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BdMO Online Forum • View topic - IMO 2015 - Problem 3

IMO 2015 - Problem 3

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IMO 2015 - Problem 3

Post Number:#1  Unread postby Nirjhor » Wed Jul 15, 2015 1:48 am

Let $ABC$ be an acute triangle with $AB > AC$. Let $\Gamma $ be its cirumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HQA = 90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ = 90^{\circ}$. Assume that the points $A$, $B$, $C$, $K$ and $Q$ are all different and lie on $\Gamma$ in this order.

Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.

Proposed by Ukraine.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.
Nirjhor
 
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Re: IMO 2015 - Problem 3

Post Number:#2  Unread postby Fm Jakaria » Sat Aug 01, 2015 8:31 pm

Let $O$ be the center of $\gamma$, the circumcircle of ABC. $ AH = 2OM$, because$ O$ is the orthocenter of the medial triangle of $\triangle ABC$. So $P := AO \cap HM$ sends $OM$ to $AH$ by a homothety of ratio $2:1$. So $P$ lies on $\gamma$, and it is diametrically opposite to $ A$. As the projection of $A$ on $HP$ lies on $\gamma$; $Q$ coincides with this projection, so Q lies on $PMH$.
Let $AHF$ intersect $\gamma$ again at $R$. As $\angle HCF = \angle FCR = 90-B$, $HR$ has perpendicular bisector $BC$.
Let segment $KR$ intersect the circumcircle $\omega$ of $QKH$ again at $X$.
Note $ \angle HXK = 180 - \angle PQK = \angle PRK$. So $HX$ and $PR$ are parallel, so both is perpendicular to $AHFR$, and so both is parallel to $BC$. $HQ$ is a diameter of $\omega$, so $QX$ is perpendicular to $HX$, so to $BC$. Let us extend $QX$ to meet $PR$ at $Y$, and $BC$ at $L$. Now it is obvious that $HRYX$ is a rectangle, with $BLC$ the perpendicular bisector of $XY$.
Let the perpendicular to $QHP$ at $H$ meet $BC$ at $Z$. Let $OZ$ meet $RXK$ at $S$.
Now note that because $HX$ is parallel to $MZ$, the right triangles $MHZ$ and $HXQ$ are similar. So
$\dfrac {HX}{MH} = \dfrac{HQ}{MZ}$. Moreover, $AQFM$ is cyclic with diameter $AM$. So
$AH.HF = MH.HQ$. Combining the last two gives
$HX.(MZ) = \dfrac{1}{2}.AH.(2HF) = OM. HR$. So $ \dfrac {HX}{HR} = \dfrac{OM}{MZ}$, so
$OMZ \sim XHR \sim RYX$. This gives $\angle LZS = \angle MZO = \angle RXL$. So $XLZS$ is cyclic. Then $\angle XSZ = \angle XLZ = 90$, so $OZ$ is perpendicular to $RXK$. Since $OR = OK$, we have that $ZK = ZR = ZH$.
Note that the circumcircle of $HMF$ and $\omega$ has the common tangent $ ZH$ at $H$; because they have the respective diameters HM and HQ. This gives, as $ZH = ZK$, that $ZK$ is tangent to $\omega$. Moreover, $ ZK^2 = ZH^2 = ZF.ZM$, giving that $ZK$ is also tangent to the circumcircle of $KMF$. So give a big clap….
Fm Jakaria
 
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