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incenter of ABC lies on (ACQ) , AN = NM = MC
parmenides51   1
N Jul 10, 2022 by StarLex1
Source: 2006 Rusanovsky Lyceum Olympiad p139
On the sides $AB$ and $BC$ of the triangle $ABC$, points $N$ and $M$ are marked, respectively, such that $AN = NM = MC$. Let $Q$ be the intersection point of the segments $AM$ and $CN$. Prove that the center of the inscribed circle of triangle $ABC$ lies on the circumcircle of triangle $ACQ$.
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parmenides51
Jun 29, 2022
StarLex1
Jul 10, 2022
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incenter of ABC lies on (ACQ) , AN = NM = MC
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Source: 2006 Rusanovsky Lyceum Olympiad p139
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parmenides51
30652 posts
parmenides51
#1 h  Jun 29, 2022, 1:02 PM
Y by
On the sides $AB$ and $BC$ of the triangle $ABC$, points $N$ and $M$ are marked, respectively, such that $AN = NM = MC$. Let $Q$ be the intersection point of the segments $AM$ and $CN$. Prove that the center of the inscribed circle of triangle $ABC$ lies on the circumcircle of triangle $ACQ$.
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StarLex1
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StarLex1
#2 h  Jul 10, 2022, 5:02 PM
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$\angle{NAM} = x,\angle{NMA} = x,\angle{MNC}=a, \angle{NCM} = a$
By LoS
$\frac{MC}{\sin(x+a)} = \frac{QM}{\sin(a)}$
$\frac{NM}{\sin(x+a)} = \frac{QM}{\sin(x)}$
$\sin(a) = \sin(x)$
$a = x $ or $180-a = x$
if $180-a=x$ then $\angle{ANQ} = $ $180-x-a-a= -a$ which is absurd
$a= x$
by thales since $\frac{BN}{NA} = \frac{BM}{MC}$ then implicating it is parallel
let we extend BQ and meet AC at K then $AK=CK$ by ceva
$\frac{BA}{AK}=\frac{BC}{CK}$ implicating BK is also bisector thus Q is incenter
This post has been edited 1 time. Last edited by StarLex1, Jul 10, 2022, 5:09 PM
Reason: typo
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