In a convex qualrilateral $ABCD$, let $P$ be the intersection point of $AD$ and $BC$. Suppose that $I_1$ and $I_2$ are the incenters of triangles $PAB$ and $PDC$,respectively. Let $O$ be the circumcenter of $PAB$, and $H$ the orthocenter of $PDC$. Show that the circumcircles of triangles $AI_1B$ and $DHC$ are tangent together if and only if the circumcircles of triangles $AOB$ and $DI_2C$ are tangent together. Proposed by Hooman Fattahimoghaddam
Problem
Source: IGO 2016,Advanced level,P3
Tags: geometry, geometry proposed
drmzjoseph
13.09.2016 07:26
Typo: $P$ is the intersection of the rays $CB$ and $DA$ ...
MRF2017
13.09.2016 08:07
drmzjoseph wrote: Typo: $P$ is the intersection of the rays $CB$ and $DA$ ... Thanks,edited
andria
13.09.2016 10:05
Let $X$ be one of intersection of circumcircles of triangles $AOB$ and $CI_2D$. Let $\odot(\triangle BXC)\cap \odot(\triangle AXD)=\{X,Y\}$: $\left.\begin{array}{ccc} \angle AYB=\angle XDP+\angle XCP=\angle DXC-\angle P=\angle DI_2C-\angle P=90^{\circ}-\frac{\angle P}{2}\Longrightarrow Y\in \odot(\triangle AI_1B)\\ \\ \angle DYC=\angle XAD+\angle XBC=\angle BXA+\angle P=180^{\circ}-\angle AOB+\angle P=180^{\circ}-\angle P\Longrightarrow Y\in \odot(\triangle DHC) \end{array}\right\}\Longrightarrow Y$ is one of intersection of circumcircles of $AI_1B,DHC$ and $Y=\odot(BXD)\cap \odot(AXD)$ Similarly we can prove if $X'$ is second intrsection of circumcircles $AOB$ and $CI_2D$ and $Y'$ be the second intersection of circumcircles $DHC,AI_1B$ then $AX'Y'D,BX'Y'C$ are cyclic. Hence: $\begin{cases} AXYD,BXYC\ \text{are cyclic}\\ \\ AX'Y'D,BX'Y'C\ \text{are cyclic}\end{cases}$ So the circumcircles of $AOB$ and $DI_2C$ are tangent $\Longleftrightarrow\ X=X'\Longleftrightarrow Y=Y'\Longleftrightarrow$ circumcircles of $DHC,AI_1B$ are tangent. Q.E.D
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blackSnoopy
13.09.2016 22:30
I have another idea Idea: two circles are tangent iff the distance between centers be equal with the sum of radiuses. And hopefully the centers of these circles and their radiuses are well known. Can someone complete this idea, please? (This is urgent for me)
Mathlinkeraa
14.09.2016 00:32
blackSnoopy wrote: I have another idea Idea: two circles are tangent iff the distance between centers be equal with the sum of radiuses. And hopefully the centers of these circles and their radiuses are well known. Can someone complete this idea, please? (This is urgent for me) this is such a nice idea , can anyone solve this problem in this way ?