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Source: 2025 CTST P16

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jhz

#1 h Mar 26, 2025, 12:56 AM • 1 Y

Y by Rounak_iitr

In convex quadrilateral $ABCD, AB \perp AD, AD = DC$ . Let $E$ be a point on side $BC$ , and $F$ be a point on the extension of $DE$ such that $\angle ABF = \angle DEC>90^{\circ}$ . Let $O$ be the circumcenter of $\triangle CDE$ , and $P$ be a point on the side extension of $FO$ satisfying $FB =FP$ . Line BP intersects AC at point Q. Prove that $\angle AQB =\angle DPF.$

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jhz

#2 h Mar 26, 2025, 2:58 AM

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Reconstruct $P$ as the intersection of $\odot ABC$ and $\odot DEC$ . remain is simple calculation.

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1540 posts

#3 h Mar 26, 2025, 3:39 AM

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Sketch since its so painful.

Redefine $P$ as $(ABC) \cap (DEC)$ . Then we may show that $\angle FBP = \angle BPO$ through angle chasing, redefine $F$ as $BF \cap PO$ . Then let $H = BA \cap ED$ , note that $(BEH)$ is cyclic and tangent to $BF'$ , likewise we may angle chase that $DAHP$ is cyclic so $\angle DPH = 90^\circ$ and thus $(EPH)$ is tangent to $F'P$ , thus $F' \in EDH$ , and thus $F' = F$ . Finally, the result is another painful angle chase.

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DottedCaculator

7348 posts

DottedCaculator

#4 h Mar 26, 2025, 6:51 PM • 1 Y

Y by fake123

this is why i'm bad at geo

Claim: $P$ lies on $(CDE)$

Proof: Verify $\sqrt{FE\cdot FD+R^2}=BF-R$ , where $R$ is the radius of $(CDE)$ by trig bash

Now, construct $P'=FO\cap(CDE)$ and then angle/arc bash the required angle equality.

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904 posts

#5 h Apr 5, 2025, 11:34 PM • 2 Y

Y by Kaimiaku, Rounak_iitr

Let $G=AB\cap DE$ . Let the circumcircle of triangle $BDE$ intersects segments $AD$ and $BF$ again at $X\neq D$ and $Y\neq B$ , respectively. Since $\angle FEB=\angle CED=\angle FBG$ it follows that $FB$ is tangent to $(BEG)$ .
Claim. $P$ lies on $(CDE)$
Proof. We first show that $BY$ has equal length to the diameter of $(CDE)$ . Notice that $\angle XBY=\angle XBE+\angle EBF=\angle ADG+\angle AGD=90^{\circ}$ Since $\angle BYX=\angle BDA$ , right triangles $ABD$ and $BXY$ are similar. Then $BY=BX\cdot \frac{AD}{AB}=AD\cdot\frac{BX}{AB}=\frac{CD}{\cos(\angle ABX)}=\frac{CD}{\sin(\angle CED)}$ Which is equal to the diameter of $(CDE)$ . Notice that the power of $F$ with respect to $(CDE)$ is equal to $FD\cdot FE$ which is equal to $FB\cdot FY$ . Denote the radius of $(CDE)$ as $r$ . The power can also be written as $FO^2-r^2$ so $(FO+r)(FO-r)=FB\cdot FY=(FO+OP)\cdot(FO+OP-2r)$ Then clearly $OP=r$ , finishing the claim.
$[asy] import geometry;size(12cm);pair A=(0,0);pair B=(4,0);pair D=(0,-3);pair C=D+rotate(220)*(A-D);pair E=C+.17(B-C);pair G=intersectionpoint(line(D,E),line(A,B));pair tac=circumcenter(G,E,B);line cat=line(B,B+rotate(90)*(tac-B));pair F=intersectionpoint(cat, line(E,D));pair O=circumcenter(C,D,E);pair P=intersectionpoint(circle(A,D,G),circle(C,E,D));circle w=circle(B,D,E);point[] X=intersectionpoints(w,line(A,D));point[] Y=intersectionpoints(w,line(B,F));pair X=X[1];pair Y=Y[0];pair K=circumcenter(A,B,C);pair Z=intersectionpoint(line(B,D),line(X,Y));pair Q=intersectionpoint(line(A,C),line(B,P));filldraw(A--B--D--cycle,white+palegreen);filldraw(B--X--Y--cycle,white+palegreen);filldraw(Z--X--B--cycle, palegreen);draw(A--B--C--D--cycle);draw(F--G);draw(circle(C,D,E), dotted);draw(P--F);draw(circle(A,D,G),dashed+purple);draw(w,blue);draw(B--F);draw(circle(G,E,B),red);draw(G--A);draw(A--C,grey ); draw(B--P,grey );draw(circle(A,B,C),dashed+orange);perpendicularmark(line(A,D),line(A,B));draw(A--D,StickIntervalMarker(1,1));draw(D--C,StickIntervalMarker(1,1));markangle(B,Q,A,radius=.4cm);markangle(B,Q,A,radius=.5cm);markangle(F,P,D,radius=.4cm);markangle(F,P,D,radius=.5cm);draw(G--P--D,grey);dot("A", A,dir(60));dot("B", B,dir(20));dot("C", C,dir(-30));dot("D", D,2*dir(180));dot("E",E,1.5*dir(100));dot("F",F,dir(-20));dot("G",G,dir(160));dot("P",P,dir(180));dot("O",O,dir(250));dot("X",X,dir(160));dot("Y",Y);dot("Q",Q,dir(-25)); [/asy]$
Next we show that $(ADPG)$ and $(ABCP)$ are cyclic. $\angle PGF=\angle FPE=\angle OPE=90^{\circ}-\angle PDG\Rightarrow 90^{\circ}=\angle DPG=\angle DAG$ $\angle APC=\angle APD+\angle DPC=(\angle ADE-90^{\circ})+\angle DEB=180^{\circ}-\angle ABC$ Then to finish the problem $\angle AQB=\angle PAC+\angle ACB=\angle PGD+\angle DAC+\angle ACB=\angle FPE+\angle DCB=\angle DPF$

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115 posts

#6 h Apr 15, 2025, 2:51 AM

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Handwritten writeup. Too lazy to type it up. If anyone is nice enough to transcribe it thanks a lot! (Tbh I am posting it here cuz someone told me to)

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1896 posts

#7 h Apr 15, 2025, 3:50 AM

Y by

incomplete transcription, since i have to lock in for schoolwork lol

Let $X=AB\cap DE,$ $\{P',D\}=(XAD)\cap(CED).$ $\measuredangle XP'D=\measuredangle XAD=90^\circ.$
$\measuredangle AP'C=\measuredangle APD+\measuredangle DP'C=\measuredangle AXD+\measuredangle DEC=\measuredangle ABC\implies A,B,C,P\text{ concyclic.}$ Let $G$ so that $GX$ tangent $(XB'P)$ and $GE$ tangent to $(EBP')$
Let $\measuredangle P'XD=\alpha$ , $\measuredangle DAC=\beta$ , $\measuredangle DEP'=\gamma.$ $\measuredangle P'AD=\measuredangle P'XD=\alpha,$

This post has been edited 1 time. Last edited by aidan0626, Apr 15, 2025, 6:02 AM

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mathuz

#8 h Apr 25, 2025, 7:35 PM

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Another way to think about this problem:

We fix $E$ , while allowing $D$ to vary.

Consider an isosceles trapezoid $BB'H'H$ whose diagonals meet at $E$ (see the figure). Assume, $BH\cap B'H'=T(.)$ . The circumcenter $F$ of the trapezoid lies on the perpendicular bisectors of both $BB'$ and $HH'$ . Call this circumcircle as $(F)$ .

Now, point $D$ moves along the line $TE$ (in the part, which is inside the circle $(F)$ ). Let the projection of $D$ on $TB$ be $A$ . The circle $(D)$ that is centred at $D$ with radius $DA$ meets the diagonal $BH'$ at $C$ . This construction gives us exactly the configuration described in the original problem.

The proof contains the following steps:

(1) In the figure, $P$ is defined as the intersection of $(TDA)$ and $(F)$ . Then $(PDE)$ and $(F)$ are tangent to each other at $P$ . This follows from the fact about harmonic points. Note that $\{T,X,E,Y\}$ is harmonic, so $(F)$ is the Apollonian circle of $T$ and $E$ , i.e. locus of $M$ such that $MT:ME=XT:XE$ . This implies $\angle PED = \angle PFD + \angle PTD = \angle PYP' + \angle DPP',$ so the circles are tangent to each other at $P$ .

(2) Assume $(PDE)$ meets $BH'$ again at $C'$ . A simple calculation shows that $\frac{DC'}{DA} = \frac{r}{R}\cdot \frac{BT}{BE},$ where $r$ and $R$ are radii of $(PDE)$ and $(TDA)$ . On the other hand, we have $r\sin \angle PED = R\sin \angle PTD$ and $\frac{r}{R} = \frac{PE}{PT}.$ Now, noting that $(F)$ is the Apollonian circle of $T$ and $E$ , we get $DC'=DA$ . This yields $C'\equiv C$ .

At this stage, we have established that our constructed point $P$ coincides with the point $P$ given in the problem.

(3) Applying the Miquel's point (configuration) for the triangle $TBE$ , where the points $A,C,D$ are chosen on the sides, we further obtain that $P$ lies on $(ABC)$ .

(4) Now it remains to show that $\angle AQB = \angle FPD$ , where $Q(.)=AC\cap BP$ .

Let $O$ be the center of $(PDEC)$ . Then $\angle FPD = \angle OPD = 90^{\circ} - \angle PED$ . Thus, it suffices to show that $\angle AQB + \angle PED = 90^{\circ}$ . Indeed, we have $\angle AQB + \angle PED = \angle APB + \angle PBC + \angle PFX +\angle PTD =$ $= \angle ATP' + \angle BYP' + \angle PYH' + \angle PYP' + \angle P'TD$ $= ( \angle ATP' + \angle P'TD) + (\angle BYP'+ \angle P'YP + \angle PYH')$ $= \angle ATD + \angle BYH' = \angle ATD + \angle BHH' = 90^{\circ}.$ This finishes the proof.

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This post has been edited 1 time. Last edited by mathuz, Apr 25, 2025, 7:40 PM

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