InCtrl wrote: the wording is not exact, I will update it once the website uploads the problems (the questions are collected alongside the solutions)

I believe the problem stated perpendicular to DQ.

So from $\measuredangle PBQ=\measuredangle QCD=\measuredangle PAD=\measuredangle BAD-60^{\circ}$ and $BP=PA=CD$ and $BQ$ $=$ $QC$ $=$ $AD$ we get $\triangle QBP$ $\cong$ $\triangle DAP$ $\cong$ $\triangle QCD$ $\Longrightarrow$ $PQ=QD=DP$. Let $R$ the intersection of of the line through $P$ perpendicular to $PD$ and the line through $Q$ perpendicular to $QD$ and let $P'$ be a point in $PR$ such that $AP'\perp PR$ and $A'$ be a point such that $P'AA'P$ is a rectangle similarly defined $Q'$ and $C'$, so from $\triangle DAP$ $\cong$ $\triangle QCD$ we get $AA'=DD'$ $\Longrightarrow$ $PP'$ $=$ $AA'$ $=$ $DD'$ $=$ $QQ'$, furthermore since $PR\perp P'A$ and $QR\perp Q'C$ we get: $$PA^2-AR^2 =PP'^2-RP'^2...(1)$$$$QC^2-CR^2=QQ'^2-RQ'^2...(2)$$Then by $(1)$ and $(2)$ we get $PA^2-AR^2 =QC^2-CR^2$ $\Longrightarrow$ $PA^2+CR^2=QC^2+AR^2$, so from $PA=AB$ and $QC=CB$ we get $AB^2+CR^2=BC^2+AR^2$, hence $AR\perp BC$.

Here is a cleaner solution with complex numbers. Set $B$ at the origin, and denote by $x$ the complex number corresponding to the point $X$. Suppose that triangle $BAC$ is in that order counterclockwise, and let $\omega=e^{\frac{i\pi}{3}}$. Then, $d=a+c$, $p=a\omega$, and $q=-c\omega^2$. As $\frac{q-d}{p-d}=\frac{-c\omega^2-c-a}{a\omega-a-c}=\frac{-c\omega-a}{a\omega^2-c}=\omega$, $DPQ$ is equilateral. Now, note that the circumcenter $O$ of $DPQ$ (which is also the centroid) is $o=\frac{a+c+a\omega-c\omega^2}{3}$. Let $a'=a-o=\frac{2a-c-a\omega-c\omega^2}{3}$ and $c'=c-o=\frac{2c-a-a\omega+c\omega^2}{3}$. As $3c'\omega^2=2c\omega^2-a\omega^2-a\omega^3+c\omega^4=-2c+2c\omega+a-a\omega+a-c\omega=2a-c-a\omega+c\omega^2=3a'$, $OA=OC$. Thus, the reflection of $D$ over $O$ lies on the $B$ altitude, as desired.

Note that $\triangle{PAD}\cong \triangle{PBQ} \cong \triangle{DCQ}$ since $PA=PB=DC=AB$, $AD=BQ=CQ=BC$, and $\angle{PAD}=\angle{PBQ}=\angle{DCQ}=120^{\circ}-\angle{B}$. So $\triangle{PQD}$ is an equilateral triangle so $\angle{PDQ}=60^{\circ}$. Let E be the point on the line parallel to $CA$ through $D$ such that $BE\perp ED$. Let $A'$ and $C'$ be $BA\cap DE$ and $BC\cap DE$ respectively. Then $BPEA'$ and $BQEC'$ are cyclic since $\angle{BEA'}=90^{\circ}=\angle{BPA'}$ and similarly for $Q$ and $C'$. So $$\angle{PEQ}=\angle{PEB}+\angle{BEQ}$$$$=\angle{PA'B}+\angle{QC'B}$$$$=30^{\circ}+30^{\circ}$$$$=\angle{PDQ}.$$So $PQDE$ is cyclic. Then the antipode of $D$ with respect to $(PQD)$ lies on $BE$, so we are done.

So salty, thought about using complex on this but then decided to waste 1.5hrs on Q1 by trying Cauhy and Muirhead

This is a very nice problem! Surely my solution is not the most efficient though...

If you have more than 60 minutes to spare, this problem is also easily-coordinate-bashable.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20V.pdf p. 57... Sincerely Jean-Louis

Let $X$ and $Y$ be the feet of perpendiculars from $E$ to $CD$, $AD$, respectively. Let $E$ be the intersections of the line through $P$ perpendicular to $PD$ and the line through $Q$ perpendicular to $DQ$. Thus, in fact, $EPXDYQ$ is cyclic. Trivially, using congruent triangles $\triangle PCB\cong\triangle AQB$ and $\triangle PAD\cong\triangle QCD$ and we can angle chase and get that $\angle QAD=\angle PCQ\implies \triangle AQD\cong\triangle CPQ$, which would furthermore imply that $\triangle PQD$ is equilateral. Thus, by spiral similarity, $X$ lies on $PB$ and $Y$ lies on $BQ$. Easy to verify that $\triangle CBX\sim\triangle AYB$. We want to show that, $AB^2+CE^2=BC^2+AE^2\Longleftrightarrow AB^2-BC^2=AE^2-CE^2$ Law of Cosines on $\triangle DEA$ and $\triangle DEC$, $$AE^2=AD^2+DE^2-2\cdot AD\cdot DE\cdot \cos{\angle CDE}$$and $$CE^2=DC^2+DE^2-2\cdot CD\cdot DE\cdot \cos{\angle ADE}$$Thus, $$AE^2-CE^2=BC^2-AB^2-2\cdot (AD\cdot DY-CD\cdot DX),$$hence$$BC^2-AB^2=AD^2-AY\cdot AD-CD^2+CD\cdot CX.$$Since $\triangle CBX\sim\triangle AYB$, we have$$\frac{AY}{CX}=\frac{AB}{BC},$$thus indeed $AC\perp BE$. $\square$

My step-by-step solution in great detail, very beginner-friendly. Enjoy!

Jesus, this problem is beautiful.

Not really clean but anyway... Let $AB = x$, $BC = y$. Let perpendicular from $P$ to $PD$ and perpendicular from $Q$ to $QD$ meet at $S$. Let $BP$ and $BQ$ meet $CD$ and $AD$ at $K$ and $T$. Claim1 : $DPQ$ is equilateral. Proof : with just simple angle chasing and Noting $DA = QB = QC$ and $DC = PB = PA$ we have $DPA$, $PQB$ and $QDC$ are congruent. Claim2 : $DTQSPK$ is cyclic. Proof : $\angle PKC = \angle 60 = \angle PSD$ so $PSDK$ is cyclic. $\angle DPS = \angle DQS = \angle 90$ so $DPSQ$ is cyclic. $\angle QTA = \angle 60 = \angle DSQ$ so $DSQT$ is cyclic. Note that for proving $AC\perp BS$ we will prove $SC^2 + BA^2 = SA^2 + BC^2$. $SC^2 + BA^2 = SD^2 + 2x^2 - 2SD.x.\cos{CDS} = SD^2 + 2x^2 - 2SD.x.\frac{DK}{DS} = SD^2 + 2x^2 - 2x.DK = SD^2 + 2x(x-DK) = SD^2 + 2x(CK)$ $SA^2 + BC^2 = SD^2 + 2y^2 - 2SD.y.\cos{ADS} = SD^2 + 2y^2 - 2SD.y.\frac{DT}{DS} = SD^2 + 2y^2 - 2y.DT = SD^2 + 2y(y-DT) = SD^2 + 2y(AT)$ so we need to prove $x.CK = y.AT$ or points $C$ and $A$ have same power w.r.t circle $DPQ$. Let $O$ be center of $DPQ$, we need to prove $AO = CO$ witch is true because $AOD$ and $COQ$ are congruent. we're Done.

We complex bash to prove that $DPQ$ is equilateral. Let $c=-a$ and $d=-b$. Then, if $\omega=cis\left(\frac{\pi}{3}\right)$ $$p=(a-b)\omega+b,$$and $$q=(b-c)\omega+c=(a+b)\omega-a.$$From this, we conclude that: $$(q-p)\omega+p=(a\omega+b\omega-a-(a\omega-b\omega+b))\omega+(a-b)\omega+b=$$$$=2b\omega^2-2b\omega+b=-b=d,$$and thus $DPQ$ is equilateral. Now, since $\angle ABQ=60^{\circ}-\angle QBP=\angle PBC$, $AB=PB$ and $QB=CB$ we get $\triangle QBA\equiv\triangle CBP$, and therefore $QA=CP$. From this, we get that $\triangle DAQ\equiv\triangle QCP$. If $O$ is the circumcenter of $DPQ$, there's a rotation centered at $O$ that takes $\triangle QBA$ to $\triangle CBP$, and therefore $AO=CO$. Let $E$ be the point of concurrency of the perpendicular lines and $M$ the midpoint of $AC$. By taking the homothety centered at $D$ that takes $M\to B$, we see that $O\to E$, and we get the desired result. $\blacksquare$

straightforward complex, but I will do it synthetically First, restate the problem as follows (slightly loosening the "inside" condition): Restated problem wrote: In $\triangle ABC$, equilateral triangles $ABP$ and $CAQ$ are constructed, where vertices are labeled in counterclockwise order, with $\overline{AP}$ and $\overline{AQ}$ lying inside $\angle BAC$. Let $A'$ be the point such that $ABA'C$ is a parallelogram. Let the perpendicular to $\overline{A'P}$ at $P$ and the perpendicular to $\overline{A'Q}$ at $Q$ concur at $X$. Then $\overline{AX} \perp \overline{BC}$. WLOG let $AB<AC$. Let $\overline{AQ} \cap \overline{A'B}=E$ and $\overline{AP} \cap \overline{A'C}=F$. Then $\angle AEB=\angle CAE=60^\circ$ and likewise $\angle AFC=60^\circ$, so $ABPE$ and $ACQF$ are cyclic. Then $\angle QEA'=60^\circ$, and $\angle QFA'=180^\circ-\angle QFC=\angle QAC=60^\circ$, hence $A'QEF$ is cyclic, Similarly (though with $60^\circ$ replaced by $120^\circ$, due to configuration things) $A'PEF$ is cyclic, hence $\angle QPA'=\angle PQA'=60^\circ$, so $\triangle A'PQ$ is equilateral. Now $X$ is simply the antipode of $A'$ in $(A'PQEF)$, so it suffices to prove that $(A'PQEF)$ is symmetric with respect to the perpendicular bisector of $\overline{BC}$, i.e. the powers of $B$ and $C$ are equal. This is equivalent to $$BE\cdot BA'=CF\cdot CA' \iff BE\cdot AC=CF\cdot AB \iff \frac{BE}{BA}=\frac{CF}{CA}.$$This since $\angle AEB=AFC=60^\circ$ and $\angle ABE=\angle ACF=\angle B+\angle C$, hence $\triangle ABE \sim \triangle ACF$. $\blacksquare$

FabrizioFelen wrote: So from $\measuredangle PBQ=\measuredangle QCD=\measuredangle PAD=\measuredangle BAD-60^{\circ}$ and $BP=PA=CD$ and $BQ$ $=$ $QC$ $=$ $AD$ we get $\triangle QBP$ $\cong$ $\triangle DAP$ $\cong$ $\triangle QCD$ $\Longrightarrow$ $PQ=QD=DP$. Let $R$ the intersection of of the line through $P$ perpendicular to $PD$ and the line through $Q$ perpendicular to $QD$ and let $P'$ be a point in $PR$ such that $AP'\perp PR$ and $A'$ be a point such that $P'AA'P$ is a rectangle similarly defined $Q'$ and $C'$, so from $\triangle DAP$ $\cong$ $\triangle QCD$ we get $AA'=DD'$ $\Longrightarrow$ $PP'$ $=$ $AA'$ $=$ $DD'$ $=$ $QQ'$, furthermore since $PR\perp P'A$ and $QR\perp Q'C$ we get: $$PA^2-AR^2 =PP'^2-RP'^2...(1)$$$$QC^2-CR^2=QQ'^2-RQ'^2...(2)$$Then by $(1)$ and $(2)$ we get $PA^2-AR^2 =QC^2-CR^2$ $\Longrightarrow$ $PA^2+CR^2=QC^2+AR^2$, so from $PA=AB$ and $QC=CB$ we get $AB^2+CR^2=BC^2+AR^2$, hence $AR\perp BC$. Why AB²+CR²=BC²+AR² hence AR perpendicular BC? I still didnt get R on altitude from B of triangle ABC. Would you explain, please?

Drop altitudes from A, R to BC. You will see by the pythagorean theorem that the only way for the equality to be true is if AR share the same perpendicular line to BC