Let $ABCD$ be a parallelogram. A line through $C$ crosses the side $AB$ at an interior point $X$, and the line $AD$ at $Y$. The tangents of the circle $AXY$ at $X$ and $Y$, respectively, cross at $T$. Prove that the circumcircles of triangles $ABD$ and $TXY$ intersect at two points, one lying on the line $AT$ and the other one lying on the line $CT$.
Problem
Source: RMM Extralist 2021 G1
Tags: RMM Shortlist, geometry, Angle Chasing
VicKmath7
19.09.2023 17:31
Nice angle-chasing exercise.
Part 1: $(TXY)$, $(ABD)$ and $CT$ concur.
Let $P=(XBC)\cap (YDC)$; we claim this is the concurrency point. Indeed:
1. $\angle PDA=\angle PCY=\angle PBA$, so $P \in (ABD)$.
2. $\angle YPX=\angle YPC-\angle XPC=\angle ADC-(180^{o}-\angle XBC)=180^{o}-2\angle BAD=\angle XTY$, so $P \in (TXY)$.
3. $\angle XPC=180^{o}-\angle ABC=\angle BAD=180^{o}-\angle TPX$, so $P \in CT$.
Part 2. $(TXY)$, $(ABD)$ and $AT$ concur.
Let $S=(BDC) \cap XY$ and $(XSB) \cap (DSY)=Q$; we claim $Q$ is the concurrency point. Indeed, this is true, as:
1. $\angle ADQ=\angle YSQ=\angle ABQ$, so $Q\in (ABD)$.
2. $\angle XQY=\angle YQS-\angle XQS=180^{o}-\angle ADS-\angle ABS=2\angle BAD=180^{o}-\angle XTY$, so $Q \in (XTY)$.
3. $\angle AQX=\angle AQB-\angle XQB=180^{o}-\angle ADB-\angle BDC=\angle BAD=\angle TQX$, so $Q \in AT$.
IAmTheHazard
19.09.2023 23:53
Worst "synthetic solution" ever. These tend to pop up when I do geo during school... We first show that one of the intersection points lies on $\overline{AT}$. Let $\overline{CD}$ intersect $(ABD)$ again at $E$, so $ABED$ is an isosceles trapezoid, and let $\overline{BC}$ intersect $(ABD)$ again at $F$, so $ADFB$ is an isosceles trapezoid. Let $\overline{EX}$ intersect $(ABD)$ again at $P$. I claim that $A,P,T$ are collinear. Since $\measuredangle BAD=\measuredangle XAY=\measuredangle TXY=\measuredangle XYT$, by Trig Ceva on $\triangle AXY$ it suffices to show that $$\frac{\sin\angle XAP}{\sin\angle YAP}=\frac{\sin \angle AXT}{\sin \angle AYT}=\frac{\sin \angle DYC}{\sin \angle DCY} \iff \frac{BP}{DP}=\frac{CD}{DY} \iff \frac{BP}{BA}=\frac{DP}{DY} \impliedby \triangle PBA \sim \triangle PDY.$$By Pascal on $BFPEDA$, $C$, $X$, and $\overline{AD} \cap \overline{FP}$ are collinear, hence $Y$ lies on $\overline{FP}$. Thus it follows that $\triangle PBA \sim \triangle PDY$ from $\measuredangle PBA=\measuredangle PDY$ and $\measuredangle APB=\measuredangle YPD \iff \measuredangle YPA=\measuredangle DPB \iff \widehat{FA}=\widehat{DB}$, which is true. To finish, this means that $\measuredangle XPT=\measuredangle EPA=\measuredangle BAD=\measuredangle XYT$, hence $TXYP$ is cyclic. Now we show that the other intersection point $Q$ lies on $\overline{CT}$. Let $\overline{TQ}$ intersect $(ABD)$ again at $S$. By Reim's, $\overline{AS}$ is parallel to the tangent to $(TXY)$ at $T$, which is parallel to $\overline{CXY}$. Furthermore, we have $\measuredangle XQS=\measuredangle XQT=\measuredangle XYT=\measuredangle BAD$. On the other hand, let $\overline{SC}$ intersect $(ABD)$ again at $Q'$. By Reim's, $BCXQ'$ is cyclic, hence $\measuredangle XQ'S=\measuredangle XQ'C=\measuredangle XBC=\measuredangle BAD$. This implies that $SXQQ'$ is cyclic. But $Q$ and $Q'$ both lie on $(ABD)$, but because $X$ doesn't lie on $(ABD)$ and $Q,Q' \neq S$, we must have $Q=Q'$ since circles intersect at at most two points (and $S$ is one of them). Hence $T,Q,C$ are collinear, so we're done. $\blacksquare$
BlizzardWizard
07.07.2024 22:07
Proof sketch (some computations skipped): Complex bash with $(ABD)$ unit circle. Let $f(z)$ denote $\frac z{\overline z}$. Let $x=a+s(b-a)$ and $y=a+t(d-a)$. Putting the fact that $X$, $Y$, and $C$ are collinear into barycentric coordinates tells us that $(1,1)$, $(0,s)$, and $(t,0)$ are collinear, so $(s-1)(t-1)=1$. So, we set $s=1+r$ and $t=1+1/r$. Then $x=b+br-ar$ and $y=d+d/r-a/r$. We now intersect $(TXY)$ with the unit circle. \begin{align*} f\left(\frac{z-x}{z-y}\right)&=f\left(\frac{t-x}{t-y}\right)=f\left(\frac{a-x}{a-y}\right)^2=f\left(\frac{a-b}{a-d}\right)^2=\frac{b^2}{d^2}\\ d^2(z-x)(\overline z-\overline y)&=b^2(z-y)(\overline z-\overline x)\\ d^2(z-(b+br-ar))\left(\frac1z-\left(\frac1d+\frac1{dr}-\frac1{ar}\right)\right)&=b^2\left(z-\left(d+\frac dr-\frac ar\right)\right)\left(\frac1z-\left(\frac1b+\frac rb-\frac ra\right)\right)\\ d(z-(b+br-ar))(adr-(ar+a-d)z)&=b(rz-(dr+d-a))(ab-(a+ar-br)z)\\ \end{align*}Computing just the $z^2$ and $1$ coefficients, we get \[(br-d)(br+d-ar-a)z^2+(\text{something})z+a(dr-b)(bdr+bd-adr-ab)\]from which we can easily guess and confirm that one root is $z_1=\frac{a(dr-b)}{br-d}$ by computing $\frac{z_1-x}{z_1-y}$; by Vieta's, the other root is $z_2=\frac{bdr+bd-adr-ab}{br+d-ar-a}$. We now compute $t$. \begin{align*} \frac{t-x}{t-y}&=f\left(\frac{a-x}{a-y}\right)=\frac bd\\ t&=\frac{by-dx}{b-d}\\ &=\frac{b(d+d/r-a/r)-d(b+br-ar)}{b-d}\\ &=\frac{-ab+adr^2-bdr^2+bd}{r(b-d)} \end{align*} After this, it suffices to check that $t=a+z_1-az_1\overline t$ and $\frac{t-z_2}{t-c}\in\mathbb R$ to confirm that $Z_1$ and $Z_2$ lie on $AT$ and $CT$, respectively.
ezpotd
07.07.2024 23:38
The first point is just the Dumpty point, see USA TST 2008/7. We only focus on the second part of the problem. Claim: $(ABD)$ and $(TXY)$ intersect at the Miquel point formed by the circles $(ABD)$, $(XCB)$, $(YCD)$. Proof: Trivial by angles, let the desired Miquel point be $M$, then consider $\angle XMY = \angle XMC + \angle CMY = \angle XBC + \angle CDY = 2\angle A$. Now it remains to show that $MC \cap (TXY) = T$, equivalent to showing $\angle CMO$ is right where $O$ is the circumcenter of $(AXY)$. However, this is trivial by considering $\angle CMO = \angle CMY + \angle YMO = \angle A + \angle YXO = 90$, so we are done.