here's the solution I gave on the test; any deductions? also apparently this problem was trivial by spiral similarity but oh well Let $AC$ and $BD$ intersect at $X$, $AP$ meet $BC$ at $Y$, and the circumcenter of $(ABCD)$ be $O$. In addition, let $M$ and $N$ denote the midpoints of $DE$ and $DF$ respectively. Claim 1. $X$ is the midpoint of $DP$. Proof. Since $\angle DNC=\angle DXA=\angle DMA=90$, points $N, X, M$ form the Simson Line from $D$ to $ABC$ and thus are collinear. The homothety at $D$ taking $N$ to $F$ and $M$ to $E$ also takes $X$ to $P$, and since $DE=2DM$, $DP=2DX$, proving our claim. $\Box$ Claim 2. $A$ is the circumcenter of $DPE$ and $C$ is the circumcenter of $DPF$. Proof. We only prove the part with point $A$ since the other part follows by symmetry. $AB$ is the perpendicular bisector of $DE$ and Claim 1 tells us that $AC$ is the perpendicular bisector of $DP$. These lines meet at $A$, which proves the claim. $\Box$ Claim 3. $P$ is the orthocenter of $ABC$. Proof. By Claim 1, $D$ is the reflection of $P$ across $AC$, and since $BX$ is an altitude in $ABC$, it's very well known that $P$ must be the orthocenter. $\Box$ Lemma (to be used in Claim 4). Let $P'$ denote the reflection of $P$ across $BC$; then $P'D\perp OC$. Proof. First we show $XY\perp OC$. We have $\angle YCO=\frac{180-\angle COB}{2}=\frac{180-2\angle CAB}{2}=90-\angle CAB$. Then, by cyclic quadrilateral $BYXA$, $\angle CYX=\angle CAB$. Thus $\angle YCO+\angle CYX=90$ which implies $XY$ is perpendicular to $OC$. Then by homothety at $P$, $YX$ maps to $P'D$, so $XY\parallel P'D$ and $P'D\perp CO$. $\Box$ Claim 4. $R$ and $Q$ are the reflections of $P$ across $BC$ and $AB$. Proof. We only show $R$ is the reflection of $P$ over $BC$ since the other case follows by symmetry. It suffices to show $R=P'$. Because $RD$ is the radical axis of $(FPD)$ and $(ABC)$, we conclude $R$ is the point on $(ABC)$ such that $RD\perp CO$. But it's well known that $P'$ lies on $(ABC)$ and by the Lemma, $P'D\perp CO$. Thus $R=P'$. $\Box$ Finally, we can finish the problem. By Claim 4, $BR=BP$ and $BP=BQ$ so $BR=BQ$. Since the perpendicular bisectors of $DF$ and $DE$ intersect at $B$, $B$ is the circumcenter of $DEF$ and $BF=BE$. Also, we know $B, P, D$ are collinear and because $R, F$ are just the reflections of $P$ and $D$ across $BC$, we have that $B,R, F$ are collinear. For similar reasons, $B,Q,E$ are collinear. Therefore, $FR=BF-BR=BE-BQ=QE$ which completes the proof.

Hello, Are there config issues on this problem?

Did ya'll get that they were isoceles trapezoids?

Let $X$, $Y$, be the feet from $D$ to $\overline{BA}$, $\overline{BC}$, and let $Z = \overline{BD} \cap \overline{AC}$. By Simson theorem, the points $X$, $Y$, $Z$ are collinear. Consequently, the point $P$ is the reflection of $D$ over $Z$, and so we conclude $P$ is the orthocenter of $\triangle ABC$. [asy][asy] size(10cm); pair B = dir(100); pair A = dir(210); pair C = dir(330); pair D = -A*C/B; pair X = foot(D, B, A); pair Y = foot(D, B, C); pair Z = extension(A, C, B, D); pair E = 2*X-D; pair F = 2*Y-D; pair P = A+B+C; pair Q = -A*B/C; pair R = -B*C/A; draw(unitcircle, red); draw(A--B--C--D--cycle, red); draw(A--C, red); draw(B--P, red); draw(A--X, red); draw(X--Y, orange); draw(E--F, orange); draw(F--D--E, orange); draw(circumcircle(P, F, D), orange); draw(circumcircle(P, E, D), orange); draw(A--R, red+dotted); draw(C--Q, red+dotted); draw(R--F, lightcyan); draw(P--D, lightcyan); draw(Q--E, lightcyan); dot("$B$", B, dir(B)); dot("$A$", A, dir(180)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(80)); dot("$Z$", Z, dir(315)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$P$", P, dir(135)); dot("$Q$", Q, dir(110)); dot("$R$", R, dir(80)); /* TSQ Source: !size(10cm); B = dir 100 A = dir 210 R180 C = dir 330 D = -A*C/B X = foot D B A Y = foot D B C R80 Z = extension A C B D R315 E = 2*X-D F = 2*Y-D P = A+B+C R135 Q = -A*B/C R110 R = -B*C/A R80 unitcircle 0.1 lightred / red A--B--C--D--cycle red A--C red B--P red A--X red X--Y orange E--F orange F--D--E orange circumcircle P F D 0.1 yellow / orange circumcircle P E D 0.1 yellow / orange A--R red dotted C--Q red dotted R--F lightcyan P--D lightcyan Q--E lightcyan */ [/asy][/asy] Suppose now we extend ray $CP$ to meet $\omega$ again at $Q'$. Then $\overline{BA}$ is the perpendicular bisector of both $\overline{PQ'}$ and $\overline{DE}$; consequently, $PQ'ED$ is an isosceles trapezoid. In particular, it is cyclic, and so $Q' = Q$. In the same way $R$ is the second intersection of ray $\overline{AP}$ with $\omega$. Now, because of the two isosceles trapezoids we have found, we conclude \[ EQ = PD = FR \]as desired.

I complex bashed. The entire problem. Apparently complex bash is not the inteded solution. Whoops!

Wait were there configuration issues for this one?

cosinechicken wrote: Wait were there configuration issues for this one? i dont think so

Th3Numb3rThr33 wrote: Let $ABCD$ be a quadrilateral inscribed in circle $\omega$ with $\overline{AC} \perp \overline{BD}$. Let $E$ and $F$ be the reflections of $D$ over lines $BA$ and $BC$, respectively, and let $P$ be the intersection of lines $BD$ and $EF$. Suppose that the circumcircle of $\triangle EPD$ meets $\omega$ at $D$ and $Q$, and the circumcircle of $\triangle FPD$ meets $\omega$ at $D$ and $R$. Show that $EQ = FR$. Note that $\overline{QR}$ is the image of the simson's line of point $D$ in $\triangle ABC$ under dilation by factor $2$. Thus, $P$ is the orthocenter of $\triangle ABC$. Consequently, $\odot(EPD), \odot(FPD)$ meet $\omega$ again at the reflections of $P$ in $\overline{BA}, \overline{BC}$ respectively. Thus, $EQ=FR=PD$ and we're done. $\blacksquare$

Once we get that $A$ and $C$ are the centers (some straightforward angle chasing), we complex bash

5 pages of regular coordinates should do the trick

AforApple wrote: 5 pages of regular coordinates should do the trick You are my inspiration.

How many points do you get if you got that A, and C were the circumcenters, and that EQB, and FRB were collinear and solved it using that without any real proof, just based on the diagram???

Does this work?

Wait I let Q' and R' be such that AQ'=AD and CR'=CD, found they were the intersections of BE and BF with omega, respectively, then proved AQ'PD was cyclic (and so was CR'PD similarly) so Q'=Q and R'=R, then proved BQR was isosceles (which finished b/c BEF is isosceles, BE=BD=BF). It was way longer than these Simson line proofs...

Are there any configuration issues if you used spiral symmetry?

Sorry for bump but time to ignore boards and start oly after 1 whole year again lol. Lets start with a few easy ones first $T=DB\cap AC$. Consider homothety $\dfrac{1}{2}$ at $D$. As $DE\perp AB$ and $DF\perp BC$, by Simson's Theorem, $\overline{\text{foot}(D,AB)-\text{foot}(D,BC)-\text{foot}(D,AC)}\implies\overline{DE\cap AB-DF\cap BC-T}\implies P\mapsto T$. So mapping back we get $D$ as reflection of $D$ over $T$. So, $AD=AP$ and initially $AD=AE\implies A$ center of $\odot(EPD)\implies AP=AQ$. Also, $\measuredangle(AB,CP)=\measuredangle(AB,BP)+\measuredangle BPC=\measuredangle ABP+\measuredangle CDP=\measuredangle ABD+\measuredangle CDT=\measuredangle ACD+\measuredangle CDT=\measuredangle TCD+\measuredangle CDT=\measuredangle CTD=90^{\circ}$. So, $AB\perp PC$ and $AB\perp DE\implies DE\parallel PQ$ which along with $DEPQ$ cyclic $\implies DEPQ$ isoceles trapezium $\implies EQ=DP$. Similarly $FR=DP$ and ggwp.

I think this is the only solution that uses a series of synthetic claims followed by a (short) complex bash. At least I finally solved a jmo p3

Edit: whoops did not find the isosceles trapezoid :skull:

First we get the lemma that $A$ is circumcentre of $\triangle DEP$ and $C$ the circumcentre pf $\triangle DFP$. This follows from $AC$ bisecting $DP$, which is a consequence of $M,N,O$ collinear. This can then be proved by Complex Numbers Coordinates, setting the circumcircle of $ABCD$ as the unit circle. Continuing the bash and removing most unnecessary points, we can get the conclusion eventually. Full proof here: https://infinityintegral.substack.com/p/usajmo-2018-contest-review

Claim. $A$ is the center of $EPQD$, and $C$ is the center of $FPRD$. Proof. Simson line! Set $S = \overline{AC} \cap \overline{BD}$. Note that $\overline{EPF}$ is the image of the Simson line of $D$ with respect to $ABC$ upon a homothety of ratio $2$ at $D$. Thus $DS = SP$, and $A$ lies on the perpendicular bisector of $\overline{PD}$. It follows $A$ is the circumcenter. $\blacksquare$ Now the key is that $PQDE$ and $FPRD$ are isosceles trapezoids. Claim. $PQC$ and $PAR$ are collinear. Proof. Straight angle chasing: $$\angle PCA = 90^\circ - \frac 12 \angle QAD = \angle QCA$$because $AD=AQ$. $\blacksquare$ To finish, $\overline{AB} \perp \overline{CP}$ because $\angle BAC + \angle ACD = 90^\circ$. So then $PQDE$ and $FPRD$ are isosceles trapezoids, and $EQ=PD=FR$.

Note that $P$ is the orthocenter of $\triangle ABC$ through Simson Line properties. Reflect $P$ over $AB$ to $Q'$. Then we know $Q'$ lies on $(ABC)$. $PQ'EC$ is an isosceles trapezoid, as $PQ \parallel DE$ and the quadrilateral is symmetric about $AB$, hence it is cyclic. As a result, we know $Q'=(ABC) \cap (EPD)=Q$, and similarly $PRFD$ is also an isosceles trapezoid. Consequently, \[EQ=PD=FR. \quad \blacksquare\]

How is this a j3???? Let $X=AC\cap BD$ and $E’$, $F’$ be the feet from $D$ to $AB$, $BC$. Simson lines give $E’, X, F’$ are collinear. The key claim is that $EBQ$ and $FBR$ are collinear. This is true as \[\measuredangle DQE= \measuredangle DPE = \measuredangle DXE’ = \measuredangle DAE’ = \measuredangle DQB\]Now, the finish is easy as $BE=BD=BF$ and $\angle QDB=\angle BEP=\angle BFP=\angle RDB$ so $BR=BQ$

Claim: $A$ and $C$ are the centers of $(DPE)$ and $(DPF)$, respectively Proof: Let $X=BD\cap AC$, $E'=ED\cap AB$, $F'=FD\cap BC$. Note that $E'XF'$ is the Simson line of $\triangle ABC$ wtr. $D$. Taking a homothety of scale factor $2$, $E'XF'$ maps to $EPF$, meaning that $DX=PX$. Therefore, $AC$ is the perp. bisector of $PD$ $\square$ Claim: $A,R,P$ and $C,Q,P$ are collinear Proof: As $\triangle CDR$ is isosceles, we can say: \[\angle CRD=\angle CDR,\]\[\angle RAC=\angle CAD,\]so $AC$ is the angle bisector of $\angle RAD$. However, note that $\triangle APD$ is isosceles with $PD\perp AC$. Therefore, $AC$ is the angle bisector of $\angle PAD$, implying that $A,R,P$ collinear. We can use a similar method to prove that $C,Q,P$ are collinear $\square$ Claim: $PR\parallel DF$, $PQ\parallel DE$ Proof: We can angle-chase, getting: \[\angle FDB=90-\angle DBC=90-\angle CAR=90-\angle CAP=\angle RPB,\]implying $PR\parallel DF$. We can use a similar method to prove that $PQ\parallel DE$ $\square$ As a result, we can conclude that $DEPR$ and $DRPF$ are isosceles trapezoids, so: \[EQ=PD=FR,\]as desired $\blacksquare$

Claim: $P$ is the orthocenter. Let the foot of the altitude from $D$ to $AB, AC, BC$ be $X, Y, Z$, respectively. Then, because $D \in (ABC)$, we know that $X, Y, Z$ are collinear by Simson lines. Then, since $X, Z$ are the midpoints of $DE, DF$, we know that $Y$ is the midpoint of $DP$. Since $Y$ lies on $AC$ and $BY \perp AC$, we can see that $D$ must be the reflection of the orthocenter over $AC$. However, the reflection of $D$ over $AC$ is $P$, so $P$ must be the orthocenter. Claim: $A, C$ are the centers of $(DPE), (DPF)$, respectively. Notice that $A$ lies on the perpendicular bisectors of $DE, DP$. Similarly, $C$ lies on the perpendicular bisectors of $DP, DF$. The claim follows. Now, let $Q' = BE \cap (ABC)$. Since $B$ lies on the perpendicular bisector of $DE$, we know that $BE$ is the reflection of $BD$ over $BA$. However, this means that the reflection of the orthocenter $P$ over $BA$ lies on $BE$. But since said reflection lies on $(ABC)$, we must have $Q'$ be the reflection of $P$ over $BA$. Then, this gives us that $EQ'PD$ is an isosceles trapezoid, so $Q' \in (EPD)$, so $Q' = Q$. Similarly, we can show that $FRPD$ is also an isosceles trapezoid, so we have $EQ = PD = FR$, done.

let $AB$ intersect $ED$ at $S$, let $BC$ intersect $DF$ at $T$, let $AC$ intersect $BD$ at $U$ simson lines yield that $SUT$ are collinear, and that $DU=DP$ and $AD=AP$ then, spiral similarity yields $RBF$ are collinear, and since it is essentially given that $DB=EB=FB$, $RB=BQ=BP$ however, by power of a point on $DBP$, we can see that $EBQ$ also need to be collinear, so $EQ=FR=DB+BP$

30 minute max coord bash, but also very interesting concurrencies here... $EF$ is homothety of factor $2$ at $D$ of simson line, by reflecting the orthocenter $P$ is $H$ of $ABC$, then we can show by phantom point that $Q$ and $R$ are the other corresponding reflections of the orthocenter so we are done by isosceles trapezoid sillyness

Let $X$, $Y$, $Z$ be the feet of the altitudes of $D$ to $AB$, $AC$, and $BC$. Then $X - Y - Z$ by Simson Line, and taking a homothety of scale factor $2$ at $D$ gives that $P$ is the reflection of $D$ over $AC \cap BD$. Since $E$ is the reflection of $D$ over $AB$ and $P$ the reflection of $D$ over $AC$, we get that $A$ is the circumcenter of $\triangle EPD$ and $C$ the circumcenter of $\triangle FPD$. We now angle chase to show $Q - P - C$. Note that since $AQ = AD$ we have $\angle QCA = \angle DCA = \angle PCA$ by reflection, so $Q - P - C$ as desired. Analogously, $A - P - R$. Now let $P'$ be the orthocenter of $\triangle ABC$. It is well known that the Simson line of $D$ wrt $\triangle ABC$ bisects $DP'$, and since $P' - P - B$ we have $P' = P$. So then $PQ \perp AB \perp DE$ so $PQED$ is an isosceles trapezoid, and similarly for $DPRF$. Then $EQ = DP = FR$ as desired.

Fun problem. Notice by simson lines and homothety that $DP$ is twice as long as the distance from $D$ to $AC.$ Since $A$ is on the perpendicular bisector of $ED$ and $PD,$ A is the center of $EDP.$ By similar steps $C$ is the center of $FDP.$ Then notice that since $AP=AQ$ then $AB \perp PQ.$ Then notice that $EQ$ is the reflection of $DP $ across $AB.$ By similar steps $FR$ is the reflection of $DP $ over $BC.$ Therefore, $EQ=FR.$

Alternate solution I guess: PEKKA wrote: Notice by simson lines and homothety that $DP$ is twice as long as the distance from $D$ to $AC.$ Since $A$ is on the perpendicular bisector of $ED$ and $PD,$ A is the center of $EDP.$ By similar steps $C$ is the center of $FDP.$ Now by reflection notice that $DB=DE.$ Then by PoP, $$BP \cdot BD= BQ \cdot CE$$$$\Leftrightarrow BP=BQ$$$$\Leftrightarrow DP=EQ.$$Similar steps on the symmetric case finishes as we get $FR=DP.$

Really misplaced j3, but the config has a ton of super cool facts. Let $AB \cap ED = H, BC \cap DF = J, AC \cap BD = G$. We instantly suspect $P$ is the reflection of $D$ over $G$, which obviously reduces to $DG = GP$. But $H, G, J$ collinear by Simson on $D$ wrt $\omega$ so a homothety at $D$ with factor $2$ finishes. From this it is clear that $A$ and $C$ are the centers of their respective circles from two perpendicular bisectors meting. Then, by the orthocenter reflection lemma we know that $P$ is the orthocenter of $\triangle ABC$. Claim: $QP \perp BH$. Proof: This reduces down into $CPQ$ collinear by the orthocenter condition. But $\angle QCA = \angle ACD = \angle PCA$ using that $AD = AQ$ and $DG = GP$. Now $QP$ and $ED$ are parallel (both perpendicular to $BH$) so $EQPD$ is a cyclic trapezoid i.e isoceles. So $EQ = PD$ and by similar reasoning on $PRFD$ we have $RF = PD$ which finishes.

Define $R$ and $Q$ as the second intersections of $FB$ and $EB$ with $\omega$ and we prove they lie on $(DPF)$ and $(DPE)$, respectively. Note that $BF=BE=BD$ so $B$ is the center of $(DFE).$ \[\angle(FR,FP)=\angle (FB,FP) = \angle (FD,FP)-\angle(FD,FB) = \angle (FD,FE) + \angle(BC,BD)-90°\]\[=\angle(BD,BA) + \angle(BC,BD)-90°=\angle(BD,BA)+\angle(BF,BC)-90°\]\[\angle(BD,BA) + \angle(BR,BC)-90°=\angle(BD,BC) +\angle(BC,BA) + \angle(BR,BC)-90°\]\[=\angle(BR,BA)+\angle(BD,BC)-90°=\angle(BR,BA)+\angle(BD,BC)+\angle(AC,BD)\]\[=\angle(AC,BC) + (BR,BA)= \overset{\frown}{BA} + \overset{\frown}{AR}=\overset{\frown}{BR} = \angle(DR,DP)\] Thus, $R\in (PDF)$ and the same goes for $Q.$ Now, by power of a point, we have \[BF\times BR = BD\times BP = BE\times BQ\]Since $BE=BF,$ we conclude that $BQ=BR.$ Now, we just sum up the previous equations and we're done. [asy][asy] import graph; size(10cm); 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 = -109.01891805577127, xmax = 136.5726710795202, ymin = -58.9565859556744, ymax = 69.57083692983869; /* image dimensions */ pen zzwwff = rgb(0.6,0.4,1); pen qqqqcc = rgb(0,0,0.8); pen qqzzcc = rgb(0,0.6,0.8); pen qqzzff = rgb(0,0.6,1); draw(circle((-9.158790074470586,26.83754466229451), 27.006768294922715), linewidth(0.8) + zzwwff); draw(circle((-29.460645813118013,9.027429915252844), 51.18937554048644), linewidth(0.8) + qqzzcc); /* draw figures */ draw((-29.460645813118017,9.027429915252842)--(7.486741638479252,5.570391403599448), linewidth(0.8) + qqqqcc); draw((-29.460645813118017,9.027429915252842)--(-8.76746779324757,-0.16638839583354423), linewidth(0.8) + qqqqcc); draw((7.486741638479252,5.570391403599448)--(-8.76746779324757,-0.16638839583354423), linewidth(0.8) + qqqqcc); draw((-3.7647883832019486,53.30016508224222)--(-8.76746779324757,-0.16638839583354423), linewidth(0.8) + qqqqcc); draw(circle((7.486741638479251,5.57039140359945), 49.03802833764072), linewidth(0.8) + qqzzff); draw((-45.09161737845497,-39.717055887231304)--(17.780877594372598,28.740132276026916), linewidth(0.8) + qqqqcc); draw((28.687470053733957,-38.64790048907621)--(-36.14227595223765,27.95871530942519), linewidth(0.8) + qqqqcc); draw((-45.09161737845497,-39.717055887231304)--(28.687470053733957,-38.64790048907621), linewidth(0.8) + qqqqcc); draw((-12.423791818945164,-39.24365640157027)--(-8.76746779324757,-0.16638839583354423), linewidth(0.8) + qqqqcc); /* dots and labels */ dot((-29.460645813118017,9.027429915252842),linewidth(4pt) + dotstyle); label("$A$", (-35.376178456504235,7.043982553102595), NE * labelscalefactor); dot((-8.76746779324757,-0.16638839583354423),linewidth(4pt) + dotstyle); label("$B$", (-18.52866491610588,-4.071902669428267), NE * labelscalefactor); dot((7.486741638479252,5.570391403599448),linewidth(4pt) + dotstyle); label("$C$", (10.997905206241752,-0.7718742439894177), NE * labelscalefactor); dot((-3.7647883832019486,53.30016508224222),linewidth(4pt) + dotstyle); label("$D$", (-5.154865507748422,57.586523174297604), NE * labelscalefactor); dot((-45.09161737845497,-39.717055887231304),linewidth(4pt) + dotstyle); label("$E$", (-49.79209210447396,-43.672243774694465), NE * labelscalefactor); dot((28.687470053733957,-38.64790048907621),linewidth(4pt) + dotstyle); label("$F$", (28.887532986252374,-43.672243774694465), NE * labelscalefactor); dot((-12.423791818945164,-39.24365640157027),linewidth(4pt) + dotstyle); label("$P$", (-14.881265077462935,-45.23541513411286), NE * labelscalefactor); dot((17.780877594372598,28.740132276026916),linewidth(4pt) + dotstyle); label("$Q$", (20.89799048255831,30.83892435758272), NE * labelscalefactor); dot((-36.14227595223765,27.95871530942519),linewidth(4pt) + dotstyle); label("$R$", (-41.28149248097376,29.62312441136841), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]