On side $BC$ of an acute triangle $ABC$ are marked points $D$ and $E$ so that $BD = CE$. On the arc $DE$ of the circumscribed circle of triangle $ADE$ that does not contain the point $A$, there are points $P$ and $Q$ such that $AB = PC$ and $AC = BQ$. Prove that $AP=AQ$.
Problem
Source: All-Russian 2022 10.2
Tags: geometry
NJAX
19.04.2022 21:16
I think it is a very easy problem for All-Russian MO. My Solution Let $O$ be the circumcenter of triangle $ADE$. Then $OB=OC.$ So $\vartriangle OAB = \vartriangle OPC$ because $OP=OA$ , $AB=CP$ and $OB=OC.$ The same with $\vartriangle OAC= \vartriangle OQB$ thats why , $\angle ABQ= \angle ABO + \angle OBQ = \angle OCP + \angle OCA =\angle ACP $. So $\vartriangle ABQ= \vartriangle ACP \Rightarrow AP=AQ$.
kentok
21.09.2022 09:59
NJAX wrote: I think it is a very easy problem for All-Russian MO. My Solution Let $O$ be the circumcenter of triangle $ADE$. Then $OB=OC.$ So $\vartriangle OAB = \vartriangle OPC$ because $OP=OA$ , $AB=CP$ and $OB=OC.$ The same with $\vartriangle OAC= \vartriangle OQB$ thats why , $\angle ABQ= \angle ABO + \angle OBQ = \angle OCP + \angle OCA =\angle ACP $. So $\vartriangle ABQ= \vartriangle ACP \Rightarrow AP=AQ$. Why from $O$ be the circumcircle of triangle $ADE$ then $OB=OC$?
kentok
21.09.2022 10:00
kentok wrote: NJAX wrote: I think it is a very easy problem for All-Russian MO. My Solution Let $O$ be the circumcenter of triangle $ADE$. Then $OB=OC.$ So $\vartriangle OAB = \vartriangle OPC$ because $OP=OA$ , $AB=CP$ and $OB=OC.$ The same with $\vartriangle OAC= \vartriangle OQB$ thats why , $\angle ABQ= \angle ABO + \angle OBQ = \angle OCP + \angle OCA =\angle ACP $. So $\vartriangle ABQ= \vartriangle ACP \Rightarrow AP=AQ$. Why from $O$ be the circumcircle of triangle $ADE$ then $OB=OC$? Oh I understand
DimPlak
25.09.2022 15:10
Another question. Prove that circles (B,AC) , (C,AB) and circumcicle of triangle ADE conccur.
Shewalala
06.02.2023 19:56
kentok wrote: kentok wrote: NJAX wrote: I think it is a very easy problem for All-Russian MO. My Solution Let $O$ be the circumcenter of triangle $ADE$. Then $OB=OC.$ So $\vartriangle OAB = \vartriangle OPC$ because $OP=OA$ , $AB=CP$ and $OB=OC.$ The same with $\vartriangle OAC= \vartriangle OQB$ thats why , $\angle ABQ= \angle ABO + \angle OBQ = \angle OCP + \angle OCA =\angle ACP $. So $\vartriangle ABQ= \vartriangle ACP \Rightarrow AP=AQ$. Why from $O$ be the circumcircle of triangle $ADE$ then $OB=OC$? Oh I understand can you explain to me why
SerdarBozdag
16.03.2023 12:42
It is enough to prove that: $A-X-T$ and $A-Y-Z$ are given. $XYZT$ is cyclic with center $O$ and its radius is $R$. $AX-AY-AZ-AT$ are given. Then $TZ$ can take only one value. Proof: By PoP, $AO$ is given (constant, certain, not sure about the term). Also by Cos Theorem $\angle AOT, \angle AOZ$ are given. Thus $\angle TOZ$ is given $TO=TZ=R$. Thus by Cos Theorem, $TZ$ is given. So $\triangle ATZ$ is determined by the given facts. If we apply it for $BAQ$ and $CPA$ we are done.
CrazyInMath
21.10.2023 17:33
Let $X$ be the point on $(ABC)$ such that $AX\parallel BC$, then $AXCB$ is an isocseles trapezoid. So $P$ is the reflection of $X$ through $CO$, $Q$ is the reflection of $X$ through $BO$. We only need $\measuredangle PXA=\measuredangle AXQ$. And we know that $\measuredangle (PX,CO)=\measuredangle (QX,BO)$ So we only need $\measuredangle (CO,AX)=\measuredangle (AX,BO)$, which is correct since $AX\parallel BC$.
sirosali
16.12.2023 17:22
#Shewalala If the middle of DE is M. Then MD=ME. We had BD=CE. So MB=MC So OB=OC