In a plane: 1. An ellipse with foci $F_1$, $F_2$ lies inside a circle $\omega$. Construct a chord $AB$ of $\omega$. touching the ellipse and such that $A$, $B$, $F_1$, and $F_2$ are concyclic. 2. Let a point $P$ lie inside an acute angled triangle $ABC$, and $A'$, $B'$, $C'$ be the projections of $P$ to $BC$, $CA$, $AB$ respectively. Prove that the diameter of circle $A'B'C'$ equals $CP$ if and only if the circle $ABP$ passes through the circumcenter of $ABC$. Proposed by Alexey Zaslavsky
Problem
Source: 2023 BMoEG III p6 https://artofproblemsolving.com/community/c594864h3181962p28994736
Tags: conics, ellipse, construction, geometry, bmoeg
NO_SQUARES
29.11.2023 19:10
For second Let $O$ be circumcenter of triangle $ABC$. Note that since $CP$ is the diameter of the circle $\omega = (A' PB')$, condition the diameter of the circle $\gamma = (A'B'C')$ and $CP$ are equals, is equivalent to the equality of the circles $\omega, \gamma$ (*), which is equivalent to the fact that the angles $A'C'B'$ and $A'PB'$ are either equal or in the sum of $180^{\circ}$. But since $\angle A' PB' = \angle A'C'B' + \angle C'B' P + \angle C'A' P \not = \angle A'C'B'$, the condition (*) is equivalent to $\angle A'C'B' + \angle A'PB'= 180^{\circ} = \angle ACB + \angle A'PB'$, so the condition (*) is equivalent to the equality $\angle ACB = \angle A'C'B'$. Since $\angle PC'B' = \angle PAC, \angle PC'A' = \angle PBC$, the condition (*) is equivalent to the condition $\angle CAP + \angle CBP = \angle ACB$, and since $\angle APB = \angle ACB + \angle CAP + \angle CBP$, then we get that the condition (*) is equivalent to the equality of $\angle APB = 2\angle ACB = \angle AOB$, which is equivalent to the $A,P,O,B$ are concyclic, which we wanted to prove.
starchan
30.11.2023 12:03
Here's a synthetic solution for the first part:
We get rid of the ugly case when $F_1, F_2$ have the same power wrt the circle by marking $P$ on the ellipse with $PF_1 = PF_2$ and drawing the line through $P$ parallel to $F_1F_2$. In this case $F_1F_2BA$ is trivially an isosceles trapezoid, so we are done anyway. Let's solve the general version now.
Pick any point $X$ on the circle and mark the second intersection of $(F_1F_2X)$ with the circle, say it is $Y$. We now mark $Z = XY \cap F_1F_2$. Using Menelaus/Ceva configuration, we now mark $Z'$ such that $(ZZ';F_1F_2) = -1$. Next, we draw the circle with diameter $ZZ'$ and mark one of its intersections with the ellipse, at say $P$. Now note that since $(ZZ';F_1F_2) = -1$ and $ZP \perp Z'P$, line $PZ$ is the external angle bisector of $\angle F_1PF_2$, that is, it is the tangent to the ellipse at $P$. Let $ZP$ meet the circle again at two points $A, B$. Now note that \[ZA \cdot ZB = ZX \cdot ZY = ZF_1 \cdot ZF_2\]and hence $A, B, F_1, F_2$ are concylic; as desired. $\square$
There's a couple of things above that need careful explanation, for example why $Z$ is not say, the midpoint of $F_1F_2$. In this case $Z'$ does not exist and the proof falls apart. Fortunately one can show without too much effort that $Z$ lies outside the circle (and thus segment $F_1F_2$). Since it is not the point at infinity (the equal power condition) everything exists and the construction works.
PROF65
17.03.2024 16:49
$a)$ it suffices to have a point of the desired tangent to the ellipse since the construction of tangent with a ruler through a point is well known. consider a circle through $F_1$ & $F_2$ that cuts $\omega$ at $C,D$ then $CD\cap F_1F_2$ is the radical center of the three circles which is on the tangent . $b)$ Lemma : Let $H$ be the orthocenter of $ABC$; $A_1,B_1,C_1$ the symmetric points of $P$ with resp. $BC,CA,AB$ $P$ is point on the circle of $BCH \iff $ the circumcircle of symmetrical triangle of P wrt $ABC$ pass through $A$. Proof: let $A_2,B_2,C_2$ be the pedal triangle of $P$, $\angle C_1A_1P=\angle C_2A_2P=\angle C_2BP= \angle ABP$ idem $\angle PA_1B_1 =\angle PCA $ thus $ \angle C_1A_1B_1=\angle ABP+ \angle PCA=\angle CPB+\angle BAC $ so $P\in (BCH)\iff\angle BPC=\angle CAB \iff \angle C_1A_1B_1=2 \angle BAC= \angle C_1AB_1\iff A\in (C_1A_1B_1)$ back to the problem Let $P'$ the isogonal conjugate of $P$ and $ A_1B_1C_1$ the symmetrical triangle of $P'$ wrt $ABC$ then: $P\in (ABO)\iff P'\in ABH\iff C\in (A_1B_1C_1)$ (by applying the lemma ) $\iff $ $2 PA$ is the diameter of $(A_1B_1C_1)\iff AP $ is the diameter of the pedal circle. Best regards . RH HAS
NO_SQUARES
03.09.2024 22:25
Second one was also on Oral Moscow Geometry Olympiad 2024 (10-11.1)