Let $ABCD$ be a cyclic quadrilateral with $AB<CD$, and let $P$ be the point of intersection of the lines $AD$ and $BC$.The circumcircle of the triangle $PCD$ intersects the line $AB$ at the points $Q$ and $R$. Let $S$ and $T$ be the points where the tangents from $P$ to the circumcircle of $ABCD$ touch that circle. (a) Prove that $PQ=PR$. (b) Prove that $QRST$ is a cyclic quadrilateral.
Problem
Source: 2015 CentroAmerican Math Olympiad #3
Tags: OMCC, geometry, cyclic quadrilateral, circumcircle
Math_CYCR
28.06.2015 00:50
a) Proof: Trivial, Angle Chasing b) Proof: Easy to see: $PA*PD=PT^2=PS^2$ Since $\bigtriangleup$$PQA$~$\bigtriangleup$$PDQ$, $PQ^2=PA*PD$ Since $\bigtriangleup$$PRA$~$\bigtriangleup$$PDR$, $PR^2=PA*PD$ Now: $PA*PD=PQ^2=PR^2=PS^2=PT^2$ Then: $PQ=PR=PS=PT$ Now take the circumference with center $P$ and radius $PQ$. Since here is easy to see that $Q$ , $R$ , $S$ and $T$ are concyclic. $Q.E.D.$
djmathman
28.06.2015 00:52
Darn, sniped. Oh well; these solutions at least are non-isomorphic.
(a)Note that $AB$ is antiparallel to $CD$. Let $H_1$ and $H_2$ be the feet of $P$ to $\overline{AB}$ and $\overline{CD}$ respectively. Because of the antiparallel nature of those two segments, $PH_1$ and $PH_2$ are isogonal to each other. Now examine $\triangle PCD$; let $O$ be the circumcenter of this triangle. By the well-known fact that $AO$ and $AH$ are isogonal, we get that $PO$ is isogonal to $PH_2$. This implies that $P$, $H_1$, and $O$ are collinear. Combining this with $PO\perp QR$ means that $PO$ is the perpendicular bisector of $\overline{QR}$; in other words, $PQ=PR$.
(b)Let $X=AB\cap CD$ and $Z=AC\cap BD$. WLOG let $XB<XA$; the reverse case is handled similarly. By Brokard on $ABCD$, $\triangle PXZ$ is self-polar, meaning that $XZ$ is the polar of $P$ with respect to $(ABCD)$. However, since $S$ and $T$ are the intersections of the tangents from $P$ to this circle, $ST$ is also said polar. Hence $ST$, $AB$, and $CD$ are collinear at $X$. Now \[XT\cdot XS=XC\cdot XD=XQ\cdot XR,\] so $QRST$ is cyclic as desired. (This part can also be handled with a radical center argument, but I think the details are a little finicky. Also, it's important to note that if $AB\parallel CD$, then $X$ is the point at infinity, so $ST$ is also parallel to $AB$. This implies that $QRST$ is an isosceles trapezoid by symmetry and stuff.)
anantmudgal09
10.09.2015 23:44
a.) Note that if $O$ is the circumcentre of $\triangle PCD$ then $PO$ is perpendicular to $AB$ and so $PQ=PR$. b.) Let $X$ be the intersection point of $AB,CD$ Then by Brokard's theorem on self polarity of the diagonal triangle in a complete quadrilateral we get that $ST,AB,CD$ concur at $T$. By Power of a point, $XQ.XR=XD.XC=XS.XT$. so $T,S,Q,R$ are concyclic.(with circumcentre $P$.)
Packito
12.06.2016 00:11
Let $\omega$ be the circumcircle of $PCD$ Lets consider inversion ($\Psi$) with center $P$ and radius $PS$. By power of point we get that $PA*PD=PB*PC=PS^2$ So $\Psi(D)=A$ and $\Psi(C)=B$ Since $C$ and $D$ lie on $\omega \Rightarrow \Psi(\omega)=AB$ But since $Q$ and $R$ lie on $AB$ and on $\omega \Rightarrow \Psi(Q)=Q$ and $\Psi(R)=R$ So $Q$ and $R$ lie on the circle with center $P$ and radius $PS$ Then $PS=PT=PR=PQ$ $Q.E.D$
WolfusA
29.11.2018 20:12
anantmudgal09 wrote: b.) Let $X$ be the intersection point of $AB,CD$ Then by Brokard's theorem on self polarity of the diagonal triangle in a complete quadrilateral we get that $ST,AB,CD$ concur at $T$. What if $AB||CD$, and there's no intersection of these lines?
UI_MathZ_25
24.12.2022 08:55
Part a) Notice that $\angle PBR =\angle ABC = 180^{\circ} - \angle ADC = 180^{\circ} - \angle PDC = \angle PRC$, so by case $AA$, $\triangle PBR \sim \triangle PRC$, which implies that $\angle PRB = \angle PCR$, so $PR$ is tangent to $\odot(\triangle RBC)$. Analogously $PQ$ is tangent to $\odot(\triangle QAD)$. It turns out that $PQ^{2} = PA \cdot PD$ y $PR^{2} = PB \cdot PC $ In addition, by Pop at $P$ wrt $\odot(ABCD)$, $PA \cdot PD = PB \cdot PC \Rightarrow PQ^{2} = PR^{2} \Rightarrow PQ = PR$ $\blacksquare$ Part b) Again by PoP at $P$ wrt $\odot(ABCD)$, $PS^{2} = PT^{2} = PA \cdot PD = PQ^{2} = PR^{2} $ Therefore, $PQ = PR = PS = PT$, then $P$ is the center of $\odot(QRST)$ and the conclusion follows $\blacksquare$
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