Claim: $BP = BQ$. Proof: Repeatedly using ratio lemma and LoS, we have $$\frac{BQ}{BA} = \frac{\sin{\angle BAQ}}{\sin{\angle BQA}} = \frac{\sin{\angle IBD}}{\sin{\angle IBA}} = \frac{ID}{IA}\div\frac{BD}{BA} = \sin{\dfrac{A}{2}}\div \frac{\sin{\angle BAD}}{\sin{\angle BDA}} = \sin{\dfrac{A}{2}}\div \frac{\sin{\frac{A}{2}}}{\cos{\frac{A}{2}}} = \cos{\frac{A}{2}} = \frac{BP}{BA}$$ as desired. Now, we have $$\angle AXP = 180^\circ - \angle PAX - \angle APX = 180^\circ - (90^\circ + \frac{1}{4}\angle A) - (90^\circ - \angle BPQ) = \angle BPQ - \frac{1}{4}\angle A = 90^\circ - \frac{1}{2}\angle PBQ - \frac{1}{4}\angle A = 90^\circ - \frac{1}{2}(90^\circ - \frac{1}{2}\angle A) - \frac{1}{4}\angle A = 45^\circ$$ as desired.

this title is good enough that i will support the legitimacy of this thread by posting my solution

Magnum opus or magnum bogus? You decide... We use cartesian coordinates, with $\overline{AP}$ as the $x$-axis and $\overline{AI}$ as the $y$-axis, so $A=(0,0)$. Let $D=(0,d)$ with $d>0$, and WLOG let the inradius of $ABC$ be $1$, so $I=(0,d+1)$. Suppose WLOG that $\overline{AB}$ has positive slope (so $\overline{AC}$ will have negative slope); then from similar triangles and Pythagorean theorem we find that the equations of $\overline{AB}$ and $\overline{AC}$ are $y=x\sqrt{d^2+2d}$ and $y=-x\sqrt{d^2+2d}$ respectively. The perpendicular to $\overline{AC}$ through $D$ has slope $\tfrac{1}{\sqrt{d^2+2d}}$, so its equation is $y=\tfrac{1}{\sqrt{d^2+2d}}x+d$. To compute $B$, we intersect this line with $y=x\sqrt{d^2+2d}$: $$x\sqrt{d^2+2d}=\frac{1}{\sqrt{d^2+2d}}x+d \implies x(d^2+2d-1)=d\sqrt{d^2+2d} \implies B=\left(\frac{d\sqrt{d^2+2d}}{d^2+2d-1},\frac{d^2(d+2)}{d^2+2d-1}\right).$$Since $P$ is the foot from $B$ to the $x$-axis, $P=(\tfrac{d\sqrt{d^2+2d}}{d^2+2d-1},0)$. We have $\angle ABI=\tfrac{1}{2}\angle B$ and $\angle ABD=90^\circ-\angle A$, so $\angle BAQ=\angle DBI=\angle A+\tfrac{1}{2}\angle B-90^\circ$. Then since $\angle PAB=90^\circ-\tfrac{1}{2}\angle A$, it follows that $\angle PAQ=\tfrac{1}{2}(\angle A+\angle B)$. On the other hand, we have $\angle AIB=180^\circ-\tfrac{1}{2}(\angle A+\angle B)$. Therefore, the reflection of $\overline{BI}$ over $y=x$ produces a line parallel to $\overline{AQ}$, so the slope of $\overline{AQ}$ is the reciprocal of the slope of $\overline{BI}$. We now compute the latter; it is: $$\frac{\frac{d^2(d+2)}{d^2+2d-1}-(d+1)}{\frac{d\sqrt{d^2+2d}}{d^2+2d-1}-0}=\frac{d^2(d+2)-(d+1)(d^2+2d-1)}{d\sqrt{d^2+2d}}=-\frac{d^2+d-1}{d\sqrt{d^2+2d}},$$so $\overline{AQ}$ has equation $y=-\tfrac{d\sqrt{d^2+2d}}{d^2+d-1}x$. We can now intersect this with $y=\tfrac{1}{\sqrt{d^2+2d}}x+d$ to find $Q$: $$-\frac{d\sqrt{d^2+2d}}{d^2+d-1}x=\frac{1}{\sqrt{d^2+2d}}x+d \implies -(d^2(d+2)+(d^2+d-1))x=d(d^2+d-1)\sqrt{d^2+2d},$$which yields $$Q=\left(-\frac{d(d^2+d-1)\sqrt{d^2+2d}}{(d+1)(d^2+2d-1)},\frac{d^3(d+2)}{(d+1)(d^2+2d-1)}\right).$$ I now claim that $\angle AXP=45^\circ$ is equivalent to $\angle APQ=45^\circ-\tfrac{1}{4}\angle A$. Indeed, this follows from $\angle PAX=90^\circ+\tfrac{1}{4}\angle A$ and summing the angles of $\triangle PAX$. On the other hand, note that $\angle PAB=90^\circ-\tfrac{1}{2}\angle A$, so we just have to show that $\tan(2\angle APQ)=\tan \angle PAB$ (it's obvious that $\angle PAQ$ is acute, since $Q$ lies in the first quadrant). Since $Q$ is also to the "left" of $P$, $\tan \angle APQ$ is the negative of the slope of $\overline{PQ}$. We now compute this slope; it is \begin{align*} \frac{\frac{d^3(d+2)}{(d+1)(d^2+2d-1)}-0}{-\frac{d(d^2+d-1)\sqrt{d^2+2d}}{(d+1)(d^2+2d-1)}-\frac{d\sqrt{d^2+2d}}{d^2+2d-1}}&=-\frac{d^3(d+2)}{(d(d^2+d-1)+d(d+1))\sqrt{d^2+2d}}\\ &=-\frac{d^3(d+2)}{(d^3+2d^2)\sqrt{d^2+2d}}\\ &=-\sqrt{\frac{d}{d+2}} \implies \tan \angle APQ=\sqrt{\frac{d}{d+2}}. \end{align*} We now use the tangent double angle formula (), which states that $\tan 2\theta=\tfrac{2\tan \theta}{1-\tan^2\theta}$. This implies that $$\tan(2\angle APQ)=\frac{2\sqrt{\frac{d}{d+2}}}{1-\frac{d}{d+2}}=\frac{2\sqrt{d^2+2d}}{d+2-d}=\sqrt{d^2+2d},$$which is the slope of $\overline{AB}$, which is precisely $\tan \angle PAB$ by definition. This finishes the problem. $\blacksquare$ Remark: I am told that this problem was back-constructed from trigonometry. Also for the record, the quote was not about this problem but in general, looking at historical trends. It seems that the trend is showing no signs of stopping though.

Let $E = BD \cap AC$ and the incircle of $ABC$ touch $BC$ at $T$. Compute $$\angle PBQ = \angle ADE = 90^{\circ} - \angle IAC = \frac{\angle B}{2} + \frac{\angle C}{2}.$$Now, it suffices to show $$\angle APX = 135^{\circ} - \angle PAX = 45^{\circ} - \frac{\angle A}{4}$$or $$\angle BPQ = 90^{\circ} - \angle APX = 45^{\circ} + \frac{\angle A}{4}.$$But we have $$\angle BPQ + \angle BQP = 180^{\circ} - \angle PBQ = 90^{\circ} + \frac{\angle A}{2} = 2 \left(45^{\circ} + \frac{\angle A}{4} \right)$$meaning the desired conclusion is equivalent to $BP = BQ$. First, observe that $$BP = AB \cdot \sin PAB = AB \cdot \sin \left( \frac{B}{2} + \frac{C}{2} \right)$$while LoS in $ABQ$ yields $$BQ = \frac{AB}{\sin AQB} \cdot \sin BAQ = AB \cdot \frac{\sin IBQ}{\sin ABI}.$$Thus, proving $$\sin ABI = \frac{\sin IBQ}{\sin \left( \frac{B}{2} + \frac{C}{2} \right)}$$finishes. Indeed, looking at triangles $BDI$ and $BIT$ gives $$\frac{\sin IBQ}{\sin \left( \frac{B}{2} + \frac{C}{2} \right)} = \frac{\sin IBD}{\sin ADE} = \frac{\sin IBD}{\sin IDB} = \frac{ID}{IB} = \frac{TI}{BI} = \sin IBT = \sin ABI$$as required. $\blacksquare$ Remarks: This solution is written backwards in an attempt to preserve some naturality and a sense of motivation. Although this problem is fairly peculiar and strangely rigid, one potential starting point is to notice that $BD \perp AC$ corresponds to $\angle PBQ = \angle BDI = \frac{\angle B}{2} + \frac{\angle C}{2}$. Regardless, I still believe finding any source of motivation for this approach is quite difficult.

TRIG BAD Let $EFK$ be the intouch triangle of $ABC$. Note that basic angle chasing gives that it suffices to show that $DE\parallel PQ$ as $\angle AEP=45^{\circ} -\frac a4$. Thus it also suffices to show that $\triangle BPQ$ and $\triangle IDE$ are homothetic as $BQ\parallel IE$ as both are perpendicular to $AC$ and $PB\parallel AI$. Then, if $\mathcal H$ is a homothety sending $BQ$ to $IE$ and $A\mapsto A'$, we show $A'D\parallel AP$. By the given conditions, we note that $A'$ is the point on the line through $I$ parallel to $AB$ such that $(IEA')$ is tangent to $BI$. Then, if $K'=AE\cap (EFK)$, then $\angle IEA=180 ^{\circ} - \angle BIA'=\angle ABI=180^{\circ} -\frac b2$ so as $\angle KFI=\frac b2$ we obtain that $KK'\parallel EF$. We then show that $A'D \parallel AI$ since $AP\perp AI$ or equivalently $A'D$ is tangent to the incircle. All of this allows us to reduce the problem to the following claim. Claim: Let $ABC$ be a triangle with intouch triangle $EFK$ and incircle $\omega$. $K'$ is on $\omega$ with $KK'\parallel EF$, and $D=AI\cap \omega$. If $BD\perp AC$, then the three following lines concur: The tangent to $\omega$ at $D$. Line $K'E$. The line through $I$ parallel to $AB$. Proof. Let $A'=EK'\cap DD$, $M$ be the midpoint of $EK'$, and $N$ is the midpoint of $FK$ so $IM\perp EK'$ and $IN\perp FK$ so $DIA'M$ is cyclic. Then, \[\angle AIA'=\angle DIA'=\angle DMA=\angle DME=\angle FND\]where we used the cyclic quad and reflection across $AI$. But \[\angle IND=\angle BDI=\angle BAI+\angle IBA=\frac a2 + 90^{\circ} -a=90^{\circ} -\frac a2.\]Thus $\angle DMI=90^{\circ} -\angle DIM=\frac a2$ so $\angle BAI=\frac a2=\angle IND=\angle IMD=\angle AIA'$ and $IA'\perp AB$, finishing. $\blacksquare$ Thus in the original problem, $A'D$ is a tangent to $\omega$ which as established above finishes. Remark: This problem took me $5$ hours to do sobs! I think there is a quicker solution with trig, though.

will post a very short coordinate bash with $A(0,0)$ and $P(1,0)$ soon

Delete $X$ and let $X=(I)\cap AB$. Then, $p-q=\frac{(x-1)(x^2+1)^2}{x(x^4+1)}$, as desired.

[asy][asy] // USA TST 2024/2 size(8cm); defaultpen(0.8); defaultpen(fontsize(9pt)); dotfactor*=0.6; pen lightbluedraw,bluedraw,purpledraw,pinkdraw; lightbluedraw = RGB(85,187,255); bluedraw = RGB(0,102,255); purpledraw = RGB(170,34,255); pinkdraw = RGB(255,17,255); pair A,B,D,P,I,Q,QQ,E,C,X; real a1 = 7; A = dir(180+a1); B = dir(90); D = dir(180-a1); P = (A-D)/2; I = (distance(A,B)*D-distance(A,P)*A)/(distance(A,B)-distance(A,P)); QQ = extension(B,D,A,A+I-B); Q = B+(QQ-B)*distance(B,A)^2/distance(B,QQ)^2; E = foot(I,A,B); C = extension(A,2foot(E,A,I)-E,B,2foot(E,B,I)-E); X = intersectionpoints(P--11*Q-10*P,A--I+intersectionpoints(A--C,circle(A,distance(A,I)))[0]-A)[0]; path c,d,e; d = circumcircle(A,B,Q); e = circle(I,distance(I,E)); draw(d,lightbluedraw); draw(e,purpledraw+dotted); draw(A--B,bluedraw); draw(B--C--A,bluedraw+dashed); draw(A--P,purpledraw); draw(A--I,purpledraw); draw(B--P,purpledraw); draw(B--QQ,pinkdraw); draw(B--I,lightbluedraw); draw(A--QQ,lightbluedraw); draw(P--X,pinkdraw); draw(A--X,purpledraw); dot("$A$",A,dir(320)); dot("$B$",B,dir(90)); dot("$D$",D,dir(130)); dot("$P$",P,dir(270)); dot("$I$",I,dir(140)); dot("$Q$",Q,dir(105)); dot("$Q^*$",QQ,dir(220)); dot("$E$",E,dir(315)); dot("$C$",C,dir(150)); dot("$X$",X,dir(120)); clip((0.3,1.5)--(-2.3,1.5)--(-2.3,-0.5)--(0.3,-0.5)--cycle); [/asy][/asy] Let $r$ be the inradius of $ABC$. Claim: $BP=BQ$. Proof: Let $R$ be the image of $Q$ under an inversion centered at $B$ with radius $BA$. The circumcircle of $ABQ$ inverts to the line through $A$ parallel to $\overline{BI}$, so $\overline{AR} \parallel \overline{BI}$. Let $B'$ be the point on $\overline{BP}$ such that $ADBB'$ is a parallelogram. Notice that $\angle CAB'=90^\circ$, so \[\angle PAB'=\angle PAC-90^\circ=180^\circ-\angle PAB-90^\circ=\angle ABP.\]Then, we have \[BR=BD \cdot \frac{IA}{ID}=AB' \cdot \frac{IA}{r}=\frac{AB'}{\sin \angle BAI}=AB' \cdot \frac{AB}{AP}=\frac{AB}{\cos \angle PAB'}=\frac{AB}{\cos \angle ABP}=\frac{BA^2}{BP}.\]However, we also have $BR=\tfrac{BA^2}{BQ}$, so $BP=BQ$. $\square$ Therefore, \begin{align*} \angle AXP&=180^\circ-\angle PAX-\angle APX \\ &=90^\circ-\angle IAX-(90^\circ-\angle BPQ) \\ &=90^\circ-\tfrac{\angle PBQ}{2}-\tfrac{\angle BAI}{2} \\ &=90^\circ-\tfrac{90^\circ-\angle PAB'}{2}-\tfrac{\angle ABP}{2} \\ &=45^\circ, \end{align*}as desired. $\square$

Stupid Problem indeed, solution same as others. Claim 1: $BP=BQ$

Now $\measuredangle AXP = 180^\circ - \measuredangle PAI - \measuredangle IAX - \measuredangle QPA = 90^\circ-\frac{\measuredangle A}{4}-45^\circ+\frac{\measuredangle A}{4}=45^\circ $

We will prove the following claim, which implies the problem by an easy angle chase. $\textbf{Claim.}$ $BP = BQ$. Let the circumcircle of $ADQ$ intersect $AB$ again at $S$. By Reim's theorem, $DS \parallel BI$. Let $BD \cap AC = T$. Note that we have \[ \frac{AI}{AD} = \frac{DI}{DT}. \]Length bash: \begin{align*} BP = BQ &\iff BP \cdot BD = BS \cdot BA \\ &\iff \frac{BS}{BA} = \frac{BP \cdot BD}{BA^2} \\ &\iff \frac{DI}{IA} = \frac{BP \cdot BD}{BA^2} \\ &\iff \frac{DT}{AD} = \frac{BP \cdot BD}{BA^2} \\ &\iff \frac{DT}{BD} = \frac{AD \cdot BP}{BA^2} \\ &\iff \frac{AT}{AB} = \frac{AD \cdot BP}{BA^2} \\ &\iff \frac{AT}{AD} = \frac{BP}{AB}, \end{align*}but this is easily verifiable from $\triangle ABP \sim \triangle DAT$. $\square$ With this claim, note that after even more angle chasing it is thus sufficient to prove that $\angle PBQ + \angle IAC = 90^\circ$, but this is very easy since $BP \parallel AD$ and $ADT$ is a right triangle. $\textbf{Remark.}$ This problem is rather dissatisfying.

Who cares about bash when you can synthetic? Solved with meql. [asy][asy] size(0,7cm); real eps = 1.15104332322704; /* trololol */ real v = 52 + eps; real v = 52 + eps; real w = 150 + eps; real w = 150 + eps; pair V = dir(v); pair W = dir(w); pair A = 2*V*W/(V+W); pair D = dir((v+w)/2); pair B = (D/(V*V)-1/D+2/W)/(1/(V*V)+1/(W*W)); pair I = 0; pair U = 2*foot(W,I,B)-W; pair C = 2*U*V/(U+V); pair P = foot(B,A,A+A*dir(90)); pair Q = B + dir(D-B)*(P-B)/dir(P-B); pair X = extension(P,Q,A,incenter(A,I,V)); pair T = extension(U,W,D,D+A-P); fill(P--A--Q--B--cycle, palered); fill(T--D--I--W--cycle, paleblue); draw(A--B--C--cycle, linewidth(1.2)); draw(incircle(A,B,C), linewidth(0.8)); draw(A--I, linewidth(0.7)); draw(P--A--Q--B--cycle, red+linewidth(0.8)); draw(T--D--I--W--cycle, blue+linewidth(0.8)); draw(P--X, linewidth(0.7)); draw(T--U, blue+linewidth(0.8)); draw(Q--D, red+linewidth(0.7)); dot("$V$", V, dir(53)); dot("$W$", W, dir(180)); dot("$A$", A, dir(77)); dot("$D$", D, dir(54)); dot("$B$", B, dir(-130)); dot("$I$", I, dir(-54)); dot("$U$", U, dir(-90)); dot("$C$", C, dir(-19)); dot("$P$", P, dir(146)); dot("$Q$", Q, dir(-70)); dot("$X$", X, dir(-12)); dot("$T$", T, dir(-141)); [/asy][/asy] Let the incircle be $\omega$ and touch $BC$ and $AB$ at point $U$ and $W$. Let the tangent to $\omega$ at $D$ meet $UW$ at $T$. Notice that $T$ is the pole of $BD$ w.r.t. $\omega$, so $IT\perp BD$. Now, we make the following critical claim. Claim. Quadrilaterals $DIWT$ and $PBQA$ are inversely similar. Proof. This follows from four angle relations. $\measuredangle IDT = \measuredangle BPA = 90^\circ$. $\measuredangle TIW = \measuredangle ABQ$. $\measuredangle DIT = \measuredangle IAC = \measuredangle BAI = \measuredangle ABP$. $\measuredangle ITW = \measuredangle QBI = \measuredangle QAB$. $\blacksquare$ The rest is just angle chasing. Notice that from the claim, we have $$\angle APQ = \angle TDW = \angle DUW = \frac{\angle AIZ}2 = 45^\circ - \frac{\angle A}4.$$Hence, \begin{align*} \angle AXP &= 180^\circ - \angle PAX - \angle APQ \\ &= 180^\circ - \left(90^\circ + \frac{\angle A}4\right) - \left(45^\circ - \frac{\angle A}4\right) = 45^\circ. \end{align*}

I have time now but I don't have my calculations. Write $\sin \angle BAP=a$, $\cos \angle BAP=b$, $\cos 2\angle BAP=c$. Set $A(0,0)$, $P(1,0)$, and $B\left(1,\frac{a}{b}\right)$. Reflect $AB$ across the $y$-axis to line $\ell$. Draw point $Z$ where $AZ=AP$, hence $Z(-b,a)$. Draw altitude $E$ from $B$ to $\ell$. From lengths you can get the coordinates of $E$ easily. Now we intersect at a point $Q'=BE\cap PZ$. (It is technically only necessary to find the slope of $BE$, hence $E$ is irrelevant.) Then we use similar triangles to find the coordinates of $I$ along the $y$-axis (it should be something like $\frac{AD}{1-b}$ if I remember correctly). Finally prove that $AQ'$ and $BI$ have reciprocal (not negative reciprocal!) slopes. This step is the magical part, and everything cancels. Now we're done by angle chase. The most complicated step is finding $Q'$, and even this expression need not be expanded (it all cancels later).

oh I guess I sillied edit: looks like I forgot to cancel one side when dividing by x resulting in an x^2 instead of x factor

I am posting the author's original solution and an inversion-based solution from Jeffrey Kwan. (The #12 solution will also appear in the final production PDF.) First solution, by author. [asy][asy] unitsize(1.0inches); real eps = 1.15104332322704; /* trololol */ real v = 52 + eps; real w = 150 + eps; pair V = dir(v); pair W = dir(w); pair A = 2*V*W/(V+W); pair D = dir((v+w)/2); pair B = (D/(V*V)-1/D+2/W)/(1/(V*V)+1/(W*W)); pair I = 0; pair U = 2*foot(W,I,B)-W; pair C = 2*U*V/(U+V); pair P = foot(B,A,A+A*dir(90)); pair Q = B + dir(D-B)*(P-B)/dir(P-B); pair X = extension(P,Q,A,incenter(A,I,V)); draw(circumcircle(V,-V,W)); draw(A--B--C--cycle); draw(A--I); draw(I--U,linewidth(0.3)); draw(B--P--A); draw(B--Q--D); draw(B--I,linewidth(0.3)); draw(P--Q--X); draw(A--X,linewidth(0.3)); pair O = circumcenter(A,Q,B); pair AA = (A-O)*dir(20)+O; pair BB = (B-O)*dir(-20)+O; draw(arc(O,BB,AA),linewidth(0.4)+dashed); dot("$A$",A,dir(45)); dot("$D$",D,dir(-30)); dot("$B$",B,dir(260)); dot("$I$",I,dir(0)); dot("$E$",U,dir(U)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(-55)); dot("$X$",X,dir(270)); [/asy][/asy] Claim: We have $BP=BQ$. Proof. For readability, we split the proof into three unconditional parts. We translate the condition $\overline{BD} \perp \overline{AC}$. It gives $\angle DBA = 90^{\circ} - A$, so that \begin{align*} \angle DBI &= \left| \frac B2 - (90^{\circ} - A) \right| = \frac{|A-C|}{2} \\ \angle BDI &= \angle DBA + \angle BAD = (90^{\circ}-A) + \frac A2 = 90^{\circ} - \frac A2. \end{align*}Hence, letting $r$ denote the inradius, we can translate $\overline{BD} \perp \overline{AC}$ into the following trig condition: \[ \sin \frac B2 = \frac{r}{BI} = \frac{DI}{BI} = \frac{\sin \angle DBI}{\sin \angle BDI} = \frac{\sin\frac{|A-C|}{2}}{\sin\left(90^{\circ}-\frac A2\right)}. \] The length of $BP$ is given from right triangle $APB$ as \[ BP = BA \cdot \sin \angle PAB = BA \cdot \sin\left( 90^{\circ}-\frac A2 \right). \] The length of $BQ$ is given from the law of sines on triangle $ABQ$. The tangency gives $\angle BAQ = \angle DBI$ and $\angle BQA = 180^{\circ} - \angle ABI = 180^{\circ} - \angle IBE$ and thus \[ BQ = BA \cdot \frac{\sin \angle BAQ}{\sin \angle AQB} = BA \cdot \frac{\sin \angle DBI}{\sin \angle ABI} = BA \cdot \frac{\sin \frac{|A-C|}{2}}{\sin \frac B2}. \] The first bullet implies the expressions in the second and third bullet for $BP$ and $BQ$ are equal, as needed. $\blacksquare$ Remark: In the above proof, one dos not actually need to compute $\angle DBI = \frac{|A-C|}{2}$. The proof works equally leaving that expression intact as $\sin \angle DBI$ in place of $\sin \frac{|A-C|}{2}$. Now we can finish by angle chasing. We have \[ \angle PBQ = \angle PBA + \angle ABD = \frac{A}{2} + 90^{\circ} - A = 90^{\circ} - \frac{A}{2}. \]Then \[ \angle BPQ = \frac{180^{\circ} - \angle PBQ}{2} = 45^{\circ} + \frac{A}{4}, \]so $\angle APQ = 90^{\circ} - \angle BPQ = 45^{\circ} - \frac{A}{4}$. Also, if we let $J$ be the incenter of $IAC$, then $\angle PAJ = 90 ^{\circ} + \frac{A}{4}$, and clearly $X$ lies on line $AJ$. Then $\angle APQ + \angle PAJ = 135 ^{\circ} < 180 ^{\circ}$, so $X$ lies on the same side of $AP$ as $Q$ and $J$ (by the parallel postulate). Therefore $\angle AXP = 180 ^{\circ} - 135 ^{\circ} = 45 ^{\circ}$, as desired. Remark: The problem was basically written backwards by starting from the $BD \perp AC$ condition, turning that into trig, and then contriving $P$ and $Q$ so that the $BD \perp AC$ condition implied $BP=BQ$. Second solution, by Jeffrey Kwan. We prove the following restatement: \begin{quote} Consider isosceles triangle $AEF$ with $AE = AF$ and incenter $D$. Let $B$ be the point on ray $AE$ such that $BD\perp AF$, and let $P$ be the projection of $B$ onto the line through $A$ parallel to $EF$. Let $I$ be the point diametrically opposite $A$ in the circumcircle of $AEF$, and let $Q$ be the point on line $BD$ such that $BI$ is tangent to the circumcircle of $AQB$. Then $\angle APQ = 45^\circ - \angle A / 4$. \end{quote} First note that $\angle DFE = 45^\circ - \angle A / 4$, so it suffices to show that $\overline{PQ}\parallel \overline{DF}$. Let $U = \overline{BD} \cap \overline{EF}$, and let $V = BI\cap (AEF)$. Observe that: $P$ and $V$ both lie on the circle with diameter $AB$, so $\angle BVP = \angle PAB = 90^\circ - \angle A / 2$. We have $\angle EVB = \angle EVI = \angle A / 2 = \angle DUF = \angle BUE$. Hence $BEUV$ is cyclic. Now $\angle BVU = \angle AEU = 90^\circ - \angle A / 2 = \angle BVP$, so $\overline{PUV}$ are collinear. [asy][asy] unitsize(0.9inches); size(600); pair A, E, F, D, B, I, P, Q, U, V; A = dir(90); E = dir(191); F = dir(349); D = incenter(A, E, F); I = dir(270); B = extension(A, E, D, foot(D, A, F)); P = foot(B, A, A + E - F); Q = extension(B, D, P, P + F - D); U = extension(B, D, E, F); V = extension(B, I, P, U); draw(A--E--F--cycle, orange); draw(circumcircle(A, E, F), red); draw(incircle(A, E, F), heavyred); draw(E--B--foot(D, A, F), orange); draw(A--P--B, heavycyan); draw(A--Q--B^^P--Q, lightblue); draw(P--V, dotted+magenta); draw(B--V, dotted+magenta); draw(circumcircle(A, P, B), magenta); draw(circumcircle(A, P, Q), purple + dashed); dot("$A$", A, dir(60)); dot("$B$", B, dir(B)); dot("$E$", E, dir(160)); dot("$F$", F, dir(F)); dot("$D$", D, dir(330)); dot("$I$", I, dir(270)); dot("$P$", P, dir(135)); dot("$Q$", Q, dir(300)); dot("$U$", U, dir(270)); dot("$V$", V, dir(300)); dot(foot(D, A, F)); [/asy][/asy] From the tangency condition, we have that $\angle AQB = 180^\circ - \angle ABI$, which implies that \[\angle AQU + \angle APU = \angle AQB + \angle APV = (180^\circ - \angle ABI) + \angle ABI = 180^\circ,\]and so $APUQ$ is cyclic. Finally, note that $D$ is the orthocenter of $\triangle AUF$, which implies that \[\angle APQ = \angle AUQ = \angle AUD = \angle AFD = \angle DFE.\]This forces $\overline{PQ}\parallel \overline{DF}$, as desired.

revenge. kill all geometry. $Z=PX\cap AC$ $E=BD\cap AC$ $D'\in AB$ and $DD'\parallel BI$ $R$ where $BPAR$ is a rectangle $F$ is the point of tangency of incircle to $AB$ Claim: $BR=BD'$. Proof: Because \[\frac{BD'}{BA}=\frac{ID}{IA}=\frac{IF}{IA}=\frac{BR}{BA}.\] Claim: $BP=BQ$. Proof: $\angle BDD'=\angle IBD=\angle BAQ$ so $D,D',Q,A$ concyclic. And $\angle BDR=\angle ADE=90^{\circ}-\frac{1}{2}\angle BAC=\angle BAP$. So \[\frac{BQ}{BA}=\frac{BD'}{BD}=\frac{BR}{BD}=\frac{BP}{BA}.\] Claim: $\angle AXP=45^{\circ}$. Proof: Because $\angle PBQ=180^{\circ}-\angle PAE=90^{\circ}-\frac{1}{2}\angle BAC$ then $\angle BQP=45^{\circ}+\frac{1}{4}\angle BAC$ so \[\angle AXP=\frac{1}{4}\angle BAC+\angle QZE=\frac{1}{4}\angle BAC+90^{\circ}-45^{\circ}-\frac{1}{4}\angle BAC=45^{\circ}.\]

Let $S$ be the foot of the perpendicular line from $B$ to $AI$. Note that $\angle SBD = \angle SAC = \angle SAB$, so $\triangle SBD \sim \triangle SAB$. Therefore, \[ \frac{BQ}{BA} = \frac{\sin \angle BAQ}{\sin \angle BQA} = \frac{\sin \angle IBD}{\sin \angle IBA} = \frac{ID}{AD} \cdot \frac{\sin \angle BDI}{\sin \angle BAI} = \frac{ID}{AD} \cdot \frac{AS}{BS} = \frac{BP}{AB}, \]where the last equality is because if we let the incircle touch $AB$ at $E$ then $\triangle AEI \sim \triangle ASB$. Therefore, $BQ = BP$, and so \[ \angle APX = 90^\circ - \angle APQ = \frac{1}{2}\angle PBD = \frac{1}{2}\angle BDS = \frac{1}{2}\angle ABS. \]Finally, \[ \angle AXP = 90^\circ - \frac{1}{2} \angle BAS - \angle APX = 90^\circ - \frac{1}{2} \angle BAS - \frac{1}{2}\angle ABS = 45^\circ. \]

Average Trig bash be like

Most angles can be determined using simple angle chase. If we suppose $\angle AXP = \theta$, we can find \[\angle BPQ = \theta + \frac A4, \quad \angle BQP = 90 - \theta + \frac A4,\] so it suffices to prove $BP=BQ$. Notice $BP = AB \cos \tfrac A2$, and Law of Sines gives us \[BQ = AB \cdot \frac{\sin \angle BAQ}{\sin \angle BQA} = AB \cdot \frac{\sin \angle BDI}{\sin \angle ABI} = AB \cdot \frac{ID}{BD} \cdot \frac{AB}{AI} = \frac{AB^2}{BD} \sin \frac A2.\] Thus we require \[\frac{BD}{AB} = \tan \frac A2 \iff \frac{\sin \tfrac A2}{\sin \left(90+\tfrac A2\right)} = \tan \frac A2. \quad \blacksquare\]

i will try to present a motivated solution Part 1- "Reverse angle chase": $BP=BQ$ is sufficent. We have $$\angle AXP=45\iff \angle APQ=45-\alpha/4\iff \angle DPB=45+\alpha/4.$$However, $$\angle PBQ=180-\angle PAE=180+(90+\alpha/2)=90-\alpha/2,$$which shows the claim. Part 2- Reduction to relationship of angles in $\triangle ABC$: Now, we have by law of sines on $\triangle AQB$ that $$BQ=AB\cdot\frac{\sin(\alpha-\gamma)/2}{\sin\beta/2}$$and $$BP=AB\cdot\cos\alpha/2.$$Thus, it suffices to show that $$\sin(\alpha-\gamma)/2=\sin\beta/2\cos\alpha/2.$$ Part 3- Interpretation of $BE\perp AC$: The idea is to interpret the perpendicularity condition by "double-counting" $AI$, one time without using $E$ and one time using the perpendicularity condition. In the above work, the most "strange" thing is the angle $(\alpha-\gamma)/2$, but this is just $\angle DBI$. Thus, using the condition $BE\perp AC$ to angle chase, we use law of sines on $\triangle DBI$ to get $$BI=r\cdot\frac{\sin90-\alpha/2}{\sin(\alpha-\gamma)/2}$$so then $$AI=\frac{\sin\beta/2}{\sin\alpha/2}r\cdot\frac{\sin90-\alpha/2}{\sin(\alpha-\gamma)/2}.$$ However, we can also compute unconditionally that $$AI=\frac{r}{\sin\alpha/2}.$$Thus, setting these equal gives $$\sin(\alpha-\gamma)/2=\sin\beta/2\cos\alpha/2,$$as desired. remark: the main difficulty of this problem is interpreting $BE\perp AC$, which is what I was stuck on in-test. There are actually many ways to do this, but most of them end up with messy trig expressions. The reason that this particular calcuation is so nice is that the expression we said was equivalent to the conclusion had the angle $(\alpha-\gamma)/2=\angle BAQ=\angle DBI$ in it. Thus, we want to use this angle, and law of sines on $\triangle DBI$ is a very natural choice for this.

It follows easily by angle chasing and trig bashing Step 1 (angle Chasing): We Claim $\measuredangle AXP = \dfrac{\pi}{4} \iff BP = BQ$ Proof: In which follows we let $ A = \measuredangle BAC, B = \measuredangle CAB, C = \measuredangle ABC.$ Note that $ \measuredangle PAX = \measuredangle PAI + \measuredangle IAX = \dfrac{\pi }{4} + \dfrac{A}{4}$ and so \[ \measuredangle AXP = \dfrac{\pi}{4} \iff \measuredangle XPA = \dfrac{\pi - A}{4} \iff \measuredangle BPQ = \dfrac{\pi + A}{4} \] and as $\measuredangle QBP= \dfrac{\pi - A}{2}$ the conclusion follows as we wanted. Step 2 : Rewriting $BP = BQ$ using trig, and then trig bashing the condition on $D.$ We now compute both $BP$ and $BQ$ using trig. Easy angle chasing shows $ \measuredangle ABP = \dfrac{A}{2}$ and $ \measuredangle AQB = \pi - \frac{B}{2}, \measuredangle BAQ = \frac{A - C}{2} $ and so \[ BP = BA \cdot \cos \left( \frac{A}{2} \right); ~~ BQ = BA \cdot \dfrac{ \sin \left( \dfrac{A-C}{2} \right) }{\sin \left( \frac{B}{2} \right) } \]and therefore $BP = BQ$ if, and only if \[ \sin \left( \dfrac{A-C}{2} \right) = \cos \left( \frac{A}{2} \right) \cdot \sin \left( \frac{B}{2} \right) ~~ (*)\] we now show that if $BD\perp AC$ then $(*)$ is true. Let $ \omega$ be the incircle of $ \triangle ABC$ and let $P,Q,R$ be the foots of $D$ at $BC,CA, AB,$ respectively. Also let $M$ be the midpoint of arc $QPR$ of $\omega $ and let $U$ be the point in $\omega$ such that $MU \parallel AC.$ Also let $p,q,r$ denote the angles of triangle $\triangle PQR.$ First note that as $p = \frac{\pi + A}{2} $ and so on, it follows that $(*)$ is equivalent to \[ \sin (r -p ) = \sin p \cdot \cos q ~~(***) \] We now show that $BD \perp AC$ implies $(***)$. We first claim that $B, U,D$ are collinear; indeed, if $BD$ cuts $\omega$ again at $U'$ then $U'M \perp U'D \perp AC$ and so $U'M \parallel AC$ and $U' = U$. So indeed, $B,U,D$ are collinear and then quadrilateral $PURD$ is harmonic from which follows \[ \dfrac{DP}{DR} = \dfrac{UP}{UR} \iff DP \cdot UR = DR \cdot UP ~~(**) \]but now straightforward angle chasing shows \[ \measuredangle DPR = \frac{p}{2}; ~\measuredangle PRD = \frac{-p + 2r}{2} \]\[ \measuredangle UPR = \frac{\pi - 3p}{2} = \frac{-2p + q +r}{2}, ~ \measuredangle PRU = \frac{2p-q+r}{2} \] and therefore by sine rule and $(**)$ we have that problem condition is equivalent to \[ \sin \left(\frac{p+2r}{2} \right) \cdot \sin \left(\frac{-2p + q +r}{2} \right) = \sin \left( \frac{p}{2} \right) \cdot \sin \left( \frac{2p+q - r}{2} \right) \] which by product-to-sum formula is equivalent to \[ \cos \left( \frac{ 3p -q +r }{2} \right) - \cos \left( \frac{ -p +q + 3r }{2} \right) = \cos \left( \frac{ p +q - r }{2} \right) - \cos \left( \frac{ 3p +q-r }{2} \right) \] now using that $\cos x = - \sin (x - \frac{\pi}{2}) = \sin (x + \frac{\pi}{2})$ and that $p + q + r = \pi$ the above equation simplifies to \[ -\sin ( p - q) + \sin (-p +r) = - \sin (-r) + \sin (p-r) \iff \]\[ 2\sin (r-p) = \sin r + \sin (p-q) \]but by sum-to-product formula, we can rewrite the right hand side as \[ 2\sin (r-p) = 2 \sin \left( \frac{ p+q -r }{2} \right) \cdot \cos \left( \frac{q+r-p }{2} \right) = 2 \cos r \cdot \sin p \]where you used $p+q+r = \pi.$ So $(***)$ is proved and we are done.

Let $\angle BAI = \alpha$ and $\angle ABI = \beta$. We will show that $\angle AXP = 45^{\circ}$ $\Leftrightarrow$ BP = BQ. From angle chase we get that $\angle ABQ = 90 - 2\alpha$, $\angle DBI = \beta + 2\alpha - 90$, $\angle BAD = \alpha$, $\angle IAX = \frac{\alpha}{2}$, $\angle PAB = 90 - \alpha$, $\angle PBA = \alpha$ $\Rightarrow$ $\angle PBQ = \alpha + 90 - 2\alpha = 90 - \alpha$. Let BP = BQ $\Rightarrow$ $\angle BPQ = \angle BQP = 45 + \frac{\alpha}{2}$ $\Rightarrow$ $\angle QPA = 90 - 45 - \frac{\alpha}{2} = 45 - \frac{\alpha}{2}$ $\Rightarrow$ $\angle PXA = 180 - \angle XPA - \angle PAX = 180 -(45 - \frac{\alpha}{2}) - \angle PAI - \angle IAX = 135 + \frac{\alpha}{2} - 90 - \frac{\alpha}{2} = 45^{\circ}$, we do similarly in the other direction $\Rightarrow$ BP = BQ $\Leftrightarrow$ $\angle AXP = 45^{\circ}$. Now it is only left to prove that BP = BQ. From $\triangle ABP$ we get that $\frac{BP}{AB} = \cos \alpha$. From law of sines on $\triangle ABQ$ we get that $\frac{BQ}{AB} = \frac{\sin \angle BAQ}{\sin \angle BQA}$ $\Rightarrow$ since we want to show that BP = BQ, we want $\frac{BQ}{AB} = \frac{BP}{AB}$, so we want to prove that $\frac{\sin \angle BAQ}{\sin \angle BQA} = \cos \alpha$. Now by angle chase we have that $\frac{\sin \angle BAQ}{\sin \angle BQA} = \frac{\sin \angle IBD}{\sin (180 - \angle ABI)} = \frac{\sin \angle IBD}{\sin \angle IBA}$. Now from law of sines on $\triangle BDI$ we have that $\frac{DI}{\sin \angle DBI} = \frac{BD}{\sin \angle BID}$ $\Rightarrow$ $\angle DBI = \frac{DI \cdot \sin \angle BID}{BD}$. Now from law of sines on $\triangle ABI$ we have that $\frac{AI}{\sin \angle IBA} = \frac{AB}{\sin \angle AIB}$ $\Rightarrow$ $\sin \angle IBA = \frac{AI \cdot \sin \angle AIB}{AB}$. From these two things we just got, we have that $\frac{\sin \angle IBD}{\sin \angle IBA} = \frac{DI}{AI} \cdot \frac{AB}{BD} = \frac{r}{AI} \cdot \frac{AB}{BD} = \frac{4R \sin \alpha \cdot \sin \beta \cdot \sin \gamma}{4R \sin \beta \cdot \sin \gamma} \cdot \frac{AB}{BD} = \sin \alpha \cdot \frac{AB}{BD} = \sin \alpha \cdot \frac{\sin \angle ADB}{\sin \angle BAD} = \sin \alpha \cdot \frac{\sin (90 + \alpha)}{\sin \alpha} = \cos (-\alpha) = \cos \alpha$, which is exactly what we wanted to prove $\Rightarrow$ BP = BQ and $\angle AXP = 45^{\circ}$. We are ready.

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Let $H_B$ be the intersection of $BD$ with $AC$. Let the perpendicular from $I$ onto $AC$ and $AB$ be $E$ and $F$, respectively. Let $AQ$ intersect $BI$ in $R$. Claim: $BP=BQ$ Note that $\triangle RBQ\sim \triangle RAB$. By the Law of Sines, \[\frac{BQ}{BA}=\frac{QR}{RB}=\frac{\sin(\angle DBI)}{\sin(\angle BQR)}=\frac{\sin(\angle DBI)}{\sin(\angle ABI)}=\frac{\frac{H_BE}{IB}}{\frac{IF}{IB}}=\frac{H_BE}{IF}\]and \[\frac{BP}{BA}=\cos(\angle PBA)=\cos(\angle BAI)=\cos(\angle IAE)=\frac{AE}{AI}=\frac{H_BE}{ID}=\frac{H_BE}{IF}\]so $BP=BQ$. Let $\angle AXP=x$. Note that $\angle PAX=90^\circ+\tfrac14\angle A$, so $\angle APX=90^\circ-\tfrac14\angle A-x$, implying that $\angle BPQ=\tfrac14\angle A+x$. We have $\angle BQP=\tfrac14\angle A+x$, so $\angle PBQ=180^\circ-(2x+\tfrac12 A)$. Since $PAH_BB$ is cyclic, we have $\angle PAH_B=2x+\tfrac12 \angle A$. On the other hand, $\angle PAH_B=90^\circ+\tfrac12 \angle A$ so this shows that $x=45^\circ$ as desired.