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 $\Delta ABC$.
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Tags: geometry, incenter, trigonometry, power of a point, radical axis, geometry proposed
oneplusone
30.09.2010 12:53
Let the excircle touch $BC$ T $D$. Then $BB'=BD,CC'=CD$ so $\triangle BB'P\equiv \triangle BDP$ and similarly for the other pair. Then just angle chase to get $\angle DPC'=\angle BCA$, so $PDCC'$ is cyclic and similarly for the symmetrical one. If $I$ is the incenter, then $\angle BCM=\angle DC'P=\frac12 \angle DQP=\frac12 \angle DBA=\angle CBI$. Similarly $\angle CBM=\angle BCI$ so $\triangle BCM\equiv \triangle CBI$ and it should be trivial from here.
vladimir92
13.02.2011 20:40
I think it's a nice and easy problem, But I add my solution since it proves some interesting facts in the configuration.
let $B_1$ be the orthogonal projection of $A$ into $I_aB$ and denote $(\omega)$ the $A$-excircle, since $P$ lies on the polar of $A$ with respect to $(\omega)$, $A$ lies also on the polar of $P$ with repsect to $(\omega)$, immediatly, $AB_1$ is the polar of $P$ with respect to $(\omega)$. let $D$ be the orthogonal projection of $I_a$ into $BC$, and call $S \equiv (AB_1)\cap(BC)$ we have $(AB_1)||(IB)$ and $A$,$I$and $I_a$ are collinear, and $IBCI_a$ is cyclic, so $\angle{ASC}=\angle{IBC}=\angle{AI_aC}$, so $I_a\in \odot(ACS)$ it follow from Simson that , $B_1$ lies on $C'D$ that is the polar of $C$ with respect to $(\omega)$, hence $C$ lies on the polar $B_1$ with respect to $(\omega)$ as a result, $(CP) \perp (I_aB)$, similary $(BQ)\perp(CI_a)$ and so $M$ is the orthocenter of $\triangle{CI_aB}$ which lead to $D\equiv(I_aM)\cap(BC)$, let the incircle of $\triangle{ABC}$ touche $BC$ at $D_1$. It's know that $CD=BD_1$, moreover we have: $\angle{MCD}=\angle{IBD_1}$, so $\triangle{IBD_1}\cong \triangle{MCD}$ from which follow that $MD=ID_1$. which end the proof.
siavosh
12.04.2012 11:47
Let the excircle touch $BC$ at $D$. we know that $\triangle BDP = \triangle BB'P$ and $\triangle DCQ= \triangle CC'Q$ $\angle BDP =\angle BB'P=\angle CC'Q=\angle CDQ = \frac{b+C}{2} \Rightarrow \angle DPI_{a}=C+\frac{A+B}{2} , \angle DQI_{a}=B+\frac{A+C}{2} $ and we get $BB'QD$ is cycle (1) $\angle I_{a}DB=90$ so $\angle PI_{a}D=\frac{B}{2} ,\angle QI_{a}D=\frac{C}{2}$ (2) (1),(2)gives us that $ \angle PDI_{a}=\angle QDI_{a}= \frac{A}{2} $ so $DI_{a}$ is bisector of $\angle PDQ$ now by Ceva's theorem we get that $CP,BQ,I_{a}D $ are co-liner so $\angle MDB= 90$ $DC=CC' \Rightarrow \angle CDC'=\angle CC'D=\frac{C}{2}$ let incircle touch $BC$ at $E$ $BB'QC$ is cycle so $\angle DBM=\angle DB'Q=\angle DC'C=\angle ICD$ now we have : $BD=CE$ $\angle MDB=\angle IEC$ $\angle MBD=\angle ICD$ we get $MD=IE$ we are down .
jayme
12.04.2012 13:45
Dear Mathlinkers, why this reference gives the same result. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2026669 Sincerely Jean-Louis
jayme
20.05.2014 16:02
Dear Mathlinkers, in order to have a more comprehensive look, the result which lead to 90 can be obtain immediately with an extaversion of http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=571358 Sincerely Jean-Louis
utkarshgupta
31.01.2015 14:08
Sorry for reviving this old thread but the problem is intimidating 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 $\implies \angle CPI_A = 90$ $\implies \angle TCP = \frac{\angle B}{2}$ $\implies MT = CT \tan {\frac{B}{2}}$ $\implies MT = (s-b)\tan {\frac{B}{2}} = r$
Mahdi_Mashayekhi
07.04.2022 16:22
Claim : $CC'I_aP$ and $BB'I_aQ$ are cyclic. Proof : $\angle CI_aP = \angle CI_aB = \angle 90 - \frac{\angle A}{2} = \angle AI_aB' = \angle AC'B' = \angle CC'P$. we prove the other one with same approach. $\angle BCM = \angle BCP = \angle BQP = \angle BQB' = \angle BI_aB' = \frac{\angle B}{2}$ and $\angle CBM = \angle CBQ = \angle CPQ = \angle CPC' = \angle CI_aC' = \frac{\angle C}{2}$ so triangles $BIM$ and $CMB$ are congruent so $I,M$ have same distance from $BC$ so ength of the perpendicular from $M$ to $BC$ is equal to $r$.