \begin{align*} \measuredangle(AO, PQ) &= \measuredangle(AO, AC) + \measuredangle(AC, AQ) + \measuredangle(AQ, PQ)\\ &=-\measuredangle(CX, BC) + \measuredangle(AC, AQ) - \measuredangle(OC, CX) \\ &= \measuredangle(AC, AQ) + \measuredangle(BC, OC) \\ &= 90^{\circ} \end{align*}as desired.

Here is a quick and easy solution. Clearly $ABPR$ and $ACQR$ are inscribed; next, meet $AP$ and $(ACQR)$ at $X$, and $AQ$ and $(ABPR)$ at $Y$. Since $\measuredangle AXQ = \measuredangle ARQ = \measuredangle ARB = \measuredangle AYP$, $PQXY$ is inscribed. So, if $\rho$ is the radius of $(BCR)$, \[ PO^2-PA^2 = PR\cdot PC + \rho^2 - PA^2 = PA\cdot PX +\rho^2 - PA^2 = PA\cdot AX + \rho^2 = \mathcal P(A,(PQXY)), \]where $\mathcal P$ denotes power. Since the last expression is the same for $P,Q$, this means $PO^2-PA^2 = QO^2-QA^2$, done!

Denote $\omega_b \equiv \odot(ABP)$ and $\omega_c \equiv \odot(ACQ).$ Let $AP$ meet $\omega_c$ for a second time at $X.$ Let $AQ$ meet $\omega_b$ for a second time at $Y.$ Because $A$ is the center of spiral similarity sending $\overline{PB} \mapsto \overline{CQ}$, we know that $R$ is the second intersection point of $\omega_b$ and $\omega_c.$ By Reim's theorem for $\omega_b, \omega_c$ cut by $PX, BQ,$ it follows that $PB \parallel XQ.$ Because $\triangle APB$ is isoceles, $PB$ is parallel to the tangent to $\omega_b$ at $A.$ Therefore, by the converse of Reim's theorem for $\omega_b$ cut by $PX, YQ,$ it follows that $PQXY$ is cyclic. Thus, there exists $r \in \mathbb{C}$ such that \[ r^2 = \text{pow}(A, \odot(PQXY)) = AP \cdot AX = AQ \cdot AY. \][asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = 974.3508556900372, xmax = 1046.0579652024412, ymin = 1011.7387089646945, ymax = 1035.0218454829537; /* image dimensions */ pen evevff = rgb(0.8980392156862745,0.8980392156862745,1.); pen evfuev = rgb(0.8980392156862745,0.9568627450980393,0.8980392156862745); pen qqzzqq = rgb(0.,0.6,0.); filldraw((1002.5480877317876,1028.5468363406196)--(990.6757045460399,1018.2042280191417)--(998.363012344087,1013.367634339197)--cycle, evevff, linewidth(0.8) + blue); filldraw((1002.5480877317876,1028.5468363406196)--(1016.3654794880058,1013.4008492232264)--(1022.4320417351443,1023.5518774921607)--cycle, evfuev, linewidth(0.8) + qqzzqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((1002.5480877317876,1028.5468363406196)--(990.6757045460399,1018.2042280191417), linewidth(0.8) + blue); draw((990.6757045460399,1018.2042280191417)--(998.363012344087,1013.367634339197), linewidth(0.8) + blue); draw((998.363012344087,1013.367634339197)--(1002.5480877317876,1028.5468363406196), linewidth(0.8) + blue); draw((1002.5480877317876,1028.5468363406196)--(1016.3654794880058,1013.4008492232264), linewidth(0.8) + qqzzqq); draw((1016.3654794880058,1013.4008492232264)--(1022.4320417351443,1023.5518774921607), linewidth(0.8) + qqzzqq); draw((1022.4320417351443,1023.5518774921607)--(1002.5480877317876,1028.5468363406196), linewidth(0.8) + qqzzqq); draw((998.363012344087,1013.367634339197)--(1022.4320417351443,1023.5518774921607), linewidth(0.8) + red); draw((1016.3654794880058,1013.4008492232264)--(990.6757045460399,1018.2042280191417), linewidth(0.8) + red); draw(circle((1011.7378103749238,1023.0547794798424), 10.705778384581844), linewidth(0.8)); draw(circle((998.1695210212386,1021.5875190512957), 8.222161733523125), linewidth(0.8)); draw((1002.5480877317876,1028.5468363406196)--(1007.5572715124796,1032.9105790739395), linewidth(0.8)); draw((1002.5480877317876,1028.5468363406196)--(997.599658737153,1029.7899089804253), linewidth(0.8)); /* dots and labels */ dot((1002.5480877317876,1028.5468363406196),linewidth(3.pt) + dotstyle); label("$A$", (1002.1745359708872,1029.1587799000256), N * labelscalefactor); dot((998.363012344087,1013.367634339197),linewidth(3.pt) + dotstyle); label("$B$", (998.0439421049653,1011.7415989196447), NE * labelscalefactor); dot((1016.3654794880058,1013.4008492232264),linewidth(3.pt) + dotstyle); label("$C$", (1016.3837788696585,1012.0068226742816), NE * labelscalefactor); dot((990.6757045460399,1018.2042280191417),linewidth(3.pt) + dotstyle); label("$P$", (989.5183963657025,1017.5600723245166), N * labelscalefactor); dot((1022.4320417351443,1023.5518774921607),linewidth(3.pt) + dotstyle); label("$Q$", (1022.9927290551335,1023.475082740517), E * labelscalefactor); dot((1003.9346138590062,1015.7251263534112),linewidth(3.pt) + dotstyle); label("$R$", (1003.7276392644737,1014.1216867336337), N * labelscalefactor); label("$\omega_c$", (1011.849575658217,1033.9971654909085), NE * labelscalefactor); label("$\omega_b$", (992.2524532204959,1028.4734976027294), N * labelscalefactor); dot((1007.5572715124796,1032.9105790739395),linewidth(3.pt) + dotstyle); label("$X$", (1007.0982038590661,1033.2893737659476), N * labelscalefactor); dot((997.599658737153,1029.7899089804253),linewidth(3.pt) + dotstyle); label("$Y$", (997.316957584563,1030.3814356843386), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Define a new circle $\gamma \equiv \odot(A, r).$ Because \[ PR \cdot PC = PA \cdot PX = PA(PA + AX) = PA^2 - r^2, \]we see that $P$ lies on the radical axis of $\odot(BCR)$ and $\gamma.$ Similarly, $Q$ lies on the radical axis of $\odot(BCR)$ and $\gamma.$ Therefore $PQ \perp AO.$

Skimming through (not very carefully), I haven't seen this solution yet. Note that $A$ is the spiral similarity that sends $PB$ to $CQ$, so it rotates $PC$ to $BQ$ implying $PC=BQ$. Also, $APBR, AQCR$ are cyclic. Let $N$ be the intersection of $(PRQ), (BRC)$. Then $N$ sends $BQ$ to $CP$, but $BQ=CP$, so $N$ is the midpoints of arcs $PQ$ and $BC$ in the circles. Note that $\measuredangle BRA = \measuredangle BPA = \measuredangle AQC = \measuredangle ARC$ so $AR$ bisects $BRC$ and $AR$, $NO$ concur on $(BRC)$. Therefore, $\measuredangle BPA = \measuredangle BRA = \measuredangle BNO$, so (since $AP=AB, ON=OB$), by (a working SSA similarity), $\triangle BON\sim \triangle BAP$. Therefore, a spiral similarity sends $NO$ to $PA$, and $AO$ to $PN$. Thus, the angle between $AO$ and $PQ$ is \[\measuredangle (AO, PN)+\measuredangle (PN, PQ)=\measuredangle ABP+\measuredangle NPQ = 90^{\circ}\] Oh jk this is the second solution in post 3

Much shorter solution. Let $\theta = \angle BAP$. Note that $\triangle ABQ\stackrel{+}{\sim}\triangle ACP$ thus $\angle(BQ, CP) = \theta)$. Therefore $\angle BRC = 2\theta$. Let $C'$ be the reflection of $C$ across $AQ$. Note that $\triangle ABP\stackrel{+}{\sim}\triangle AC'Q$ thus $\triangle ABC'\stackrel{+}{\sim}\triangle APQ$ which means $\angle(PQ,BC') = \theta\hdots (\spadesuit)$. Moreover, $\triangle CAC'\stackrel{+}{\sim}\triangle COB$ thus $\triangle CAO\stackrel{+}{\sim}\triangle CC'B$. Thus $\angle(BC', AO) = \angle(CA,CC') = 90^{\circ}-\theta$. Combining this with $(\spadesuit)$ gives the conclusion.

its easiest p6 i seen ! solution: its obvious $APBR , BRCQ$ are cyclic let $T , S$ be intersection of this circles with lines $AP$ and $AQ$ respectively. its obvious $PQST$ is cyclic quadrilateral and with this observations according power of point theorem we have: $QR.QB-PR.PC=QO^2-PO^2=AQ^2-AP^2$ problem solved.

We have $\angle BOC = 2(\pi - \angle BRC) = 2\angle BAP$. Let $(ABC)$ be the unit circle, and let $q = a + (c-a)z$ and $p = a + \frac{b-a}{z}$ for some complex number $z$ on the unit circle. The point $O$ is given by \[\frac{b-o}{c-o} = z^2 \implies o = \frac{b-z^2c}{1-z^2}.\]Then \[\frac{a-o}{p-q} = \frac{a - \frac{b-z^2c}{1-z^2}}{\frac{b-a}{z} - z(c-a)} = \frac{-z}{1-z^2}\]which is purely imaginary.

Triple Jacobi's theorem Let $S$ be a point outside $\triangle ABC$ and $\angle SBC = \angle ABP=\angle SCB$. By Jacobi, $AS,BQ,CP$ are concurrent at $R$ and $S$ lies on $\odot(BCR)$. Hence, $\angle OBC = \angle OCB=90^{\circ}-\angle BAP$. Let $B',C'$ be the foot of perpendicular from $B,C$ to $AP,AQ$ respectively. By Jacobi, $AO,BC',CB'$ are concurrent.Then Jacobi $\triangle AB'C'$, $AO \perp B'C'$ but $B'C' \parallel PQ$ ($\frac{AB'}{AP} = \frac{AC'}{AQ} \implies \triangle AB'C' \sim \triangle APQ$). Therefore, $AO \bot PQ$.

Let $\theta:=\angle PAB=\angle CAQ$. Note that $A$ is the center of spiral similarity $PB\mapsto CQ$, so it also maps $PC\mapsto BQ$. Thus $\angle (\overline{PC},\overline{BQ})=\theta$, in particular $\angle BRC =\pi-\theta$. Hence $\angle BOC = 2(\pi-\angle BRC) = 2\theta$. This eliminates $R$ entirely! Indeed, $O$ is now located on the perpendicular bisector of $\overline{BC}$ with $\angle BOC=2\theta$. Now use complex numbers with $A=0$. Let $t=e^{i\theta}$. Then \begin{align*} p=b/t,\quad q=tc. \end{align*}We know $\angle BOC=2\theta$ and $BO=CO$. Hence $|\tfrac{b-o}{c-o}| = 1 = |t^2|$, so actually \begin{align*} \frac{b-o}{c-o}\div t^2 = 1 \implies o=\frac{t^2c-b}{t^2-1}. \end{align*}Finally, confirm \begin{align*} \frac{o-a}{p-q} = \frac{t^2c-b}{(t^2-1)(b/t-tc)} = \frac{t}{1-t^2} \in i\mathbb{R}.

See my solution to this problem on my Youtube channel here: https://www.youtube.com/watch?v=Hz8kyLKP0_k

Here is a barycentric coordinate solution. Observe that $A$ is the miquel point of $PBQC$. This means that we must have $(APBR)$ and $(QARC)$ are cyclic. Since \[\angle ARP = \angle ABP = \angle ACQ = \angle ARQ\]this means $A$ lies on the angle bisector of $\angle BRC$. We now proceed with barycentric coordinates. Let our reference triangle be $\triangle BRC$, so $R = (1:0:0), B = (0:1:0), C = (0:0:1)$. For convenience, let $BC = a, CR = b, BR = c$. Since $A$ lies on the angle bisector, let $A = (t:b:c)$, for some number $t$. Now, the equation for $(ARB)$ is \[-a^{2}yz - b^{2}xz - c^{2}xy + (wz)(x+y+z)\]because $(ARB)$ passes through $B,R$. Plugging in $A$ gives \[-a^{2}bc - b^{2}tc - c^{2}tb + wc(t+b+c) = 0\Rightarrow w = \frac{a^{2}b + b^{2}t + cbt}{t+b+c}\]Now, since $P = (ARB)\cap RC$, if $P = (p:0:1-p)$, then we can plug this into our equation for $(ARB)$ to get \[-b^{2}p(1-p) + \frac{a^{2}b + b^{2}t + cbt}{t+b+c}\cdot (1-p) = 0 \Rightarrow p = \frac{a^{2} + bt + ct}{bt + b^{2} + bc}\]Therefore, by symmetry, \[P = \left(\frac{a^{2} + bt + ct}{b^{2}+bt + bc} : 0 : \frac{b^{2} + bc - ct - a^{2}}{bt + b^{2} + bc}\right), Q = \left(\frac{a^{2} + bt + ct}{c^{2} + ct + bc} : \frac{c^{2} + bc -bt - a^{2}}{c^{2} + ct + bc} : 0 \right)\]To show that $AO\perp PQ$, we can use generalized perpendicularity (Theorem 7.25 of EGMO). Assume that $O$ was the origin. We have \[\overrightarrow{PQ} = (a^{2} + bt + ct)\cdot \frac{c-b}{bc(b+c+t)}\vec{R} + \frac{bt + a^{2} - bc - c^{2}}{c(b+c+t)}\vec{B} + \frac{b^{2} + bc - ct - a^{2}}{c(b+c+t)}\vec{C}\]Since $O$ is the origin, we have $A = \frac{t}{t+b+c}\vec{R} + \frac{b}{b+c+t}\vec{B} + \frac{c}{b+c+t}\vec{C}$. By Theorem 7.25 of EGMO, we must show the following expression is $0$: \[0 = a^{2}\left(\frac{bt+a^{2}-bc-c^{2}}{b+c+t} + \frac{b^{2} + bc -ct - a^{2}}{b+t+c}\right)+ b^{2}\left(\frac{c-b}{b(b+c+t)}(a^{2} + bt + ct) + \frac{t(b^{2} + bc - ct - a^{2})}{b(b+c+t)}\right)+ c^{2}\left(\frac{c-b}{c(b+c+t)}(a^{2} + bt + ct) + \frac{t(bt + a^{2}-bc-c^{2})}{c(b+c+t)}\right)\]\[\Rightarrow 0 = a^{2}(b^{2}-c^{2} + bt - ct) + b((c-b)(a^{2} + bt + ct) + b^{2}t + bct - ct^{2} - a^{2}t)+ c((c-b)(a^{2}+bt+ct) + bt^{2} + bct - ct^{2} - at^{2})\]\[\Rightarrow 0 = a^{2}(b^{2} - c^{2} + bt - ct) + b[a^{2}c + bct + c^{2}t - a^{2}b - ct^{2} - a^{2}t] + c[a^{2}c - a^{2}b - b^{2}t + bt^{2} + a^{2}t - bct]\]\[= a^{2}b^{2} - a^{2}c^{2} + a^{2}bt - a^{2}ct + a^{2}bc + b^{2}ct + bc^{2}t - a^{2}b^{2} - bct^{2}-a^{2}tb + a^{2}c^{2} - a^{2}bc - b^{2}ct + bct^{2} + a^{2}ct - bc^{2}t = 0\]Therefore, this expression is $0$, which means $AO\perp PQ$.

k12byda5h wrote: Triple Jacobi's theorem Let $S$ be a point outside $\triangle ABC$ and $\angle SBC = \angle ABP=\angle SCB$. By Jacobi, $AS,BQ,CP$ are concurrent at $R$ and $S$ lies on $\odot(BCR)$. Hence, $\angle OBC = \angle OCB=90^{\circ}-\angle BAP$. Let $B',C'$ be the foot of perpendicular from $B,C$ to $AP,AQ$ respectively. By Jacobi, $AO,BC',CB'$ are concurrent.Then Jacobi $\triangle AB'C'$, $AO \perp B'C'$ but $B'C' \parallel PQ$ ($\frac{AB'}{AP} = \frac{AC'}{AQ} \implies \triangle AB'C' \sim \triangle APQ$). Therefore, $AO \bot PQ$. "Then Jacobi $\triangle AB'C'$, $AO \perp B'C'$" Why'd have this perpendicularity?

Quote: "Then Jacobi $\triangle AB'C'$, $AO \perp B'C'$" Why'd have this perpendicularity? Let $BC' \cap CB' = X$.By Jacobi, $AO,BC',CB'$ are concurrent $\implies A,O,X$ are collinear. Then Jacobi $\triangle AB'C'$ with point $\infty_{\perp B'C'},C,B$. We get $AOX \perp B'C'$

Let $R'=\overline{PB}\cap\overline{QC}$. By spiral sim $A,B,P,R$ and $A,C,R,Q$ are cyclic. Under an inversion at $A$ with radius $\sqrt{AP\cdot AQ}$ and a reflection about the angle bisector of $\angle PAQ$, $R$ goes to $R'$. Since $\angle APR'=\angle AQR'$, if we let $R''$ be the point such that $R'PR''Q$ is a parallelogram, $\overline{AR''},\overline{AR'}$ are isogonal in $\angle PAQ$, so $R-A-R''$. Let $O_1$ be the center of $(APR)$ and let $O_2$ be the center of $(AQR)$. Then the perpendicular from $P$ to $\overline{AO_2}$, the perpendicular from $Q$ to $\overline{AO_1}$, and the perpendicular from $R$ to $\overline{O_1O_2}$ meet at $R''$. Thus $\triangle AO_1O_2,\triangle RPQ$ are orthologic, as desired.

A very, very funny solution. Let $\angle BAP = \angle CAQ = \theta$. Since $\triangle APB\sim \triangle ACQ$, we have $\triangle APC\sim \triangle ABQ$, so $\angle (PC, BQ) = \theta$ (the spiral similarity at $A$ sending $APC$ to $ABQ$ has a rotation of angle $\theta$). Thus, $\angle BRC = 180^\circ - \theta$, so $\angle BOC = 2\theta$. It suffices to show that $\overrightarrow{AO}\cdot \overrightarrow{PQ} = 0$. Note that $\angle(PA, OC) = \measuredangle (PA, AB) + \measuredangle (AB, BC) + \measuredangle (BC, CO) = \theta - \angle B +90^\circ - \theta = 90^\circ - \angle B$. Similarly, $\angle(QA, OB) = 90^\circ - \angle C$. Now, we compute \begin{align*} \overrightarrow{AO}\cdot \overrightarrow{AP} &= \left(\overrightarrow{AC} + \overrightarrow{CO}\right)\cdot \overrightarrow{AP}\\ & = AB\cdot AC \cdot \cos (\angle A + \theta) + AB\cdot CO \cdot \cos(90^\circ - \angle B)\\ & = AB\cdot AC \cdot \cos(\angle A + \theta) + AB\sin \angle B\cdot CO\cdot \\ \end{align*}which is clearly symmetric. Thus, $\overrightarrow{AO}\cdot \overrightarrow{AP} = \overrightarrow{AO}\cdot \overrightarrow{AQ}$, and we are done.

Note that $\triangle AQC$ and $\triangle ABP$ are rotations about $A$. In particular, $APBR$ and $AQCR$ are cyclic. It follows that $\angle ARP = \angle ABP = \angle ACQ = \angle ARQ$. Call this common angle $\theta$. Now let $\triangle XYZ$ be congruent to $\triangle AQC$ and $\triangle ABP$, and let $K$ and $L$ be on $YZ$ so that $\angle XKY = \angle XLZ = \theta$. Let $M$ be the foot from $X$ to $YZ$. For points $T$ in the plane, define $f(T) = \text{pow}_{(BRC)}(T) - TA^2$. We have $$f(P) - f(Q) = (PR\cdot PC - PA^2) - (QR\cdot QB - QA^2) = $$ $$= (YK\cdot YZ - YX^2) - (ZL\cdot ZY - ZX^2)$$ $$= YZ(YK - ZL) - YX^2 + ZX^2$$ $$=YZ\cdot YM - ZY\cdot ZM - YX^2 + ZX^2$$ $$=\text{pow}_{(XZ)}(Y) - \text{pow}_{(XY)}(Z) - YX^2 + ZX^2$$ If $D$ and $E$ are the midpoints of $XY$ and $XZ$, this is $$(YE^2 - EX^2) - (ZD^2 - DX^2) - YX^2 + ZX^2$$ $$=(\frac{1}{2}YX^2 + \frac{1}{2}YZ^2 - \frac{1}{4}XZ^2 - \frac{1}{4}XZ^2) - (\frac{1}{2}ZX^2 + \frac{1}{2}ZY^2 - \frac{1}{4}XY^2 - \frac{1}{4}XY^2) - YX^2 + ZX^2$$ $$=0$$ so $PQ$ is parallel to the radical axis of $(BRC)$ and the point circle at $A$, which is perpendicular to $AO$, as desired.

Clearly $\triangle APB \sim \triangle ACQ$, so by spiral similarity facts $ABPR$ and $ACQR$ are cyclic. Obviously $\overline{AR}$ bisects $\angle BRC$ as well. Thus we may view the problem with reference triangle $RBC$, restating it as follows. Restated wrote: Let $ABC$ be a triangle with circumcenter $O$ and $D$ be a point on its internal $\angle A$-bisector. Let $(ABD)$ and $(ACD)$ intersect $\overline{AC}$ and $\overline{AB}$ at $E$ and $F$ respectively. Then $\overline{OD} \perp \overline{EF}$. Let $D=(t:b:c)$. If the equation of $(ABD)$ is $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0$, then plugging in $A$ and $B$ implies $u=v=0$, and then plugging in $D$ implies $w=\tfrac{a^2b+b^2t+bct}{b+c+t}$. To compute $E$, we solve for $r_1$ given that $(r_1,0,1-r_1)$ lies on the circle: this will yield (after dividing out $1-r_1$, which would correspond to $A$ itself) $r_1=\tfrac{a^2+bt+ct}{b(b+c+t)}$. Likewise, if $F=(r_2,1-r_2,0)$ then $r_2=\tfrac{a^2+bt+ct}{c(b+c+t)}$. Now, by "strong EFFT" we just have to prove that \begin{align*} 0&=a^2\left(\frac{a^2+bt+ct}{b+c+t}-c+b-\frac{a^2+bt+ct}{b+c+t}\right)\\ &+b^2\left(\left(\frac{c}{b}-1\right)\left(\frac{a^2+bt+ct}{b+c+t}\right)+t-\frac{t}{b}\left(\frac{a^2+bt+ct}{b+c+t}\right)\right)\\ &+c^2\left(\left(1-\frac{b}{c}\right)\left(\frac{a^2+bt+ct}{b+c+t}\right)+\frac{t}{c}\left(\frac{a^2+bt+ct}{b+c+t}\right)-t\right)\\ &=a^2(b-c)(b+c+t)+b(c-b-t)(a^2+bt+ct)+b^2t(b+c+t)+c(c-b+t)(a^2+bt+ct)-c^2t(b+c+t)\\ &=(b-c)(a^2+bt+ct)(b+c+t)-(b+c)(b-c)(a^2+bt+ct)-t(b-c)(a^2+bt+ct))\\ &=(b-c)(a^2+bt+ct)(b+c+t-b-c-t)\\ &=0, \end{align*}so we're done. $\blacksquare$

Started around 8:25 CST (since Hazard is pro god at math I need to compete with him) Let $U\in PC$ satisfy $UP=UB$, define $V$ similarly. Consider ellipse $\Gamma$ with foci at $B,C$ through $U,V$ such that $AU$ and $AV$ are tangents (not sure if it'll come in handy later). Also note that there is a circle $\Omega$ centered at $A$ which is tangent to $BU,CV,RU,RV$. This entire ellipse configuration comes from djmathman's handout, yay! Hence showing $PQ\perp AO$ is basically showing $PQ$ is parallel to some radical axis. Let $K$ be the foot of the altitude from $A$ to $PC$, and define $L$ similarly. Notice that it suffices to show \[PK^2-PR\cdot PC=QL^2-QR\cdot QB.\]We're basically almost done I think. Notice that this equates to \[(PK+QL)(PK-QL)=PR\cdot PC-QR\cdot QB\implies PC(PK-QL)=PR\cdot PC-QR\cdot PC\implies PK-QL=PR-QR\]or just \[PK-PR=QL-QR\implies RK=RL\]which is obviously true. Ta-da! 22 minutes! WOOOOOOOOOOOOOO

Solved with hints First, notice that since the angle of rotation is the same, there is a spiral similarity centered at $A$ taking $BP$ to $QC$. This means that $A$ is the Miquel point of $BPCQ$, so $ARBP$ is cyclic. This means that $$\angle BOC=2\angle BRC=2\angle BAP.$$Thus, if we let $M$ denote the arc midpoint of $BC$ not containing $R$, we have that $\angle BOM=\angle BAP$. Since they are both isosceles, we have $\triangle BOM\sim\triangle BAP$ as well. Use complex numbers with unit circle $(BRC)$. Let $B=b^2,C=c^2,R=r^2$. Furthermore, we let $A=a$ as another free variable as well (the problem has four degrees of freedom, so this decision is ok). We can use $\triangle BOM\sim\triangle BAP$ to compute $p$. We have $$\frac{a-p}{a-b^2}=\frac{0-(-bc)}{0-b^2},$$which we can solve for $p$ to get $$p=\frac{ab+ac-b^2c}{b}.$$Similarly, $$q=\frac{ab+ac-bc^2}{c}.$$Thus, we have $$p-q=\frac{abc+ac^2-b^2c^2-ab^2-abc+b^2c^2}{bc}=\frac{ac^2-ab^2}{bc}.$$Finally, dividing this by $a$ gives $\frac{c^2-b^2}{bc}$, which is clearly imaginary after conjugating and multiplying top and bottom by $b^2c^2$, so we are done. Remark: There are many things to take away from this problem. First of all, two rotations of the same angle around the same point should motivate using a spiral similarity, and spiral similiarity often works well with complex numbers. Next, there is a very "tempting" setup, in fact the one I originally used and failed with, where we set $(ABC)$ as the unit circle as usual, then note that there is an "angle of rotation" of some $\theta$ to get $P$ and $Q$, and setting $a$, $b$, $c$, and $d=e^{i\theta}$ as the four "free" variables to represent the four degrees of freedom. While this quickly gives $p$ and $q$, the computation for $r$ is now very messy as it requires the general intersection formula, much to the detriment of this setup, as $R$ is required to compute the circumcenter. The motivation for setting $(BRC)$ as the unit circle is that, rather than having to go through the tedious process of computing the circumcenter, this gives us the circumcenter "for free". Then, we can make $a$ our remaining free variable. Next, the introduction of the arc midpoint allows us to turn the condition $\angle BOC=2\angle BAP$ into a similarity condition, which is actually a much stronger condition than just the angle equivalence, as it effectively encodes two degrees of freedom rather than just one. If we were to use $\angle BOC=2\angle BAP$ directly, we would actually need two equations to compute $P$ (the other one probably being $PRC$ collinear), which will be much less elegant. Finally, this problem illustrates another important point of computing points "indirectly". In this solution, rather than plugging into some formula to compute $p$ and $q$, we used the fact that $\triangle BAP$ is similar to $\triangle BOM$ to indirectly compute $P$.

We will employ complex numbers, setting $A$ as the origin, $P$ at $(0, 1)$, $B$ at $|b| = 1$, $C$ at $c$, and $Q$ at $bc$. Obviously, $APRB$ and $AQCR$ are cyclic, say by considering the pair of congruent triangles. Letting $\angle PAB = \theta$, it follows that $\angle BOC = 2\theta$. In other words, we have $\frac{o-b}{o-c} = b^2$, so $o = \frac{b(bc-1)}{b^2-1}$, and thus $$\frac o{p-q} = \frac b{b^2-1} \in i\mathbb R.$$Hence $\overline{AO} \perp \overline{PQ}$ as required.

First $PABR$ and $QACR$ are cyclic so $\angle ARB=\angle ARC$. Fix $B,R,C$ and move $A$ with degree $1$ along the internal angle bisector of $\angle BRC$. Then all $\triangle BAP$ are spirally similar at $B$, so $P$ has degree $1$. Similarly $Q$ has degree $1$, so line $PQ$ has degree $2$. Line $AO$ has degree $1$, so the conclusion has degree $3$. It suffices to check $4$ cases. If $A=R$ then $PQ$ is the reflection of $BC$ over the external $\angle A$ bisector, so taking isogonals of the desired result gives $BC$ perpendicular to the $R$ altitude, which is clear. If $AB=AR$ then $\angle CRA=\angle ARB=\angle ABR$ so $P=R$ and $AO\perp BR=PQ$. Similarly this works when $AC=AR$. Finally if $AB=AC$ then $P=C,Q=B$ so $AO\perp BC=PQ$. We finish.