Let ABC be an acute-angled triangle. Let D be a point on the segment BC, I the incentre of ABC. The circumcircle of ABD meets BI at P and the circumcircle of ACD meets CI at Q. If the area of PID and the area of QID are equal, prove that PI*QD=QI*PD.
Problem
Source: CHKMO
Tags: geometry, incenter, circumcircle
kk108
31.12.2016 11:56
May I know what CHKMO stands for ? Thanks .
noobatron3000
31.12.2016 11:58
China Hong-Kong Mathematical Olympiad
Complex2Liu
07.01.2017 13:40
noobatron3000 wrote: Let $ABC$ be an acute-angled triangle. Let $D$ be a point on the segment $BC,$ $I$ the incentre of $ABC.$ The circumcircle of $ABD$ meets $BI$ at $P$ and the circumcircle of $ACD$ meets $CI$ at $Q.$ If the area of $PID$ and the area of $QID$ are equal, prove that $PI\cdot QD=QI\cdot PD.$ Let $E$ be a point on $BC$ such that $(B,C;D,E)=-1.$ The condition that $[PID]=[QID]$ implies that $ID$ is median. Notice that $-1=(IP,IQ;ID,\infty)\stackrel{I}{=}(B,C;D,E) \implies IE\perp AD.$ Now we cliam that $D$ is the foot of $I$ on $BC.$
WLOG assume $AB>AC.$ We use barycentric coordinate. Set $D=(0:m:a-m)$ and $E=(0:t:a+b+c-t)$ where $0<m<a$ and $\tfrac{t}{a+b+c-t}=\tfrac{-m}{a-m}\implies t=\tfrac{m(a+b+c)}{2m-a}.$ Thus
http://www.mit.edu/~evanchen/handouts/bary/bary-full.pdf
we get \[0=a^2(m(a+b)-mt+mb-mt-ab+at)+b^2(at-a^2-ab+ma-a^2)+c^2(ab-at-ma),\]collect the $t$ terms we get \[(2ma-a^2+c^2-b^2)t=am(a+2b)-b(a^2+b^2+2ab-c^2)+m(b^2-c^2).\]Substitute $a(2m-a)t=am(a+b+c)$ we get \[\begin{aligned} (c^2-b^2)t&=am(b-c)+m(b^2-c^2)-b((a+b)^2-c^2)\\ &=m(b-c)(a+b+c)-b(a+b+c)(a+b-c)\\ &=(a+b+c)(mb-mc-ab-b^2+bc). \end{aligned}\]Using $(2m-a)t=m(a+b+c)$ again we get \[(c^2-b^2)m=(2m-a)(mb-mc-ab-b^2+bc),\]which is the following quadratic in $m$: \[2(b-c)m^2-m(c^2+b^2-2bc+3ab-ac)+ab(a+b-c)=0.\]Since $0<m<a$ so we conclude that $m=\tfrac{a+b-c}{2},$ as desired. It's easy to see that $\angle PIQ=90^\circ+\tfrac{1}{2}\angle A$ and $\angle QAP=90^\circ-\tfrac{1}{2}\angle A,$ so $A,I,P,Q$ are concyclic. Notice that $DI$ is median and $AI$ is isogonal to $DI$ with respect to $\angle PIQ,$ it follows that $A,P,I,Q$ is harmonic quadrilateral. Therefore $QI\cdot PD=QI\cdot AP=PI\cdot QA=PI\cdot QD.$
Wave-Particle
18.03.2017 21:28
I've been thinking over this for a bit now...does anyone have a solution without projective (still haven't learned it ) and without bary, complex Thanks!
PROF65
19.03.2017 06:30
Let $P',Q'$ be the intersections of $DQ$ with $PB$ and $DP$ with $QC$ .$\angle P'DA =\frac{1}{2}\angle C,\angle ADQ' =\frac{1}{2}\angle B$ thus $Q'IP'D$ is cyclic . we have $PP'.QD.\sin PP'D=QQ'.PD.\sin QQ'D$ then $(PI+IP').QD.\sin PP'D=(QI+IQ').PD.\sin QQ'D$ since $PID$ and $QID$ have the same area and $ \angle PP'D$ and $ \angle QQ'D$ are supplementary then $PI.QD=QI.PD$ . RH HAS P.S.
this property effectively occurs only when $D$ is the tangency point :
Denotes $ \alpha =\angle DAB, \beta =\angle DAC,\gamma =\angle DIB,\delta =\angle DIC$.the areas of $PID,QID$ are equal means $PI. \sin \gamma =QI. \sin \delta$ besides $\frac{\sin \gamma }{PD} =\frac{\sin \alpha}{DI},\frac{sin \delta}{QD}=\frac{sin \beta}{ID}\ $ so $PI. \sin \gamma .QD.\frac{\sin \beta}{sin \delta} =QI. \sin \delta .PD.\frac{\sin \alpha}{\sin \gamma}$ so if $PI.QD=QI.PD$ then $. \sin \gamma .\frac{\sin \beta}{\sin \delta} = \sin \delta .\frac{\sin \alpha}{\sin \gamma}$ ie $ \frac{\sin \gamma^2}{\sin \delta^2}=\frac{sin \alpha}{\sin \beta}$ but $\frac{sin \alpha}{\sin \beta}=\frac{BD}{CD}. \frac{AC}{AB}=\frac{\frac{BD}{ID}}{\frac{CD}{ID}} \frac{AC}{AB}=$$\frac{\frac{\sin \gamma}{\sin\frac{\angle B}{2}}}{\frac{\sin \delta}{\sin\frac{\angle C}{2}}}$$ \frac{\sin \angle B}{\sin \angle C}$ thus $ \frac{sin \alpha}{\sin \beta}=\frac{sin \gamma}{\sin \delta}.\frac{\cos\frac{\angle B}{2}}{\cos\frac{\angle C}{2}}$ implies$ \frac{sin \gamma}{\sin \delta}=\frac{\cos\frac{\angle B}{2}}{\cos\frac{\angle C}{2}}=\frac{\sin \frac{\pi-\angle B}{2}}{\sin\frac{\pi-\angle C}{2}}$ we know that $\frac{\angle B}{2}+\frac{\angle C}{2}=\pi -(\gamma +\delta)$ therefore $\gamma =\frac{\pi}{2}-\frac{\angle B}{2},\delta =\frac{\pi}{2}-\frac{\angle C}{2}$
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githjijjj
20.06.2017 09:07
As in the solution by PROF65, let $P',Q'$ be the intersections of $DQ$ with $PB$ and $DP$ with $QC$, then $Q'IP'D$ is cyclic. Since $\triangle PID$ and $\triangle QID$ have the same area, $DI$ will pass through the midpoint $M$ of $PQ$, then by Ceva's theorem we have $\frac{DP'}{P'Q}\cdot \frac{QM}{PM}\cdot \frac{PQ'}{Q'D}=1$ so $\frac{DP'}{P'Q}= \frac{PQ'}{Q'D}$ so $PQ\parallel P'Q'$. Angle chasing yields $\angle P'DI=\angle P'Q'I=\angle IQP$ and $\angle Q'DI=\angle Q'P'I=\angle IPQ$. Therefore, $$\frac{PI}{QI}=\frac{\sin \angle IQP }{\sin \angle IPQ}=\frac{\sin \angle P'DI(=\angle QDI) }{\sin \angle Q'DI(=\angle PDI)}=\frac{PD}{QD}$$since $DI$ is the $D$-median of $\triangle DPQ$. And so we are done.