Iran 2008 Geometry

by utkarshgupta, Jan 31, 2015, 12:07 PM

An easy problem as a refresher before INMO :P

Problem :

Let $I_a$ be the $A$-excenter of $\Delta ABC$ and the $A$-excircle of $\Delta ABC$ be tangent to the lines $AB,AC$ at $B',C'$, respectively. $ I_aB,I_aC$ meet $B'C'$ at $P,Q$, respectively. $M$ is the meet point of $BQ,CP$. Prove that the length of the perpendicular from $M$ to $BC$ is equal to $r$ where $r$ is the radius of incircle of $\Delta ABC$.

Solution :

Let the given excircle be tangent to $BC$ at $T$.
Let $\angle ABC = \angle B$, $\angle ACB =\angle C$ and $\angle BAC =\angle A$

Then it is easy to see by simple angle chasing that
$PI_ACC'T$, $QI_ABB'T$ and $BCQP$ are cyclic.

Radical Axis of $\odot PI_ACC'T$ and $\odot QI_ABB'T$ is $TI_A$
Since $BCQP$ are cyclic, $TMI_A$ is a straight line.

$\angle MTC = \angle I_ATC =90$

Also since $M$ lies on the radical axis,
$MT \cdot MI_A = MC \cdot MP$
$\implies CTPI_A$ is cyclic
$\implies \angle CPI_A = 90$
$\implies \angle TCP = \frac{\angle B}{2}$
$\implies MT = CT \tan {\frac{B}{2}}$
$\implies MT = (s-b)\tan {\frac{B}{2}} = r$

QED

Comment

1 Comment

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here is my solution: I will written the problem: ABC is a..triangle (I) , (O) ,blah blah .we have BB'C=BIaC => BQ pedendicular CIa and analogously CP pedendicular BIa => X is othorcentre BCIa .associate IBIaC concylic with IIa is a diameter then BICX paralegrogram ..

by lebathanh, Aug 25, 2016, 3:13 AM

Stay insane,Coz it's your will, labour and pain,which takes you to the top of the mountain.

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