Let $ABC$ be a triangle. An arbitrary circle which passes through the points $B,C$ intersects the sides $AC,AB$ for the second time in $D,E$ respectively. The line $BD$ intersects the circumcircle of the triangle $AEC$ at $P{}$ and $Q{}$ and the line $CE$ intersects the circumcircle of the triangle $ABD$ at $R{}$ and $S{}$ such that $P{}$ is situated on the segment $BD{}$ and $R{}$ lies on the segment $CE.$ Prove that: The points $P,Q,R$ and $S{}$ are concyclic. The triangle $APQ$ is isosceles. Petru Braica
Problem
Source: Romania JBMO TST 2024 Day 1 P4
Tags: geometry
ratatuy
31.07.2024 16:53
Solution for $a$ $PX\cdot XQ=CX\cdot XE=BX\cdot XD=RX\cdot XS$
khanhnx
31.07.2024 17:22
a) Let $F \equiv BD \cap CE$ then $\overline{FP} \cdot \overline{FQ} = \overline{FC} \cdot \overline{FE} = \overline{FB} \cdot \overline{FD} = \overline{FR} \cdot \overline{FS}$ or $P, Q, R, S$ lie on a circle b) Let $G$ be second intersection of $(AEC)$ and $(ADB)$. We have $\angle{DGA} = \angle{DBA} = \angle{ECA} = \angle{EGA}$. We also have $\angle{PEG} = \angle{DQG}$ and $\angle{EPG} = 180^{\circ} - \angle{BAG} = 180^{\circ} - \angle{BDG} = \angle{QDG}$. Then $\angle{PGE} = \angle{QGD}$. Hence $\angle{PGA} = \angle{QGA}$ or $GA$ is bisector of $\angle{PGQ}$. This means $AP = AQ$ or $\triangle AQP$ is isosceles
FarrukhKhayitboyev
22.09.2024 14:30
a)Generalization of USAMO $1990$ $P5$ b)We will prove that $AP$ is tangent to $(PCD)$, $\angle APE= \angle ACE = \angle DBE = \angle PBE$ it means that AP is tangent to $(PBE)$ $AP^2=AE*AB=AD*AC $ so $AP$ is tangent to $(PCD)$ $\implies \angle AQP = \angle DCP = \angle APQ$, $AP=AQ$ and we are done.