Let $ABC$ be a right-angled triangle with the right angle at $B$ and circumcircle $c$. Denote by $D$ the midpoint of the shorter arc $AB$ of $c$. Let $P$ be the point on the side $AB$ such that $CP=CD$ and let $X$ and $Y$ be two distinct points on $c$ satisfying $AX=AY=PD$. Prove that $X, Y$ and $P$ are collinear. Proposed by Dominik Burek, Poland
Problem
Source: 2019 MEMO Problem T-6
Tags: geometry, circumcircle, memo, MEMO 2019
timon92
30.08.2019 16:31
This problem was proposed by Burii.
Tintarn
30.08.2019 18:50
This is straightforward with coordinates:
Let $Z$ be one point on $c$ such that $PZ$ is perpendicular to $AC$. It clearly suffices to prove that $AZ=PD$.
It's easy to see that $AZ^2=AP \cdot AB$.
We now introduce coordinates. Let $B=(0,0), C=(x,0), A=(0,y)$ so that the radius of $c$ is $r=\sqrt{\frac{x^2+y^2}{4}}$.
Then $D=\left(\frac{x}{2}-r, \frac{y}{2}\right)$ and $P=(0,z)$ with $z^2=\frac{x^2+y^2}{2}+rx-x^2$. So $AZ^2=y(y-z)$ and
\[PD^2=\left(\frac{x}{2}-r\right)^2+\left(\frac{y}{2}-z\right)^2=y(y-z)+\frac{x^2-y^2}{2}-rx+z^2=AZ^2\]as desired.
MarkBcc168
01.09.2019 12:13
I agreed that the coordinate bash above is very feasible. Here is my synthetic solution. Let $Q$ be the reflection of $P$ across $B$. Clearly $CP = CQ = CD$ thus $\angle ADC =90^{\circ}\implies$ $AD$ is tangent to $\odot(DPQ)$. Construct parallelogram $AKPD$. Notice that $KP = AD = DB$ and $\angle KPA = \angle PAD = \angle DBA$ thus $\triangle KPB\cong\triangle DBQ$ by S.A.S. Therefore $$\angle PKA = \angle PDA = \angle PQD = \angle KBP$$which means $AK$ is tangent to $\odot(KBP)$. In particular, this implies $$AP\cdot AB = AK^2 = DP^2 = AX^2 = AY^2$$thus an inversion around $\odot(A,AX)$ maps $\odot(ABC)\to\overline{XY}$ and $P\to B$. Hence $P\in XY$.
gghx
23.05.2021 14:29
Note that we want to prove $$X,P,Y\text{ collinear }\iff \angle PXA = \angle YXA \iff \angle PXA = \angle ABX \iff AX^2=AP\times AB \iff PD^2=AP\times PB$$ Now we go through the coord bash checklist. 1. Prominent right angle. Check. 2. Not so many points. Check. 3. Easy to bash. Check. (Note especially the $PD^2$ term which works very well with the distance formula). Lets go. Define $B=(0,0),A=(0,2),C=(2c,0)$ (the $2$ and $2c$ are simply to make the equation of $(ABC)$ nicer). Now $M=(c,1)$ is the midpoint of $AC$ and thus the equation of $(ABC)$ is $(x-c)^2+(y-1)^2=c^2+1$ When $y=1$, $(x-c)^2=c^2+1$ or $x=c-\sqrt{c^2+1}$. That gives $D=(c-\sqrt{c^2+1},1)$ Now the circle with radius $CD$ and center $C$ is $(x-2c)^2+y^2=(c+\sqrt{c^2+1})^2+1$ Let $x=0$ we get $y^2=2c\sqrt{c^2+1}-2c^2+2$ so $P=(0,\sqrt{2c\sqrt{c^2+1}-2c^2+2})$ Now we need to show $AP\times AB=PD^2$ which is equivalent to $$2(2-\sqrt{2c\sqrt{c^2+1}-2c^2+2})=(c-\sqrt{c^2+1})^2+(\sqrt{2c\sqrt{c^2+1}-2c^2+2}-1)^2$$This is obvious because the right hand side, after expanding out the squares, is $2c^2+1-2c\sqrt{c^2+1}+2c\sqrt{c^2+1}-2c^2+2-1-2\sqrt{2c\sqrt{c^2+1}-2c^2+2}$ which is obviously the same as the LHS. FYI did the coord in $<10$ minutes.