Let ${ABC}$ be an acute angled triangle whose shortest side is ${BC}$. Consider a variable point ${P}$ on the side ${BC}$, and let ${D}$ and ${E}$ be points on ${AB}$ and ${AC}$, respectively, such that ${BD=BP}$ and ${CP=CE}$. Prove that, as ${P}$ traces ${BC}$, the circumcircle of the triangle ${ADE}$ passes through a fixed point.
Problem
Source: 2016 JBMO Shortlist G4
Tags: geometry, JBMO
RC.
08.10.2017 12:53
The required point is the incenter of \(\Delta ABC\).
Let \(AI\) be the bisector of \(\angle A\) with \(I\) on \(\odot ADE.\) Clearly, \(ID = IE ; \angle DIE = 180^{\circ} - \angle A = 2\times (90 ^{\circ} -\angle A/2) = 2 \times \angle DPE \Rightarrow I\) is the circumcenter of \(\odot DPE \Rightarrow ID = IE = IP \Rightarrow BDIP ; CEIP\) are kites. \(\Rightarrow BI ; CI\) bisects the \(\angle B ; C\) respectively. \(\Rightarrow I\) is the incenter.
Jzhang21
02.09.2018 05:31
We claim that the fixed point is $I$, the incenter of $\triangle ABC$. Since $\triangle BDP$ is isosceles and $BI$ is $B$ angle bisector, then $BI\perp PD$ so $DI=IP$ and $\angle IDP=\angle IPD.$ Similarly, $PI=IE$ and $\angle IEP=\angle IPE.$ Hence, $DI=IP=IE$ so $\angle IDE=\angle IED.$ Note that $\angle DPE=180^{\circ}-\angle DPB-\angle EPC=180^{\circ}-(90^{\circ}-\frac{\angle B}{2})-(90^{\circ}-\frac{\angle C}{2})=\frac{\angle B+\angle C}{2}$ and $\angle DPE=\angle DPI+\angle IPE=\angle PDI+\angle IEP.$ Hence, $2\angle IED=\angle IED+\angle IDE=180^{\circ}-(\angle IDP+\angle IEP)-(\angle IPD+\angle IPE)=\angle A$. Thus, $\angle IED=\frac{\angle A}{2}=\angle IAD$ so $ADIE$ is cyclic so all $(ADE)$ goes through $I$, as desired. $\blacksquare$
Shaddoll1234
02.09.2018 06:55
WLOG suppose that $AB<AC$ Let $I$ be the incenter of $\triangle ABC,$ $\quad \quad \quad$ $CI \cap PD= F.$ Simple angle chasing gives $\angle PFI = \angle EFI = \angle A /2$, so $A, D, F, I, E$ are concyclic $\Rightarrow$ $(ADE)$ passes through $I$, which is fixed Q.E.D
Mahdi_Mashayekhi
23.12.2021 11:08
Let I be incenter of ABC. BP = BD ---> ID = IP ---> IBD and IBP are congruent ---> ∠IDA = ∠IPC the same way we can prove ∠IEA = ∠IPB. so we have : ∠IDA + ∠IEA = 180 ---> IDAE is cyclic. I is the fixed point. we're Done
Tafi_ak
01.11.2022 19:55
Let $I$ be the incenter. First solution Notice that $BDIP$ and $CEIP$ are kite which implies $ID=IP=IE$. So $(ADE)$ passes through $I$ since $AI$ is the $A-$angle bisector. Second solution Notice that $\angle BDI=\angle BPI=\angle AEI$. Third solution Notice that $AD+AE=b+c-a$ which is fixed therefore by using this lemma we can say $(ADE)$ passes through a fixed point on the $A-$ angle bisector. Fourth solution Let $Q=AI\cap BC$. It is sufficient to prove the power of $Q$ wrt $(ADE)$ is fixed. Let \[ f(\bullet)=P(\bullet, (ADE))-P(\bullet, (ABC)) \]It is known that $f$ is linear over the plane. We know $Q=\left(0,\frac{b}{b+c}, \frac{c}{b+c}\right)$. We want $f(Q)$ is fixed. By the linearity we have \begin{align*} f(Q)&=\frac{b}{b+c}f(B)+\frac{c}{b+c}f(C) \\ &=\frac{1}{b+c}\left(bf(B)+cf(C)\right) \\ &=\frac{bc}{b+c}(BD+CE) \\ &=\frac{abc}{b+c} \end{align*}which is fixed.
rafigamath
06.04.2023 07:40
$Claim:$ The fixed point is $I$ Notice that $BI$ is angle bisector of $\angle{PBD}$. $BP=DP$ , $IP=ID$ because $BI$ is perpendicular bisector of $PD$ . Likewise $IP=IE$. So we can say that $I$ is the circumcenter of triangle $DEP$. Let $\angle{B} =\angle{2X}$ and $\angle{C} =\angle{2Y}$. If $\angle {IDE}= \angle{90-X-Y}$ the problem will end. $\angle{DEP}= \angle{X+Y}$ because $\angle{DEP}=\angle {180 - (90-X)-(90-Y)}= \angle{X+Y}$. $\angle{DIE}=\angle{2(X+Y)}$ because it is the central angle .$ID=IE$ and we get that $\angle {IED}= \angle {90-X-Y}$ .