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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Cyclic Quads and Parallel Lines
gracemoon124   16
N 21 minutes ago by ohiorizzler1434
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
16 replies
gracemoon124
Aug 16, 2023
ohiorizzler1434
21 minutes ago
Radical Center on the Euler Line (USEMO 2020/3)
franzliszt   37
N 41 minutes ago by Ilikeminecraft
Source: USEMO 2020/3
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$. Let $\Gamma$ denote the circumcircle of triangle $ABC$, and $N$ the midpoint of $OH$. The tangents to $\Gamma$ at $B$ and $C$, and the line through $H$ perpendicular to line $AN$, determine a triangle whose circumcircle we denote by $\omega_A$. Define $\omega_B$ and $\omega_C$ similarly.
Prove that the common chords of $\omega_A$,$\omega_B$ and $\omega_C$ are concurrent on line $OH$.

Proposed by Anant Mudgal
37 replies
+1 w
franzliszt
Oct 24, 2020
Ilikeminecraft
41 minutes ago
Functional equation with powers
tapir1729   13
N 42 minutes ago by ihategeo_1969
Source: TSTST 2024, problem 6
Determine whether there exists a function $f: \mathbb{Z}_{> 0} \rightarrow \mathbb{Z}_{> 0}$ such that for all positive integers $m$ and $n$,
\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]Jaedon Whyte
13 replies
tapir1729
Jun 24, 2024
ihategeo_1969
42 minutes ago
Powers of a Prime
numbertheorist17   34
N an hour ago by KevinYang2.71
Source: USA TSTST 2014, Problem 6
Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*}
		ca &- db \\
		ca^2 &- db^2 \\
		ca^3 &- db^3 \\
		ca^4 &- db^4 \\
&\vdots
	\end{align*} and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$.
34 replies
numbertheorist17
Jul 16, 2014
KevinYang2.71
an hour ago
No more topics!
Queue geo
vincentwant   6
N May 4, 2025 by ihategeo_1969
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $Y, Z$ be the feet of the altitudes from $B, C$ to $AC, AB$ respectively. Let $D$ be the midpoint of $BC$. Let $\omega_1$ be the circle with diameter $AD$. Let $Q\neq A$ be the intersection of $(ABC)$ and $\omega$. Let $H$ be the orthocenter of $ABC$. Let $K$ be the intersection of $AQ$ and $BC$. Let $l_1,l_2$ be the lines through $Q$ tangent to $\omega,(AYZ)$ respectively. Let $I$ be the intersection of $l_1$ and $KH$. Let $P$ be the intersection of $l_2$ and $YZ$. Let $l$ be the line through $I$ parallel to $HD$ and let $O'$ be the reflection of $O$ across $l$. Prove that $O'P$ is tangent to $(KPQ)$.
6 replies
vincentwant
Apr 30, 2025
ihategeo_1969
May 4, 2025
Queue geo
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vincentwant
1380 posts
#1
Y by
Let $ABC$ be an acute scalene triangle with circumcenter $O$. Let $Y, Z$ be the feet of the altitudes from $B, C$ to $AC, AB$ respectively. Let $D$ be the midpoint of $BC$. Let $\omega_1$ be the circle with diameter $AD$. Let $Q\neq A$ be the intersection of $(ABC)$ and $\omega$. Let $H$ be the orthocenter of $ABC$. Let $K$ be the intersection of $AQ$ and $BC$. Let $l_1,l_2$ be the lines through $Q$ tangent to $\omega,(AYZ)$ respectively. Let $I$ be the intersection of $l_1$ and $KH$. Let $P$ be the intersection of $l_2$ and $YZ$. Let $l$ be the line through $I$ parallel to $HD$ and let $O'$ be the reflection of $O$ across $l$. Prove that $O'P$ is tangent to $(KPQ)$.
This post has been edited 1 time. Last edited by vincentwant, Apr 30, 2025, 3:55 PM
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hukilau17
288 posts
#2 • 8 Y
Y by vincentwant, Yiyj, kamatadu, lpieleanu, EpicBird08, Ilikeminecraft, Sedro, Amkan2022
I assume $\omega$ is the same thing as $\omega_1$?

Complex bash with $\triangle ABC$ in the unit circle, so that
$$|a|=|b|=|c|=1$$$$o=0$$$$y=\frac{b^2+ab+bc-ac}{2b}$$$$z=\frac{c^2+ac+bc-ab}{2c}$$$$d=\frac{b+c}2$$$$q = \frac{a+d}{a\overline{d}+1} = \frac{bc(2a+b+c)}{ab+ac+2bc}$$$$h = a+b+c$$$$k = \frac{aq(b+c) - bc(a+q)}{aq-bc} = \frac{a^2b+a^2c+ab^2-2abc+ac^2-b^2c-bc^2}{2(a^2-bc)}$$Now we find the equation of $\ell_1$. This is
$$\frac{w-q}{\frac{a+d}2-q} \in i\mathbb{R}$$$$\frac{4\left[w(ab+ac+2bc) - bc(2a+b+c)\right]}{2a^2b+2a^2c+ab^2-2abc+ac^2-2b^2c-2bc^2} = \frac{4a\left[bc\overline{w}(2a+b+c) - (ab+ac+2bc)\right]}{2a^2b+2a^2c-ab^2+2abc-ac^2-2b^2c-2bc^2}$$$$\left[w(ab+ac+2bc) - bc(2a+b+c)\right](2a^2b+2a^2c-ab^2+2abc-ac^2-2b^2c-2bc^2) = a\left[bc\overline{w}(2a+b+c) - (ab+ac+2bc)\right](2a^2b+2a^2c+ab^2-2abc+ac^2-2b^2c-2bc^2)$$$$\left[w(ab+ac+2bc) - bc(2a+b+c)\right](2a-b-c)(ab+ac+2bc) = a\left[bc\overline{w}(2a+b+c) - (ab+ac+2bc)\right](2a+b+c)(ab+ac-2bc)$$$$\overline{w} = \frac{(2a+b+c)(ab+ac+2bc)(a^2b+a^2c-4abc+b^2c+bc^2) + w(2a-b-c)(ab+ac+2bc)^2}{abc(ab+ac-2bc)(2a+b+c)^2}$$The equation of line $KH$, meanwhile, is
$$\frac{h-w}{h-k} \in \mathbb{R}$$$$\frac{a+b+c-w}{a+b+c-\frac{a^2b+a^2c+ab^2-2abc+ac^2-b^2c-bc^2}{2(a^2-bc)}} \in \mathbb{R}$$$$\frac{2(a^2-bc)(a+b+c-w)}{2a^3+a^2b+a^2c-ab^2-ac^2-b^2c-bc^2} \in \mathbb{R}$$$$\frac{2(a^2-bc)(a+b+c-w)}{(a+b)(a+c)(2a-b-c)} = \frac{2(a^2-bc)(ab+ac+bc-abc\overline{w})}{(a+b)(a+c)(ab+ac-2bc)}$$$$(ab+ac-2bc)(a+b+c-w) = (2a-b-c)(ab+ac+bc-abc\overline{w})$$$$\overline{w} = \frac{(a^2b+a^2c-2ab^2-2ac^2+b^2c+bc^2) + w(ab+ac-2bc)}{abc(2a-b-c)}$$We intersect these to find the coordinate of $I$ (which we denote as $j$):
$$\frac{(2a+b+c)(ab+ac+2bc)(a^2b+a^2c-4abc+b^2c+bc^2) + j(2a-b-c)(ab+ac+2bc)^2}{abc(ab+ac-2bc)(2a+b+c)^2} = \frac{(a^2b+a^2c-2ab^2-2ac^2+b^2c+bc^2) + j(ab+ac-2bc)}{abc(2a-b-c)}$$$$(2a-b-c)(2a+b+c)(ab+ac+2bc)(a^2b+a^2c-4abc+b^2c+bc^2) + j(2a-b-c)^2(ab+ac+2bc)^2 = (ab+ac-2bc)(2a+b+c)^2(a^2b+a^2c-2ab^2-2ac^2+b^2c+bc^2) + j(2a+b+c)^2(ab+ac-2bc)^2$$Solving for $j$ gives
$$j = \frac{(2a-b-c)(2a+b+c)(ab+ac+2bc)(a^2b+a^2c-4abc+b^2c+bc^2) - (ab+ac-2bc)(2a+b+c)^2(a^2b+a^2c-2ab^2-2ac^2+b^2c+bc^2)}{(2a+b+c)^2(ab+ac-2bc)^2 - (2a-b-c)^2(ab+ac+2bc)^2}$$The denominator factors as
$$\left[(2a+b+c)(ab+ac-2bc) + (2a-b-c)(ab+ac+2bc)\right]\left[(2a+b+c)(ab+ac-2bc) - (2a-b-c)(ab+ac+2bc)\right]$$which we simplify to
$$(4a^2b+4a^2c-4b^2c-4bc^2)(2ab^2-4abc+2ac^2) = 8a(b-c)^2(b+c)(a^2-bc)$$Meanwhile we factor out $2a+b+c$ in the numerator, so that
$$j = \frac{(2a+b+c)\left[(2a-b-c)(ab+ac+2bc)(a^2b+a^2c-4abc+b^2c+bc^2) - (2a+b+c)(ab+ac-2bc)(a^2b+a^2c-2ab^2-2ac^2+b^2c+bc^2)\right]}{8a(b-c)^2(b+c)(a^2-bc)}$$We write the expression in brackets as
$$(a^2b+a^2c-4abc+b^2c+bc^2)\left[(2a-b-c)(ab+ac+2bc) - (2a+b+c)(ab+ac-2bc)\right] + (2ab^2-4abc+2ac^2)(2a+b+c)(ab+ac-2bc)$$which we simplify to
$$2a(b-c)^2\left[(2a+b+c)(ab+ac-2bc) - (a^2b+a^2c-4abc+b^2c+bc^2)\right] = 2a(b-c)^2(a^2b+a^2c+ab^2+2abc+ac^2-3b^2c-3bc^2)$$So factoring out $2a(b-c)^2(b+c)$ (note that $b+c\neq 0$ since the triangle is acute) we have
$$j = \frac{(2a+b+c)(a^2+ab+ac-3bc)}{4(a^2-bc)}$$Now we find the equation of $\ell_2$. This is
$$\frac{w-q}{\frac{a+h}2 - q} \in i\mathbb{R}$$$$\frac{2\left[w(ab+ac+2bc) - bc(2a+b+c)\right]}{a(b+c)(2a+b+c)} = -\frac{2a\left[bc\overline{w}(2a+b+c) - (ab+ac+2bc)\right]}{(b+c)(ab+ac+2bc)}$$$$w(ab+ac+2bc)^2 - bc(2a+b+c)(ab+ac+2bc) = -a^2bc\overline{w}(2a+b+c)^2 + a^2(2a+b+c)(ab+ac+2bc)$$$$\overline{w} = \frac{(ab+ac+2bc)\left[(a^2+bc)(2a+b+c) - w(ab+ac+2bc)\right]}{a^2bc(2a+b+c)^2}$$The equation of line $YZ$, meanwhile, is
$$\frac{w-z}{y-z} \in \mathbb{R}$$$$\frac{b(2cw-c^2-ac-bc+ab)}{a(b+c)(b-c)} = -\frac{2abc\overline{w}-ab-ac-bc+c^2}{(b+c)(b-c)}$$$$2bcw-bc^2-abc-b^2c+ab^2 = -2a^2bc\overline{w}+a^2b+a^2c+abc-ac^2$$$$\overline{w} = \frac{(a^2b+a^2c-ab^2+2abc-ac^2+b^2c+bc^2) - 2bcw}{2a^2bc}$$We intersect these to get
$$\frac{(ab+ac+2bc)\left[(a^2+bc)(2a+b+c) - p(ab+ac+2bc)\right]}{a^2bc(2a+b+c)^2} = \frac{(a^2b+a^2c-ab^2+2abc-ac^2+b^2c+bc^2) - 2bcp}{2a^2bc}$$$$2(a^2+bc)(2a+b+c)(ab+ac+2bc) - 2p(ab+ac+2bc)^2 = (2a+b+c)^2(a^2b+a^2c-ab^2+2abc-ac^2+b^2c+bc^2) - 2bcp(2a+b+c)^2$$$$p = \frac{(2a+b+c)\left[2(a^2+bc)(ab+ac+2bc) - (2a+b+c)(a^2b+a^2c-ab^2+2abc-ac^2+b^2c+bc^2)\right]}{2(ab+ac+2bc)^2 - 2bc(2a+b+c)^2}$$$$p = \frac{(2a+b+c)(a^2b^2-2a^2bc+a^2c^2+ab^3-ab^2c-abc^2+ac^3-b^3c+2b^2c^2-bc^3)}{2(a^2b^2-2a^2bc+a^2c^2-b^3c+2b^2c^2-bc^3)}$$$$p = \frac{(b-c)^2(2a+b+c)(a^2+ab+ac-bc)}{2(b-c)^2(a^2-bc)} = \frac{(2a+b+c)(a^2+ab+ac-bc)}{2(a^2-bc)}$$Now we find the equation of $\ell$. This is
$$\frac{j-w}{h-d} \in \mathbb{R}$$$$\frac{(2a+b+c)(a^2+ab+ac-3bc) - 4w(a^2-bc)}{2(a^2-bc)(2a+b+c)} = \frac{(ab+ac+2bc)(3a^2-ab-ac-bc) - 4abc\overline{w}(a^2-bc)}{2(a^2-bc)(ab+ac+2bc)}$$$$(2a+b+c)(ab+ac+2bc)(a^2+ab+ac-3bc) - 4w(a^2-bc)(ab+ac+2bc) = (2a+b+c)(ab+ac+2bc)(3a^2-ab-ac-bc) - 4abc\overline{w}(a^2-bc)(2a+b+c)$$$$\overline{w} = \frac{(a-b)(a-c)(2a+b+c)(ab+ac+2bc) + 2w(a^2-bc)(ab+ac+2bc)}{2abc(a^2-bc)(2a+b+c)}$$Then plugging in $w = o$, we find
$$\overline{o'} = \frac{(a-b)(a-c)(2a+b+c)(ab+ac+2bc)}{2abc(a^2-bc)(2a+b+c)} = \frac{(a-b)(a-c)(ab+ac+2bc)}{2abc(a^2-bc)}$$and so
$$o' = -\frac{(a-b)(a-c)(2a+b+c)}{2(a^2-bc)}$$Then we find the vectors
$$p - o' = \frac{(2a+b+c)\left[(a^2+ab+ac-bc) + (a-b)(a-c)\right]}{2(a^2-bc)} = \frac{a^2(2a+b+c)}{a^2-bc}$$$$p - k = \frac{(2a+b+c)(a^2+ab+ac-bc) - (a^2b+a^2c+ab^2-2abc+ac^2-b^2c-bc^2)}{2(a^2-bc)} = \frac{a(a+b)(a+c)}{a^2-bc}$$$$p - q = \frac{(2a+b+c)\left[(ab+ac+2bc)(a^2+ab+ac-bc) - 2bc(a^2-bc)\right]}{2(a^2-bc)(ab+ac+2bc)} = \frac{a(a+b)(a+c)(b+c)(2a+b+c)}{2(a^2-bc)(ab+ac+2bc)}$$$$k-q = (p-q) - (p-k) = \frac{a(a+b)(a+c)\left[(b+c)(2a+b+c) - 2(ab+ac+2bc)\right]}{2(a^2-bc)(ab+ac+2bc)} = \frac{a(a+b)(a+c)(b-c)^2}{2(a^2-bc)(ab+ac+2bc)}$$And then
$$\frac{(p-o')(k-q)}{(p-k)(p-q)} = \frac{\frac{a^3(a+b)(a+c)(b-c)^2(2a+b+c)}{2(a^2-bc)^2(ab+ac+2bc)}}{\frac{a^2(a+b)^2(a+c)^2(b+c)(2a+b+c)}{2(a^2-bc)^2(ab+ac+2bc)}} = \frac{a(b-c)^2}{(a+b)(a+c)(b+c)}$$All the factors we canceled out are nonzero. ($a+b,a+c$ are nonzero since $\triangle ABC$ is acute; $a^2-bc$ is nonzero since $\triangle ABC$ is scalene; and $2a+b+c,ab+ac+2bc$ are nonzero since otherwise $a=-\frac{b+c}2=-\frac{2bc}{b+c}$ which implies $a=b=c$.) Moreover, this is real, so $O'P$ is tangent to $(KPQ)$ as desired. $\blacksquare$
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MathLuis
1524 posts
#3 • 1 Y
Y by vincentwant
Sneaky sneaky sneaky.
$I$ is actually the midpoint of $KH$ because if you let $AH$ hit $BC, (ABC)$ at $X, H_A$ respectively then $H$ is also orthocenter of $\triangle AKD$ so it becomes clear that $\omega, (HK)$ are orthogonal from orthocenter config or existence of Humpty point.
Now the next thing is to let $AA'$ diameter on $(ABC)$ and $AA_1CB$ isosceles trapezoid and also notice that since $-1=(K, X; B, C)$ we have projecting at $A$ that $-1=(Q, H; Z, Y)$ and as a result $PH$ is also tangent to $(AH)$ so $PH \parallel BC$ and so if you reflect $P$ over $I$ to be $P'$ then $P'$ lies on $BC$ but also remember that $KI=IQ$ so in fact since we can see that $PI$ is the perpendicular bisector of $QH$ we have that $l$ is actually just the perpendicular bisector of $KQ$ and thus reflecting we want to prove $(KP'Q)$ is tangent to $OP'$ however as $-1=(K, X; B, C)$ we project from $A$ to get $-1=(Q, H_A; B, C)$ and so since we have from the angles that $\measuredangle KQP'=\measuredangle ZKQ=\measuredangle ABQ$ which shows that $QP'$ is tangent to $(ABC)$ so we also have $PH_A$ is tangent to $(ABC)$ from the harmonic and this means that $P'QODH_A$ is cyclic with diameter $P'O$ and thus $\measuredangle OP'Q=\measuredangle ODQ=\measuredangle XHD=\measuredangle XKQ$ which gives our tangency as desired thus we are done :cool:.
This post has been edited 1 time. Last edited by MathLuis, Apr 30, 2025, 8:37 PM
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Ilikeminecraft
617 posts
#4 • 1 Y
Y by vincentwant
We prove a series of collinearities. Let $P'$ be the point on $BC$ such that $PP'\parallel TG.$
Claim: $TGPP'$ is an isosceles trapezoid, and thus cyclic.
Proof: Let $X$ be the circumcenter of $TGM,$ and thus, the midpoint of $TM.$

Note that $\angle IGH = \angle GAM = \angle GHT.$ Hence, $I$ is midpoint of $TH.$

Two times homothety centered at $T$ sends $I, X$ to $H, M.$ Thus, $IX\parallel GM.$

Note that $\angle XGM = \angle XMG = \angle GAH$ which implies $GX$ is tangent to $(AEF).$

Finally, note that $\angle IXG = \angle XGM = \angle GMX = \angle IXP',$ so $IX$ bisects $\angle FXP'$. However, $TG\perp GM\parallel IX,$ implying our result.

Hence, it suffices to show that $OP'$ is tangent to $(GPP'T)$.
Claim: $P'G$ is tangent to $(ABC)$
Proof: First note that $TGEC$ is cyclic since $G$ is the miquel point of quadrilateral $BFEC.$ Hence, $\angle TGP' = \angle GP'P = \angle ATE = \angle ACG$ which finishes.

Cyclic implies that $\angle OGP' = 90 = \angle OMP'.$ Thus, $P'GOM$ is cyclic. Thus, $\angle OP'M = \angle OGM = \angle OGH' = \angle OH'G = \angle ACG = \angle ATE = \angle TGP',$ which finishes.
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vincentwant
1380 posts
#5
Y by
@above I recommend that you provide your own diagram because you renamed the points
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Ilikeminecraft
617 posts
#6
Y by
[asy]
            unitsize(4.5cm);
import geometry;
        
            pair O = (0, 0);
            pair A = (-10/17, 3*sqrt(21)/17);
            pair B = (-12/13, -5/13);
            pair C = (12/13, -5/13);
            pair D = foot(A, B, C);
            pair EE = foot(B, C, A);
            pair F = foot(C, A, B);
            pair G = intersectionpoints(circumcircle(A, EE, F), circumcircle(A,B,C))[0];
            pair H = extension(A, D, EE, B);
            pair M = (B + C)/2;
            pair T = extension(EE, F, B, C);
            pair I = (T + H)/2;
            pair X = (T + M)/2;
            pair P = extension(G, X,T, EE);
            pair Pp = intersectionpoint(tangent(circle(A, B, C), G), line(B, C));
            pair Hp = -A;
        
            draw(A -- B -- C -- cycle); 
            draw(A -- M);
            draw(A -- D);
            draw(T -- H);
            draw(T -- B); 
            draw(M -- Hp -- A, dashed);
        
            draw(T -- B -- EE);
            draw(F -- EE);
            draw(T -- EE);
    
            draw(G -- I, red);
            draw(circumcircle(A, G, M), red);
    
            draw(G -- X, deepgreen);
            draw(circumcircle(A,EE,F), deepgreen);
    
            draw(I -- X, purple);
            draw(G -- M, purple);
            
            draw(Pp -- G, deepmagenta);
            draw(unitcircle, deepmagenta);
    
            draw(Pp -- P, orange);
            draw(T -- A, orange);
            
            draw(Pp -- O, blue);
            draw(circle(T, G, P), blue);
        
            dot(A);
            dot(B);
            dot(C);
            dot(D);
            dot(EE);
            dot(F);
            dot(H);
            dot(M);
            dot(T);
            dot(G);
            dot(I);
            dot(X);
            dot(P);
            dot(Pp);
            dot((0, 0));
            dot(Hp);
        
            label("$A$", A, N);
            label("$B$", B, SW);
            label("$C$", C, SE);
            label("$D$", D, NE);
            label("$E$", EE, NE);
            label("$F$", F, W);
            label("$G$", G, W);
            label("$H$", H, SE);
            label("$M$", M, NE);
            label("$T$", T, W);
            label("$I$", I, N);
            label("$X$", X, S);
            label("$P$", P, S);
            label("$P'$", Pp, S);
            label("$O$", O, S);
            label("$H'$", Hp, SE);
[/asy]
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ihategeo_1969
233 posts
#7
Y by
Sorry I change the point names a bit.
Quote:
Let $\triangle ABC$ be a triangle with orthic triangle $\triangle DEF$. Let $H$, $Q_A$, $X_A$, $M$ be orthocenter, $A$-Queue point, $A$-ex point, midpoint of $\overline{BC}$. Let $T$ be the intersection of tangent from $Q_A$ at $(ADM)$ and $\overline{X_AH}$; and let $P$ be intersection of tangent from $Q_A$ at $(AEF)$ and $\overline{FE}$. If $\ell$ is the line through $T$ parallel to $\overline{HM}$ and $O'$ is reflection of $O$ over $\ell$ then prove that $\overline{O'P}$ is tangent to $(X_APQ_A)$.
We start with some claims.

Claim: $T$ is circumcenter of $(X_AQ_AHD)$.
Proof: Say $T'$ is circumcenter of $(X_AQ_AHD)$ then see that $\angle T'Q_AM=90^\circ-\angle Q_AX_AH=90^\circ-\angle Q_ADH=\angle Q_AAM$. So $T' \equiv T$. $\square$

So see that $\ell$ is $\perp$ bisector of $\overline{X_AQ_A}$ and so let $P'$ be reflection of $P$ over $\ell$. We want to prove that $\overline{OP'}$ is tangent to $(X_AP'PQ_A)$.

Claim: $P'\in \overline{BC}$ and $\overline{P'Q_A}$ is tangent to $(ABC)$.
Proof: The first thing is just by \[\angle Q_AX_AD=180^\circ-\angle Q_AHD=\angle Q_AHA=180^\circ-\angle PQ_AA=\angle PQ_AX_A=\angle Q_AX_AP'\]And now the second thing is just \[\angle P'Q_AA=180^\circ-\angle P'Q_AX_A=180^\circ-\angle Q_AX_AF=180 ^\circ-\angle Q_ABA\]And so we are done. $\square$

Now if $H'=\overline{AH} \cap (ABC)$ then $(Q_A,H';B,C)=-1$ and hence see that \[\angle OP'Q_A=90^\circ-\angle Q_AAH'=\angle AX_AD=\angle Q_AX_AP'\]And done.
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