EDIT: Mostly synthetic

[asy][asy] import graph;import olympiad; size(8.5cm); pen fonted=fontsize(7.5); defaultpen(fonted); pair B=(0,0),A=(3,8),C=(10,0),D,Bp,Cp,O,Pp,P,Q; P=circumcenter(A,B,C); pair Db=rotate(-90,B)*P; pair Dc=rotate(90,C)*P; D=extension(B,Db,C,Dc); Bp=reflect(A,C)*B; Cp=reflect(A,B)*C; Q=(P+D)/2; O=circumcenter(D,Bp,Cp); Pp=2*A-P; draw(A--C--B--A^^C--Cp^^B--Bp,magenta);draw(D--C--B--D--Bp--C^^D--Cp--B^^Cp--Bp^^P--D^^C--P--B^^P--Pp--D^^A--O,heavycyan);draw(circumcircle(A,B,C)^^circumcircle(D,Bp,Cp),green); dot(A);dot(B);dot(C);dot(D);dot(Bp);dot(Cp);dot(P);dot(Q);dot(O);dot(Pp); label("$A$",A,N);label("$B$",B,SW);label("$C$",C,SE);label("$D$",D,S);label("$B'$",Bp,NE);label("$C'$",Cp,NW);label("$O$",O,ESE);label("$P'$",Pp,NNW);label("$P$",P,NE);label("$Q$",Q,E); [/asy][/asy]If $AB=AC$, then the result is obvious by symmetry, so WLOG assume $AB<AC$. Let $P$ be the circumcenter of $\triangle ABC$, $Q$ the midpoint of $PD$. Note that $B'C=BC=C'B$, $DC=DB$, and $\angle B'CD=2\angle C+\angle A=360^{\circ}-(2\angle B+\angle A)=360^{\circ}-\text{reflex}\angle C'BD=\angle C'BD$. So the triangles $B'CD$ and $C'BD$ are congruent, so a rotation centered at $D$ takes $CB'$ to $BC'$, hence a spiral similarity, say $\mathcal T$ centered at $D$ takes $BC$ to $B'C'$, and it takes $\triangle DBC$ to $DB'C'$. But $B,C$ lie on the circle with diameter $PD$, so $Q$ is the circumcenter of $\triangle DBC$. This gives $\mathcal T(Q)=O$. Now suppose $\mathcal T(P)=P'$. Then $\angle (PP',BC)=\angle (PP',PD)-90^{\circ}=\angle (B'C,CD)-90^{\circ}=2\angle C+\angle A-90^{\circ}=180^{\circ}-(90^{\circ}-\angle C+\angle B)=\angle (AP,BC)$. So in fact $A\in PP'$. Also we have $$\frac{PP'}{PD}=\frac{B'C}{CD}=\frac{BC}{CD}=\frac{\sin (180^{\circ}-2\angle A)}{\sin A}=2\cos\angle A$$$$\implies PP'=PD\cdot 2\cos\angle A=2 PC=2PA.$$So $A$ is the midpoint of $PP'$. Also, $O$ is the midpoint of $DP'$, so $AO||PD$. Now since $PD\perp BC$, the desired result follows. $\blacksquare$ @Einstein314 LOC$\neq $ synthetic.

Notice that under a $\sqrt{bc}$ inversion about $A$, points $B'$ and $C'$ are swapped. The point $D$ is mapped to a point $D'$ on the median from $A$ such that $HD' \perp AD'$ where $H$ is the orthocenter of triangle $ABC$. It shall suffice to proving that the circumcenter of $D'B'C'$ lies on $AO$ where $O$ is the circumcenter of triangle $ABC$. We claim that $H,D',B',C'$ are concyclic. Indeed, consider the orthic triangle $XYZ$ wrt $ABC$. Inversion about $H$ of radius $HA.HX$ that sends the circumcircle to the nine-point center. Clearly, $B'$ is mapped to $BH \cap XZ$ and $C'$ to $CH \cap XY$ and point $D'$ is mapped to $BC \cap YZ$. Using Menelaus' theorem, we see that since in triangle $XYZ$, this is equivalent to saying that the two feet of internal bisector and one of the external bisector are col-linear. The claim holds. Now, we find the circumcenter of $HB'C'$. Using the unit circle as circumcircle of $ABC$, and placing complex coordinates to $A,B,C$ as $a,b,c$ we see that the antipode of $H=a+b+c$ in the circumcircle of $D'B'C'$ is $3a-(b+c)$ and so the center is $2a$ which lies on $AO$. The result follows.

Here's a solution I found during test-solving. It may seem a little unmotivated, but this idea comes pretty naturally if you study the spiral similarity in 2016 USAMO #5. First, note that triangles $DBC'$ and $DCB'$ are congruent and in the same orientation, so $DB'C'$ is similar to $DBC$. Now, let the circumcircle of $DB'C'$ intersect $DB$ at $P$ and $DC$ at $Q$. We have that $\angle C'PB=\angle C'PD=\angle C'B'D=\angle CBD=\angle BAC=\angle C'AB$, so $P$ lies on the circumcircle of $ABC'$. Furthermore, $\angle ABP=\angle ACB\angle AC'B=\angle APB$, so $AP=AB$. Similarly, $AQ=AC$. Now, let $X$ and $Y$ be on $DB$ and $DC$ so that $AD=AX=AY$. The key lemma is that given varying points $D$ and $E$ on fixed rays $AB$ and $AC$ such that $AD-AE$ is constant. Then the circumcenter of $ADE$ lies on a fixed line parallel to the angle bisector of $\angle BAC$. The proof of this is that all circumcircles of $ADE$ share a common midpoint of arc $DAE$, call it $Z$, by spiral similarity, so the circumcenter of $ADE$ lies on the perpendicular bisector of $AZ$, which is a fixed line parallel to the angle bisector. Now, we use this lemma on rays $DB$ and $DC$. Note that since triangles $ADX$, $ABP$, $ADY$, and $ACQ$ are all isosceles, $DX-DP=XP=DB=DC=YQ=DY-DQ$, so we have that $DX-DY=DP-DQ$. Now, note that the circumcenter of $DPQ$ is $O$ and the circumcenter of $DXY$ is $A$, so the line through them is perpendicular to $BC$ by the lemma, as desired. The advantage of this solution is that you don't need to explicitly find $O$ (the parallelogram) to solve the problem. I'd be interested in any other problem that can be solved with this method.

ABCDE wrote: Oscar is drawing diagrams with trash can lids and sticks. He draws a triangle $ABC$ and a point $D$ such that $DB$ and $DC$ are tangent to the circumcircle of $ABC$. Let $B'$ be the reflection of $B$ over $AC$ and $C'$ be the reflection of $C$ over $AB$. If $O$ is the circumcenter of $DB'C'$, help Oscar prove that $AO$ is perpendicular to $BC$. James Lin From $BD=CD,BC'=B'C,\angle DBC'=\angle DCB'$ so $\triangle DBC'=\triangle DCB'\implies \triangle B'DC'$ is issoless $\implies \angle DC'B'=\angle DBC$. Let $E\equiv DB\cap (DB'C')$. From $\angle BEB'=\angle DC'B'=\angle DBC$ so $BC\parallel B'E$. But $AB=AB',\angle BAB'=2\angle BEB'$ so $A$ is the circumcentre of $\triangle BB'E\implies AO\perp B'E\implies AO\perp BC.\blacksquare$ PS. I think my proof is the most natural idea for this problem!

The obvious complex bash is not actually too difficult. We use complex numbers, with $A,B,C$ as points on the unit circle. Rotate the figure so that the real axis is the perpendicular bisector of $\overline{BC}$: i.e. $c=\overline{b}$. Then $d=\frac{2}{b+\overline{b}}$, which is real, and with the reflection formula, $$b'=a+\overline{b}-a\overline{b}, c'=a+b-ab^2.$$Translate the figure by $-d$ so that $d$ goes to $d'=0$, and $b',c'$ go to \begin{align*}b''=\frac{\overline{b}^2+ab-a\overline{b}^3-1}{b+\overline{b}} &= \frac{(b-1)(b+1)(ab^2+a-b)}{b^3\left(b+\overline{b}\right)},\\c''=\frac{b^2+a\overline{b}-ab^3-1}{b+\overline{b}} &=-\frac{(b-1)(b+1)(ab^2+a-b)}{b\left(b+\overline{b}\right)}. \end{align*}We can thus compute the circumcenter $O'$ of $\triangle D'B''C''$ as \begin{align*} o'=\frac{b''c''\left(\overline{b''}-\overline{c''}\right)}{\overline{b''}c''-b''\overline{c''}} &= \frac{-\frac{(b-1)^2(b+1)^2(ab^2+a-b)^2}{b^4}\left[\frac{(b-1)(b+1)(ab-b^2-1)}{ab}+\frac{(b-1)(b+1)(ab-b^2-1)}{ab^3}\right]}{\left(b+\overline{b}\right) \left[-\frac{(b-1)^2(b+1)^2(ab-b^2-1)(ab^2+a-b)}{ab^2} + \frac{(b-1)^2(b+1)^2(ab-b^2-1)(ab^2+a-b)}{ab^6}\right]} \\ &=\frac{(b-1)^3(b+1)^3(ab^2+a-b)^2(b^2+1)(ab-b^2-1)}{b\left(b+\overline{b}\right)(b-1)^3(b+1)^3(b^2+1)(ab-b^2-1)(ab^2+a-b)} \\ &= \frac{ab^2+a-b}{b^2+1} \end{align*}Thus it suffices to show that $AO'\perp BC$, or equivalently, that $a-o'$ is real. But $$a-o'=\frac{a(b^2+1)-(ab^2+a-b)}{b^2+1}=\frac{1}{b+\overline{b}}=\overline{(a-o')},$$and the result follows.

Once you notice that $DB'C'$ is similar to $DBC$, you directly bash with complex numbers. We will use complex numbers. As usual, we let the lowercase letter of each point denote its respective complex coordinate. Without loss of generality, assign $\odot ABC$ to be the unit circle and $b=\bar{c}$. Well known formulas immediately yield $$b'=a+c-ac\bar{b}=a+c-ac^2$$$$c'=a+b-ab\bar{c}=a+b-ab^2$$$$d=\frac{2bc}{b+c}=\frac{2}{b+c}.$$ We now claim that $\triangle DBC \sim \triangle DC'B'$. We can manipulate line by line as follows \begin{align*} ab^2+ac^2 &= ab^2+ac^2 \\ ab^2+abc-abc-ac^2+2ac^2 &= abc+ac^2-ab^2-abc+2ab^2 \\ (b+c)(ab-ac+\frac{2ac^2}{b+c}) &= (b+c)(ac-ab+\frac{2ab^2}{b+c}) \\ ab-ac+ac^2d &= ac-ab+ab^2d \\ ab+bc-abc^2-bd-ad-cd+ac^2d+d^2 &= ac+bc-ab^2c-cd-ad-bd+ab^2d+d^2 \\ (a+c-ac^2-d)(b-d) &= (a+b-ab^2-d)(c-d). \\ \frac{a+b-ab^2-d}{a+c-ac^2-d} &= \frac{b-d}{c-d}. \end{align*}So $$\frac{c'-d}{b'-d}=\frac{b-d}{c-d}$$which impljies the desired similarity. If we let $O'$ denote the circumcenter of $\triangle DBC$, then the spiral similarity $\phi(x)$ centered at $D$ which sends $BC$ to $C'B'$ also sends $O'$ to $O$. Recall that $\phi(x)=z(x-d)+d$ for some constant complex number $z$. Here, we have $\phi(b)=c'=a+b-ab^2=z(b-d)+d$. It follows that $$z=\frac{a+b-ab^2-d}{b-d}=\frac{a+b-ab^2-\frac{2}{b+c}}{b-\frac{2}{b+c}}.$$ It follows that $$\phi(x)=\frac{a+b-ab^2-\frac{2}{b+c}}{b-\frac{2}{b+c}}\cdot \left (x-\frac{2}{b+c} \right) + \frac{2}{b+c}.$$ We will now compute $o'$. Let $K$ be the circumcenter of $ABC$. Then $\angle KBD = \angle KCD = 90^{\circ}$ since $BD$ and $CD$ are tangents to $\odot ABC$. Therefore $BDCK$ is cyclic with diamter $DK$. So $O'$ is the midpoint of $DK$ which gives $$o'=\frac{d}{2}=\frac{1}{b+c}.$$ Now we can plug this in to compute $o$. We have that \begin{align*} o=\phi(o') &= \frac{a+b-ab^2-\frac{2}{b+c}}{b-\frac{2}{b+c}}\cdot \left (\frac{1}{b+c} -\frac{2}{b+c} \right) + \frac{2}{b+c} \\ &= -\frac{a+b-ab^2-\frac{2}{b+c}}{(b-\frac{2}{b+c})(b+c)} +\frac{2}{b+c}\\ &= \frac{ab^2+\frac{2}{b+c}-a-b+2b-\frac{4}{b+c}}{b^2+bc-2}\\ &= \frac{ab^2-a+b-\frac{2}{b+c}}{b^2-1}. \end{align*}Since $b=\bar{c}$, $BC$ is parallel to the imaginary axis. Therefore, it is sufficient to show that $o-a$ is real. We have \begin{align*} o-a &= \frac{ab^2-a+b-\frac{2}{b+c}}{b^2-1}-a =\frac{b-\frac{2}{b+c}}{b^2-1} =\frac{\frac{b^2+bc-2}{b+c}}{b^2-1} =\frac{1}{b+c}. \end{align*}Now since $o-a=\frac{1}{b+c}=\frac{1}{\bar{b}+\bar{c}}=\overline{o-a}$ we are done. Remark

An other approch Let $E$ be the point of intersection $B'C$ and $BC'$ it s easy to see that $\widehat{BEC}=\pi -2\widehat{A}$ i.e. $BCED$ is cyclic thus the miquel point of $BCB'C'$ which is the center of a rotation since $BC'=CB' $,is nothing but $D$ .$\widehat{EBA}+\widehat{BAO}+\widehat{OEB}= \pi$ thus $A,O,E$ are colinear and $EA$ cut $\overarc{B'C'} $ at its midpoint say $ S$ ;now if we consider the sym-inversion wrt $(A,\sqrt{bc})$ and the $A$-bisector , it swaps $B\leftrightarrow C,B'\leftrightarrow C',E \leftrightarrow H$ (the orthocenter of $ABC$) thus it swaps $(EB'C')\leftrightarrow (B'C'H)$ but $\widehat{B'HC'}=\pi -\widehat{A}$ so $O$ is on the circumcircle of $ EB'C'$ therefore it is $S$.since $AE$ is orthogonal to $(HB'C')$ then $AH$ is orthogonal to $(EB'C')$ which leads $AO\perp BC$ . R HAS

I had a much shorter complex bash... By intersecting the perpendicular from $A$ to $BC$ with the perpendicular bisector of $B'C'$ you get that $O=a+\frac{bc}{b+c}$. Now you just have to plug in and check that all the conditions are satisfied, which is true.

We uses complex numbers to bash (Thanks to V-Enhance ,he really helped me a lot with his book.) We set $(ABC)$ as a unit circle,we let $A=a,B=b,C=c$ so we get $D=\frac{2bc}{b+c},B'=\frac{ab+bc-ac}{b}$ and $C'=\frac{-ab+bc+ac}{c}$ we need to calculate the circumcenter of tri$B'C'D$,by circumcenter formula we get $O:z+\frac{-x'y'(\overline{x'}-\overline{y'})}{x'\overline{y'}-\overline{x'}y'}$ where $x'=b'-z,y'=c'-z,z=D$ So we get $x'=\frac{ab+bc-ac}{b}-\frac{2bc}{b+c}=\frac{(ab-bc+ac)(b-c)}{b(b+c)},\overline{x'}=\frac{(\frac{1}{ab}-\frac{1}{bc}+\frac{1}{ac})(\frac{1}{b}-\frac{1}{c})}{\frac{1}{b}(\frac{1}{b}+\frac{1}{c})}=\frac{(-a+b+c)(c-b)}{ac(b+c)}$ With the same method we get $y'=\frac{(ab-bc+ac)(c-b)}{c(b+c)},\overline{y'}=\frac{(-a+b+c)(b-c)}{ab(b+c)}$ Drag them in,then $O:z+\frac{-\frac{(ab-bc+ac)(b-c)}{b(b+c)}*\frac{(ab-bc+ac)(c-b)}{c(b+c)}*\frac{(b+c)(-a+b+c)(c-b)}{abc(b+c)}}{\frac{(ab-bc+ac)(b-c)(-a+b+c)(b-c)}{ab^2(b+c)^2}-\frac{(ab-bc+ac)(b-c)(-a+b+c)(b-c)}{ac^2(b+c)^2}}=\frac{2bc}{b+c}-\frac{\frac{(ab-bc+ac)(b-c)}{b^2c^2}}{\frac{c^2-b^2}{b^2c^2}}=\frac{ab+bc+ca}{b+c}$ A bit nice.To prove that $AO$ is perpendicular to $BC$,we just need to prove that $A,O,H$ are collinear where $H$ is the orthocenter of tri$ABC$ which is just $a+b+c$.So we only need to prove that tri$AOH$'s area is zero,and it is just $$\begin{vmatrix} a+b+c &\frac{ab+bc+ca}{abc} & 1 \\ \frac{ab+bc+ca}{b+c} & \frac{a+b+c}{a(b+c)} & 1 \\ a & \frac{1}{a} & 1 \\ \end{vmatrix}=0 $$$\iff$$$\begin{vmatrix} b+c &\frac{b+c}{bc} \\ \frac{bc}{b+c} & \frac{1}{b+c} \\ \end{vmatrix}=0 $$,which is obvious. Your sincerely CeuAzul

My solution: Lemma Let $ABC$ be a triangle, $B', C'$ are the symetric of $B,C$ resp around $CA$ and $AB$. $R$ is the circumcenter of $(AB'C')$. Then $AR$ pass through 9-point center of $\triangle ABC$.

Back to the problem Let $K$ be the center of $(AO'CD)$ with $O'$ is the circumcenter of $(ABC)$. Not difficult to see that $\triangle DBC'=\triangle DCB'$$\implies$ $DB'=DC'$ and $\triangle DBC\sim \triangle DC'B'$$\implies$$\triangle DKO\sim \triangle DBC'$$\implies$$\frac {OK}{BC}=\frac {DK}{DB}$$\implies$$\frac {2OK}{BC}=\frac {DO'}{DB}=\frac {2R}{BC}$$\implies$$OK=O'A$. Let $OD$ cut $O'A$ at $T$. From Lemma, we have $AK\parallel TD$ and $A$ is the midpoint of $TO'$. Since $2OK=O'T$$\implies$$DT$ is the diameter of $(O)$$\implies$$OK\parallel O'A$$\implies$$AOKO'$ is a parallegram $\implies$ $AO\perp BC.$

From $\triangle C'BD\cong \triangle B'CD$, we get $E=C'B\cap B'C$ is on the circumcircle of $DB'C'$. Now do a $\sqrt{bc}$ inversion, then $(EB'C')\leftrightarrow (HB'C')$. But the center of $(HB'C')$ is the reflection of the center of $(ABC)$ (by considering projections onto $BH,CH$), so inverting back $AH$ is orthogonal to $(DB'C')$.

Let $ T $ be the circumcenter of $ \triangle BCD. $ From $ DB $ $ = $ $ DC, $ $ BC' $ $ = $ $ CB', $ $ \measuredangle C'BD $ $ = $ $ \measuredangle B'CD $ we get $ \triangle DBC' $ $ \stackrel{+}{\cong} $ $ \triangle DCB', $ so $ \triangle DBC $ $ \cup $ $ T $ $ \stackrel{+}{\sim} $ $ \triangle DC'B' $ $ \cup $ $ O $ $ \Longrightarrow $ $ \triangle C'AC $ $ \stackrel{+}{\sim} $ $ \triangle BTD $ $ \stackrel{+}{\sim} $ $ \triangle C'OD $ $ \Longrightarrow $ $ \triangle C'AO $ $ \stackrel{+}{\sim} $ $ \triangle C'CD, $ hence we get $$ \measuredangle (AO,CD) = \measuredangle (C'O,C'D) = \measuredangle (BT,BD) = \measuredangle (DT,CD) \Longrightarrow AO \parallel DT \Longrightarrow AO \perp BC. $$

Einstein314 wrote:

EDIT: Mostly synthetic What is LOC?

tarzanjunior wrote: Einstein314 wrote:

EDIT: Mostly synthetic What is LOC? Law of Cosines

I see an extension for this nice problem. Let $ABC$ be a triangle inscribed in circle $(O)$ and $P$ is any point. $PC,PB$ cut $(O)$ again at $M,N$. $E,F$ are reflections of $B,C$ through $AM,AN$, reps. $K$ is circumcenter of triangle $PEF$. Prove that $AK\perp MN$.

Note that $\triangle DBC'$ and $\triangle DCB'$ are congruent, so $P = BC'\cap CB'$ lies on $(DB'C')$ by spiral similarity. Rename $A$ as $I_P$, the $P-$ excenter of $\triangle PBC$ (the configuration where $A$ is the incenter is analogous) and let $I$ and $O'$ be the incenter and circumcenter of $\triangle PBC$. Since the perpendicular bisectors of $PB'$ and $PC'$ are the perpendicular bisectors of $PC$ and $PB$ translated a distance $\frac{1}{2}BC$ parallel to $PC$ and $PB$ respectively, we get $O = O' + \frac{1}{2}\overrightarrow{II_P}$, so $OI_P$ is clearly perpendicular to $BC$.

Use complex numbers. Suppose $D$ lies on the real axis. Let $A=a,B=x,C=\tfrac{1}{x}$, hence $D=\tfrac{2x}{x^2+1}$. We can compute $B'=a+\tfrac{1}{x}-\tfrac{a}{x^2}$ and $C'=a+x-ax^2$, so after translating by $a$ it suffices to show that the circumcenter of points $x-ax^2$, $\tfrac{1}{x}-\tfrac{a}{x^2}$, and $\tfrac{2x}{x^2+1}-a$ is real. Consider the well-known formula for the circumcenter of an arbitrary triangle. It turns out that its denominator times $\tfrac{i}{4}$ is also the signed area of this triangle, hence the denominator is imaginary, so it suffices to show that the numerator is purely imaginary as well. After some computations of $(d-a)\overline{(d-a)}$, etc. and subtracting $2$ from each entry in the second column, we wish to show $$\begin{vmatrix}x-ax^2&ax+\frac{1}{ax}&1\\\frac{1}{x}-\frac{a}{x^2}&\frac{a}{x}+\frac{x}{a}&1\\\frac{2x}{x^2+1}-a&\left(\frac{x^2-1}{x^2+1}\right)^2+\left(a+\frac{1}{a}\right)\left(\frac{2x}{x^2+1}\right)&1\end{vmatrix} \in i\mathbb{R}.$$The second and third columns are real, so by adding its conjugate we wish to show that $$\begin{vmatrix}x+\frac{1}{x}-ax^2-\frac{1}{ax^2}&ax+\frac{1}{ax}&1\\x+\frac{1}{x}-\frac{a}{x^2}-\frac{x^2}{a}&\frac{a}{x}+\frac{x}{a}&1\\\frac{4x}{x^2+1}-a-\frac{1}{a}&\left(\frac{x^2-1}{x^2+1}\right)^2+\left(a+\frac{1}{a}\right)\left(\frac{2x}{x^2+1}\right)&1\end{vmatrix}=0.$$This follows because adding $x+\tfrac{1}{x}$ times the second column to the first column makes all terms equal to $x+\tfrac{1}{x}+a+\tfrac{1}{a}$. $\blacksquare$

Solved with popop614 First, note that \begin{align*} \measuredangle DBC' + \measuredangle B'CD &= (360 - \measuredangle C'BA -\measuredangle ABC - \measuredangle CBD) + (\measuredangle B'CA + \measuredangle ACB + \measuredangle BCD)\\ &= \measuredangle ABC' + \measuredangle CBA + \measuredangle DBC + \measuredangle B'CA + \measuredangle ACB + \measuredangle BCD\\ &= \measuredangle CBA + \measuredangle CBA + \measuredangle BAC + \measuredangle ACB + \measuredangle ACB + \measuredangle BAC\\ &= 2(\angle A + \angle B + \angle C) = 0 \end{align*}so $\measuredangle DBC' = \measuredangle DCB'$. Furthermore, since $BD = CD$ and $BC' = BC = B'C$, triangles $BDC'$ and $CDB'$ are congruent. Now, let $O'$ be the circumcenter of $BDC$, and let $L$ be the circumcenter of $ABC$. Note that $SD$ is a diameter of $BDC$ since $BD \perp SB, CD\perp SC$. We claim that $ALO'O$ is a parallogram. First, we show that $OO' \parallel AL$. We have that \[ \measuredangle ASD = \measuredangle ASB + \measuredangle BSD = 2\angle ACB + \angle BCD = \angle B'CB + \angle BCD = \angle B'CD = \angle OO'D \]where the last step comes from spiral similarity at $D$ sending $DBC$ to $DC'B'$, which also sends the circumcenters to each other. Next, we show that $OO' = AL$. We have that $AL=BL=CL=R$ and that $\angle BO'D = \widehat{BD} = \frac12 \widehat{BC} = \angle BSC$ so triangles $BSC$ and $BO'D$ are similar. We have that, by spiral similarity and normal similarity, \[ \frac{OO'}{BC} = \frac{OO'}{BC'} = \frac{O'D}{BD} = \frac{BL}{BC} \]so $OO' = BL = AL$. Finally, since $ALO'O$ is a parallelogram, $AO \parallel O'L$. Since $O'L \perp BC$, $AO \perp BC$.

Let $X$ denote the circumcenter of $\triangle BCD$, and let $Y$ denote the circumcenter of $ABC$. Note that $D$ is the Miquel point of $BCC'B$. Thus the spiral similarity centered at $D$ mapping $BC \mapsto C'B'$ also maps $X \mapsto O$. Clearly we can also note the $X$ is the midpoint of $YD$. Now as $\triangle DBC' \sim DXO'$, we can find $$\angle DOX = \angle DC'B = \angle DC'B' - \angle BC'B' = 180 - a - 2c = \angle AYX$$and thus $AY \parallel OX$. Now we will show that $AY = XO$, so that we can conclude that $AOXY$ is a parallelogram. To see this note that, \begin{align*} \frac{DB}{BC'} = \frac{\sin A}{\sin 2A} \end{align*}So then we find, \begin{align*} \frac{DX}{XO} = \frac{\sin A}{\sin 2A} \end{align*}Now noting that $$2DX = DY = \frac{CY}{ \sin (90 - A)}$$we must have, \begin{align*} XO &= \frac{\sin 2A}{\sin A} \cdot \frac{CY}{2 \cos A}\\ &= \frac{ 2 \sin A \cos A}{\sin A} \cdot \frac{CY}{ 2 \cos A}\\ &= CY\\ &= AY \end{align*}Now since $AOXY$ is a parallelogram, then it is easy to see that as $YD \perp BC$, so we have $AO \perp BC$ as desired. Remark: The complex solution is way too simple for this to be a realistic problem $2$.

Shreyasharma wrote: Remark: The complex solution is way too simple for this to be a realistic problem $2$. harder than average

Suppose $P = O_{(ABC)}$ and $Q = O_{(BCD)}$. Then SAS gives us $\triangle DBC' \cong \triangle DCB'$, so there exists spiral similarity at $D$, which also tells us that $\triangle DQO$ is similar to both triangles. We have $AP \parallel OQ$, as \[\angle OPQ + \angle QPA = 180 - \angle C'BD + \angle QPB + \angle BPA = 180.\] We have $AP = OQ$, as \[OQ = BC' \cdot \frac{QD}{BD} = BC \cdot \frac{\frac{BC}{2 \sin 2A}}{\frac{BC}{2 \cos A}} = \frac{BC}{2 \sin A} = AP.\] Hence $AOQP$ is a parallelogram, so $AO \parallel PQ \perp BC$. $\blacksquare$

Note that $\overline{AH}$ is the radical axis of $(AB'C)$ and $(ABC')$, so it suffices to show that $O$ lies on this radical axis. Since these two circles are congruent, it suffices to show that $O$ is equidistant to the centers of these circles. Let $O_B$ and $O_C$ denote the circumcenters of $(AB'C)$ and $(ABC')$, respectively. Since $DB=DC$, $BC'=CB'=BC$, and \[ \measuredangle DBC' = 2\measuredangle B + \measuredangle A = 2\measuredangle C + \measuredangle A = \measuredangle DCB',\]we have that $\triangle DBC' \cong \triangle DCB'$. Furthermore, since arcs $BC'$ and $CB'$ are equal in their respective circles and $OC'=OB'$, we have $\triangle O_BB'O \cong \triangle O_CC'O$. Thus $OO_B=OO_C$, as desired.

If one inversion doesn't work, simply invert twice! We do $\sqrt{bc}$ inversion at $A$ (inversion at $A$ with radius $\sqrt{bc}$ and then reflection across $A$-bisector). This exchanges $B'$ and $C'$ and sends $D$ to $X$ where $X$ is the $A-$humpty point of $\displaystyle \triangle ABC$. So, we need to show that if $O'$ is the circumcentre of $\displaystyle \triangle ABC$, then $AO'$ is orthogonal to $(XB'C')$. Now, let $H$ be the orthocentre of $\displaystyle \triangle ABC$, and $D,E,F$ be the foot of perpendiculars from $A,B,C$ to $BC,CA,AB$. Then it is well known that $BHXC$ is cyclic and $EXHF$ is also cyclic. Therefore, $X$ is the centre of spiral symmetry which sends $B \to E$ and $C \to F$. Now, $\displaystyle \frac{BB'}{B'E} = -2 = \frac{CC'}{C'F}$ therefore, $X$ is the centre of spiral symmetry which sends $B \to B'$ and $C \to C'$, hence $B'XHC'$ is cyclic. Now, we invert at $H$ with radius $-\sqrt{HA \cdot HD}$. This sends $A \to D$, $B \to E$, $C \to F$, $B' \to M = DF \cap BH$, $C' \to N = DE \cap CH$. We are reduced to proving that the unique circle through $D, H$ which is orthogonal to the circumcircle of $DEF$, is also orthogonal to $MN$. So, considering $\triangle HBC$ point of view, we are reduced to proving the following: Reduced Problem: Let $ABC$ be a triangle, and let $AD, BE, CF$ be the altitudes. Let $DE \cap AB = M$ and $FD \cap AC = N$. Suppose that $\ell_D$ is the tangent to the nine point circle at $D$, and $\ell_{AD}$ is the perpendicular bisector of $AD$. Prove that $\ell_D, \ell_{AD}$ and $MN$ are concurrent. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7.380248779755665cm); 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 = -15.36714502980874, xmax = -0.6066474702974087, ymin = -3.1377123012127455, ymax = 3.603532783798663; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-10.02692884184011,0.702534871117716), 2.1580245117887418), linewidth(0.8) + dtsfsf); draw(circle((-10.78690026351871,0.5419871631361872), 1.0790122558943676), linewidth(0.8) + sexdts); draw((-13.262168002572636,3.2601675704947986)--(-8.039210933627574,-0.13773134315738927), linewidth(0.8) + wvvxds); draw((-13.262168002572636,3.2601675704947986)--(-12.191065499753494,-1.8100218583358825), linewidth(0.8) + dbwrru); draw((-12.191065499753494,-1.8100218583358825)--(-11.600616761711933,2.1792122645691534), linewidth(0.8) + wvvxds); draw((-11.600616761711933,2.1792122645691534)--(-11.528232630498245,-0.24203692452870967), linewidth(0.8) + wrwrwr); draw((-10.741084314865482,1.6200262845131714)--(-11.960901673538013,-0.2549717240216773), linewidth(0.8) + wrwrwr); draw((-11.859849657708017,0.42776349711616957)--(-8.039210933627574,-0.13773134315738927), linewidth(0.8) + wrwrwr); draw((-13.262168002572636,3.2601675704947986)--(-11.528232630498245,-0.24203692452870967), linewidth(0.8) + wvvxds); draw((-12.191065499753494,-1.8100218583358825)--(-10.741084314865482,1.6200262845131714), linewidth(0.8) + wvvxds); draw((-11.600616761711933,2.1792122645691534)--(-12.77044378968005,0.9325332875516134), linewidth(0.8) + dbwrru); draw((-12.77044378968005,0.9325332875516134)--(-11.528232630498245,-0.24203692452870967), linewidth(0.8) + dbwrru); draw((-13.95576643123552,0.8970976330955449)--(-9.300937268837576,1.0362554555887578), linewidth(0.8) + dbwrru); draw((-11.960901673538013,-0.2549717240216773)--(-8.039210933627574,-0.13773134315738927), linewidth(0.8) + wvvxds); /* dots and labels */ dot((-11.600616761711933,2.1792122645691534),linewidth(3.pt) + dotstyle); label("$A$", (-11.562390206140915,2.5377233599179145), NE * labelscalefactor); dot((-11.960901673538013,-0.2549717240216773),linewidth(3.pt) + dotstyle); label("$B$", (-12.26480648127959,-0.47438392464528617), NE * labelscalefactor); dot((-8.039210933627574,-0.13773134315738927),linewidth(3.pt) + dotstyle); label("$C$", (-8.001529922451796,-0.07439687908020981), NE * labelscalefactor); dot((-11.528232630498245,-0.24203692452870967),linewidth(3.pt) + dotstyle); label("$D$", (-11.542878642942618,-0.6207206486325092), NE * labelscalefactor); dot((-10.741084314865482,1.6200262845131714),linewidth(3.pt) + dotstyle); label("$E$", (-10.703881425415867,1.6816438087664667), NE * labelscalefactor); dot((-11.859849657708017,0.42776349711616957),linewidth(3.pt) + dotstyle); label("$F$", (-12.209690849296626,0.41339220087720036), NE * labelscalefactor); dot((-11.5468716851973,0.3814394551546526),linewidth(3.pt) + dotstyle); label("$H$", (-11.406297700554543,0.40363641927805216), NE * labelscalefactor); dot((-13.262168002572636,3.2601675704947986),linewidth(3.pt) + dotstyle); label("$N$", (-13.084292135608045,3.232813083031031), NE * labelscalefactor); dot((-12.77044378968005,0.9325332875516134),linewidth(3.pt) + dotstyle); label("$T$", (-13.11355948040549,0.9792275336277961), NE * labelscalefactor); dot((-12.191065499753494,-1.8100218583358825),linewidth(3.pt) + dotstyle); label("$M$", (-12.498945239659148,-1.8011702221294417), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: $MN$ is the orthic axis, which is well known to be the radical axis of $(ABC)$ and $(DEF)$. Now, if $T = \ell_D \cap \ell_{AD}$ then $TD = TA$ and so it suffices to show that $TA$ is tangent to $(ABC)$. But this is because $$\angle TAB = \angle TAD - \angle BAD = \angle TDA - \angle BAD = C. \quad \square$$

Claim: $DB'$ = $DC'$ Proof: $\angle DBC' = 360 - \angle C'BA - \angle ABC - \angle DBC = 360 - 2B - A = A+ 2C = \angle DCB + \angle ACB + \angle B'CA = \angle DCB'$. So $\triangle$ $DBC'$ $ \cong $ $\triangle$ $DCB'$. Let $H$ = $BB'$ $\cap$ $CC'$ be the ortocenter of $ABC$. Claim: $(AHBC')$ and $(AHCB')$ cyclic. Proof: $\angle$ $ BAB' = 2A = \angle CAC'$, and $\triangle$ $C'AC$, $\triangle$ $B'AB$ are isosceles so $A$ is indeed the center of the spiral similarity that sends $BB'$ to $CC'$, proving the claim. Claim: $O$ lies in the radax of $(AHBC')$ and $(AHCB')$. Proof: Let $E$ and $F$ be the center of these circles respectively. Notice that Radius $(BAC')$ = Radius of $(ABC)$ = Radius of $(CAB')$. $Pow O (BAHC') = Pow O (CAHB')$ $\iff$ $OE$ = $OF$ $\iff$ $\triangle$ $OC'E$ $\cong$ $O'BF$, but $OC'$ = $OB'$ and $EC' = FB'$, so we need $\angle OC'E = \angle OB'F$, which can be done with an easy angle chase ( i don't care about directed angles).

Let $E$ and $F$ be the midpoints of $\overline{C'B}$ and $\overline{CB'}$ respectively. Because $\triangle DBE \cong \triangle DCF$ and $\triangle DBC' \cong \triangle DCB'$, it follows that $\triangle DBC$, $\triangle DEF$ and $\triangle DC'B'$ are all similar isosceles triangles. Furthermore, because points $B$, $E$ and $C'$ form an arithmetic sequence, the centers of $(DBC)$, $(DEF)$ and $(DC'B')$ form an arithmetic sequence as well. (These circumcenters are just points $b$, $e$ and $c'$ multiplied by some constant complex number.) In fact, $DEAF$ is cyclic because $\angle EAF = 2 \angle A$ and $\angle EDF = \angle BDC = 180 - 2 \angle A$. Because $AD$ is the bisector of $\angle EAF$ (because $AE = AF$), $A$ and $D$ are antipodes in $(DEF)$. So, the center of $(DEF)$ is the midpoint of $AD$. Since the center of $(DBC)$ clearly lies on the perpendicular bisector of $\overline{BC}$, it follows that $O$ lies on the $A$-altitude.

We have $\measuredangle DBC'=\measuredangle DBC+\measuredangle CBC'=\measuredangle BAC+2\measuredangle CBA=-(\measuredangle BAC+2\measuredangle ACB)=\measuredangle CAB+2\measuredangle BCA=\measuredangle DCB+\measuredangle BCB'=\measuredangle DCB',$ and $C'B=BC=B'C$ and $BD=CD,$ thus $\triangle C'BD\cong\triangle B'CD$ and $D$ is the miquel point of $BCB'C',$ thus $(DB'C')$ passes through $BC'\cap B'C$ and we finish by 2024 HMMT Team P8.

$d=\frac{2bc}{b+c}$, $b'=\frac{ab+bc-ca}b$, and $c'=\frac{bc+ca-ab}c$. Notice $d-\frac{ab+bc+ca}{b+c}=\frac{bc-ab-ca}{b+c}$ and $b'-\frac{ab+bc+ca}{b+c}=\frac{c(bc-ab-ac)}{b(b+c)}$ which have equal magnitudes, so $o=\frac{ab+bc+ca}{b+c}=a+\frac d2$ is on the $A$-altitude.

Rename $D$ to $T$ and let $D$ be the foot from $A$ to $BC$. Note $T$ is the spiral center of the similarity sending $CB$ to $BC'$. Thus if $BC'\cap BC'=A'$ then $DA'B'C'$ is cyclic, say with circumradius $R$. Now, \[-BC'\cdot BA' = OB^2 - R^2, \qquad -CB'\cdot CA' = OC^2 - R^2.\]Subtracting this gives \[OB^2-OC^2=CB'\cdot CA'-BC'\cdot BA'=BC(CA'-BA').\]But $A$ is the $A'$ excenter of $A'BC$ so $D$ is the extouch point, so \[OB^2-OC^2=BC(CA'-BA')=BC(BD-CD)=(BD+CD)(BD-CD)=DB^2-DC^2\]so we finish by the perpendicularity lemma. Remark. After deleting $T$ you get the exact same config as 2022 G6

[asy][asy] import olympiad; import geometry; size(10cm); defaultpen(fontsize(10pt)); // Define the points of the triangle pair A = dir(120); pair B = dir(210); pair C = dir(330); // Circumcircle of triangle ABC path circ = circumcircle(A, B, C); // Define the tangency point D pair D = extension(B, B + (B - foot(B, A, C)), C, C + (C - foot(C, A, B))); // Reflections pair Bp = reflect(A, C) * B; pair Cp = reflect(A, B) * C; // Circumcenter of triangle DBpCp pair O = circumcenter(D, Bp, Cp); // Draw the figure draw(A--B--C--cycle, heavygreen); draw(circ, red); draw(D--B, blue); draw(D--C, blue); draw(B--Bp, dashed); draw(C--Cp, dashed); draw(circumcircle(D, Bp, Cp), orange); draw(A--O, purple); draw(B--C, green); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$B'$", Bp, dir(Bp)); dot("$C'$", Cp, dir(Cp)); dot("$O$", O, dir(O)); // Draw perpendicular line from A to BC to check perpendicularity draw(rightanglemark(A, foot(A, B, C), B), heavycyan); [/asy][/asy] Reflect $B$ over $AC$ to get $B'$. Similarly, reflect $C$ over $AB$ to get $C'$. Notice that $B'$ and $C'$ are such that $B', C' \in \odot (ABC)$, because reflections over the sides of a triangle map points on the circumcircle to other points on the circumcircle. Since $DB$ and $DC$ are tangents to the circumcircle of $\triangle ABC$, we have: \[ DB = DC. \]Reflect $D$ over $AC$ and $AB$ to obtain points $D_B$ and $D_C$, respectively. Notice that $D_B$ lies on the reflection of $DB$, and similarly, $D_C$ lies on the reflection of $DC$. Reflecting $B$ over $AC$ to get $B'$ and reflecting $C$ over $AB$ to get $C'$, implies $D, B', C'$ are collinear because reflections preserve collinearity. The quadrilateral $DB'C'O$ is cyclic with $O$ as the circumcenter. Hence, the perpendicular bisectors of $DB', DC'$, and $B'C'$ intersect at $O$. Consequently, $AO$ is the radical axis of the two circles, namely the circumcircles of $\triangle DB'C'$ and $\triangle ABC$. The radical axis of two intersecting circles is perpendicular to the line joining their centers. Thus, $AO$ is perpendicular to the line joining the centers of the circumcircles of $\triangle ABC$ and $\triangle DB'C'$. Since $O$ is the circumcenter of $\triangle DB'C'$, it lies on the perpendicular bisector of $B'C'$. Hence, $AO$ being the radical axis must be perpendicular to $BC$.

It is easy to note that $\angle DBC' = 180 - \angle C + \angle B = \angle DCB'$, $DC = DB, CB' = BC'$, so $DBC'$ is congruent to $DCB'$ by $SAS$. Since $DBC, DC'B'$ are both isosceles triangles with the same middle angle, they are similar, and $O$ can be obtained as the result of the spiral similarity centered at $D$ that takes $BC$ to $C'B'$ applies to the circumcenter of $DBC$. For this we use complex numbers. The circumcenter of $DBC$ is just $\frac{bc}{b + c}$. We can find $C'$ as $a + b - \frac{ab}{c}$, so the desired $O$ is given by $\frac{a + b - \frac{ab}{c} - 2\frac{bc}{b + c}}{b - 2\frac{bc}{b + c}}(-\frac{bc}{b + c}) + 2\frac{bc}{b + c}= \frac{bc}{b + c} (\frac{a + b - \frac{ab}{c} - 2\frac{bc}{b + c} + 4\frac{bc}{b + c} - 2b}{2\frac {bc}{b +c} - b}) = (\frac{bc}{b + c})\frac{a - b - \frac{ab}{c} + 2\frac{bc}{b + c}}{2\frac{bc}{b + c} - b} = (\frac{bc}{b + c})\frac{(ac - ab - bc)(b + c)+2bc^2}{c(c - b)(b)} = (\frac{1}{b + c})\frac{(bc +ab-ac)(b + c) - 2bc^2}{(b-c)} =(\frac{1}{b + c})\frac{b^2c + ab^2 - abc + bc^2 +abc-ac^2}{(b-c)} = (\frac{1}{b + c}) \frac{b^2c + ab^2 - bc^2 -ac^2}{(b-c)}=\frac{bc +ab +ac}{b +c} $. We then desire to show that $\frac{a-o}{b- c}$ is negative conjugating, which is equivalent to showing $\frac{-\frac{bc}{b + c}}{b - c}$ is negative conjugating, which is obvious.