Iranian Geometry Olympiad (3)

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Source: IGO 2016,Advanced level,P3

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237 posts

#1 h Sep 13, 2016, 3:06 AM • 4 Y

Y by Adventure10, Mango247, TheHimMan, Rounak_iitr

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

This post has been edited 1 time. Last edited by MRF2017, Sep 13, 2016, 5:06 AM
Reason: A typo edited!

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#2 h Sep 13, 2016, 4:26 AM • 2 Y

Y by Adventure10, Mango247

Typo: $P$ is the intersection of the rays $CB$ and $DA$ ...

This post has been edited 1 time. Last edited by drmzjoseph, Sep 13, 2016, 4:26 AM

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#3 h Sep 13, 2016, 5:07 AM • 2 Y

Y by Adventure10, Mango247

drmzjoseph wrote:

Typo: $P$ is the intersection of the rays $CB$ and $DA$ ...

Thanks,edited

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andria

#4 h Sep 13, 2016, 7:05 AM • 9 Y

Y by Smkh, Siddharth03, hakN, Infinityfun, Adventure10, Mango247, Stuffybear, Om245, TheHimMan

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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#5 h Sep 13, 2016, 7:30 PM • 1 Y

Y by Adventure10

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)

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#6 h Sep 13, 2016, 9:32 PM • 2 Y

Y by Adventure10, Mango247

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 ?

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