a_507_bc wrote: Let $ABCD$ be a parallelogram and let $AX$ and $AY$ be the altitudes from $A$ to $CB, CD$, respectively. A line $\ell \perp XY$ bisects $AX$ and meets $AB, BC$ at $K, L$. Similarly, a line $d \perp XY$ bisects $AY$ and meets $DA, DC$ at $P, Q$. Show that the circumcircles of $\triangle BKL$ and $\triangle DPQ$ are tangent to each other. Let $M$ and $N$ be midpoints of $AX$, $BY$ and $T=(ABX) \cap (ADY)$. Note that $\angle ATB=90^\circ=\angle ATD$, so $T \in BD$. Making homotety at $A$ with coefficient $2$ we can easily obtain that $AMKTPN$ is cyclic. By this reason $T$ is Miquel point in both quadrilaterals $BXMK$ and $DYNP$, so $T \in (BKL), (DPQ)$. Now for proving tangent it's enough to check that $\angle KTP=\angle KBT+\angle PDT$, which is obviously, since both sides are equal to $180^\circ -\angle BAD$. $\quad \square$

very nice problem Let O be the instrection point of diagonals AC and BD , Let M and N be the midpoints of segments AX and AY respectively , first we prove that points A.P.K.M.N.O are concylic , AMN=AYX=ACX=AON so AMNO is cyclic , and by simple angle chasing you can see that DPQ=ACY=AXY=ANM so APMN is cyclic and doing the same work for the other side you can see that AKMN is cyclic so we prove the first part let this points cirle be W , Let R be the second instrection of W and line BD , by doing some easy angle chasing you can see that PDRQ is cyclic and also BKLR is cyclic to so R is the tangency point and the proof is complete. Comment: you can see that R is the instrection point of BD ,LP,KQ because of congruent of 2 triangles DPQ , BKL

Beautiful Problem, despite being easy for its position Let diagonals $AC$ and $BD$ meet at $Z$, let $T$ be the foot of $A$ onto $BD$, and let $M$ and $N$ be the midpoints of $AX$ and $AY$. Notice that $Z$ is the center of $(ACXY)$ so $\angle AMZ=\angle ANZ=90^{\circ}$. Then $K$ must be the $N$ antipode in $(AMNZ)$ as it lies on $AB$ and $l$. Similarly we get $P$ is the $M$ antipode in $(AMNZ)$, and thus $AMNPKTZ$ is cyclic with diameter $AZ$. Now $(DPQT)$ is cyclic as $$\angle DQP=180^{\circ}-\angle DPQ-\angle ADC=\angle APN-\angle MPN=\angle APM=\angle PAZ=\angle DTP$$and also $(BKTL)$ is cyclic. A trivial angle chase shows that $T$ lies on segments $KQ$ and $PL$. Now the result follows from homothety as $$\frac{TK}{TQ}=\frac{TB}{TD}=\frac{TL}{TP}$$

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Denote the intersection of diagonals $AC$ , $BD$ as $Z$ and midpoint of $AX$ , $AY$ in order $N$ , $M$ It's easy to prove $LAKC$, $APQC$, $AKNZMP$ are cyclic. Since $K$,$L$,$Q$,$P$ are unique so denote $S$ as the intersection of $(AKNZMP)$ with $BD$ we prove that $K$,$S$,$Q$ and $P$,$S$,$L$ Are collinear. It is enough to prove $$\frac{BS}{BK}=\frac{DS}{DQ}$$which is clear since $$\frac{BS}{BK}=\frac{BA}{BZ}=\frac{DC}{DZ}=\frac{DS}{DQ}$$so $\angle AKS$= $\angle DPS$ = $\angle SLB$ so $BKSL$ Is cyclic. Similarly $PSQD$ is cyclic and since $\angle PDS$ + $\angle BLS$=$\angle PSB$ the result follows.

Solved with Om245. Pretty good problem and not too hard either. We denote by $T$ the intersection of the diagonals $AC$ and $BD$. Now, it is obvious that quadrilateral $AXCY$ is cyclic, and by considering the homothety centered at $A$ with scale factor $\frac{1}{2}$, it follows that $AMTN$ is also cyclic. We can in fact take this one step further. Claim : Points $P$ and $K$ lie on circle $(AMN)$. Proof : Simply note that, \[\measuredangle APN = \measuredangle DPQ = \measuredangle PDQ +\measuredangle DQP = \measuredangle ADC + 90 + \measuredangle QYX = \measuredangle ADC + 90 + \measuredangle CAX \]\[=\measuredangle ADC + \measuredangle ACX = \measuredangle CBA + \measuredangle ACB = \measuredangle CAB = \measuredangle ACY = \measuredangle AXY = \measuredangle AMN\]which implies that point $P$ indeed lies on circle $(AMN)$. A similar proof shows that $K$ also lies on $(AMN)$ as desired. Now, let $R = \overline{BD} \cap (AMN)$ . Our claim is that this is the desired tangency point. We first show that this point actually lies on both the desired circles. Note that, \[\measuredangle PRD = \measuredangle PRT = \measuredangle PAT = \measuredangle DAC = \measuredangle XCA = \measuredangle XYA = \measuredangle PQD\]which implies that $R$ indeed lies on circle $(QPD)$. Similarly we can show that point $R$ also lies on circle $(BKL)$. Now, we can also note the following. Claim : Points $P$ , $R$ , $L$ and $Q$ , $R$ , $K$ are collinear. Proof : This is easy to see since, \[\measuredangle DRQ = \measuredangle DPQ = \measuredangle APN = \measuredangle AMN = \measuredangle AXY = \measuredangle XLM = \measuredangle BLK = \measuredangle BRK\]so points $Q$ , $R$ and $K$ are collinear. A similar proof shows that points $P$ , $R$ and $L$ are also collinear. Now, we finish off by observing that $CQRL$ is cyclic since, \[\measuredangle RQD = \measuredangle RPA = \measuredangle RKB = \measuredangle RLC\]and thus, \[\measuredangle QDR + \measuredangle RBL = \measuredangle DCB = \measuredangle QCL = \measuredangle QRL \]which is sufficient (this is actually easy to prove) to implying that circles $(DPQ)$ and $(BKL)$ are also in fact tangent at $R$ which is what we wished to show.

Nice result $F,P,A,K,E$ are cyclic. $\angle PAE = \angle PFE$ and $\angle KAF = \angle KEF$. Reflections of $Y$ and $X$ across $FQ$ and $EL$ coincide on $XY$, call it $G$. Call the intersection of $(AEF), (EXG), (GYF)$ as $H$. $\angle HDC + \angle HBC = \angle HAY + \angle HAX = 180 - \angle BCD \implies H \in BD$. $\angle PQH + \angle KLH = \angle ADH + \angle ABH = 180 - \angle BAD = \angle PHK$ as desired.

Good problem!

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CamScanner 08-10-2024 19.39_compressed.pdf (183kb)