equal angles related to the touchpoints of the incircle lead to isosceles

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equal angles related to the touchpoints of the incircle lead to isosceles

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Isosceles Triangle

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Source: V.A. Yasinsky Geometry Olympiad 2018 X-XI advanced p5 [Ukraine]

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#1 h Oct 1, 2018, 9:06 PM • 1 Y

Y by Adventure10

The inscribed circle of the triangle $ABC$ touches its sides $AB, BC, CA$ , at points $K,N, M$ respectively. It is known that $\angle ANM = \angle CKM$ . Prove that the triangle $ABC$ is isosceles.

(Vyacheslav Yasinsky)

This post has been edited 1 time. Last edited by parmenides51, Nov 7, 2022, 9:32 PM

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#2 h Nov 10, 2018, 12:39 PM • 2 Y

Y by Adventure10, Mango247

We will use complex numbers, let WLOG the incircle be a unit circle,
$A=a; B=b ;C=c; K=k; N=n; M=m$
Hence, $a=\frac{2km}{k+m} ; b=\frac{2kn}{k+n} ; c=\frac{2mn}{m+n}$
Since, $\angle ANM =\angle CKM$ , Hence,
$\frac{\tfrac{a-n}{m-n}}{\tfrac{c-k}{m-k}} \in R \implies \frac{(a-n)(m-k)}{(m-n)(c-k)} = \overline{\left(\frac{(a-n)(m-k)}{(m-n)(c-k)}\right)}$ And since, $k,m,n$ lie on the unit circle, we can convert $a,b,c$ to $k,m,n$ and exploit their property of conjugates,
$\implies \frac{(a-n)(m-k)}{(m-n)(c-k)}=\frac{n(m-k)}{k(m-n)} \left(\frac{\tfrac{\tfrac{2}{km}}{\tfrac{1}{k}+\tfrac{1}{m}}-\tfrac{1}{n}}{\tfrac{\tfrac{2}{mn}}{\tfrac{1}{m}+\tfrac{1}{n}}-\tfrac{1}{k}} \right) \implies \frac{a-n}{c-k}=\frac{(m+n)(2n-m-k)}{(m+k)(2k-m-n)} \implies \frac{\tfrac{2km}{k+m}-n}{\tfrac{2mn}{m+n}-k}= \frac{(m+n)(2n-m-k)}{(m+k)(2k-m-n)}$
After simplifying this,
$4k^2m-2km^2-2kmn-2k^2n+kmn+kn^2-2kmn+m^2n+mn^2=4mn^2-2m^2n-2kmn-2kmn+km^2+k^2m-2kn^2+kmn+k^2n \implies \sum_{cyc} km(k-m) =0$ This is equivalent to, $(k-m)(m-n)(n-k)=0 \implies \text{ either } k=m ; k=n \text{ or } m=n$ which indicates after checking properly that atleast one of the three conditions satisfy and and all of these conditions lead to that $\Delta ABC$ is isosceles

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#3 h Nov 10, 2018, 4:16 PM • 3 Y

Y by Adventure10, Mango247, Mango247

parmenides51 wrote:

The inscribed circle of the triangle $ABC$ touches its sides $AB, BC, CA$ , at points $K,N, M$ respectively. It is known that $\angle ANM = \angle CKM$ . Prove that the triangle $ABC$ is isosceles.

(Vyacheslav Yasinsky)

My solution:
We know that $AN,BM,CK$ are concurrent at $I$ . Let $AN,CK$ intersect the inscribed circle at the second points are $G,H$ , respectively.
It is easy to see that $GH \parallel AC$ .
$\Rightarrow \dfrac{CH}{AG}=\dfrac{IH}{GI}=\dfrac{NH}{GK} \Rightarrow \triangle NHC \sim \triangle KGA \Rightarrow \angle HNC=\angle GKA \Rightarrow NH=KG$ . It implies $AK=NC$ .
Q.E.D

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