Prove that CMN is tangent to S1 and S2

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Prove that CMN is tangent to S1 and S2

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Source: All-Russian MO 2000

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#1 h Dec 30, 2012, 1:00 PM • 1 Y

Y by Adventure10

Let $E$ be a point on the median $CD$ of a triangle $ABC$ . The circle $\mathcal S_1$ passing through $E$ and touching $AB$ at $A$ meets the side $AC$ again at $M$ . The circle $S_2$ passing through $E$ and touching $AB$ at $B$ meets the side $BC$ at $N$ . Prove that the circumcircle of $\triangle CMN$ is tangent to both $\mathcal S_1$ and $\mathcal S_2$ .

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JustN

#2 h Dec 30, 2012, 5:34 PM • 3 Y

Y by SSaad, Adventure10, Mango247

Since $CD$ is median and $E$ is on $\mathcal{S}_1$ and $\mathcal{S}_2$ then $CD$ is radical axis of those two circles. So $CB \cdot CN = CM \cdot CA$ and so $A,B,N,M$ are cyclic and $\angle MNC = \angle BAM$
Let tangent to $\mathcal{S}_2$ intersect $CD$ at $T$ . Then $\angle TNB = \angle ABN$ and so we get $\angle TNM = \angle NCM$ which means that $NT$ is also tangent to circumcircle of $CMN$ .
Similary we get that $\mathcal{S}_1$ is tangent to $CMN$

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#3 h Sep 30, 2017, 3:22 PM • 1 Y

Y by Adventure10

Let $F$ be the other common point of $\mathcal{S}_1, \mathcal{S}_2$ . By power of point at $D$ , we see $F$ lies on $\overline{CD}$ too. Now $CM \cdot CA=CE \cdot CF=CN \cdot CB$ so $ABMN$ is cyclic. Now parallel from $C$ to $\overline{AB}$ is tangent to $\odot(CMN)$ . Consequently, the claim follows by homothety about $M, N$ .

This post has been edited 2 times. Last edited by anantmudgal09, Sep 30, 2017, 4:26 PM

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NMN12

#4 h Dec 1, 2019, 8:03 AM • 3 Y

Y by SSaad, Adventure10, Mango247

$CD$ is the radical axis of $\mathcal{S}_1$ and $\mathcal{S}_2$ . We have
$AC\cdot CM = CN \cdot BC.$ Thus $ABMN$ is cyclic. Let $MN\cap \mathcal{S}_1=K$
$\angle MKA=\angle MAB = \angle MNB$ so (CMN) and $\mathcal{S}_1$ are tangent. Analogously (CMN) and $\mathcal{S}_2$ are tangent.

This post has been edited 1 time. Last edited by NMN12, Dec 1, 2019, 8:05 AM
Reason: Improved latex symbols

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