Prove that the line $MN$ is tangent to the inscribed circle

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Prove that the line $MN$ is tangent to the inscribed circle

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Source: Peru TST

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#1 h Feb 3, 2019, 4:26 PM • 1 Y

Y by Adventure10

Let $I$ be the incenter of the $ABC$ triangle. The circumference that passes through $I$ and has center
in $A$ intersects the circumscribed circumference of the $ABC$ triangle at points $M$ and
$N$ . Prove that the line $MN$ is tangent to the inscribed circle of the $ABC$ triangle.

This post has been edited 1 time. Last edited by janssv.200603, Feb 3, 2019, 5:21 PM

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Synthetic_Potato

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Synthetic_Potato

#2 h Feb 3, 2019, 4:44 PM • 1 Y

Y by Adventure10

Let $MN$ intersect $AB,AC$ at points $P,Q$ . It suffices to show that $BPQC$ has an incircle. Thus it suffices to show that $IP$ bisects $\angle BPQ$ . Now see that from shooting lemma $AM^2=AP\cdot AB = AI^2$ . Thus, $AI$ is tangent ti $\odot (BIP)$ . Thus angle chase yields $\angle BPI = 90^\circ - \frac{\angle C}{2}$ . Also note since $AM=AN$ we have that $PQ$ is antiparallel to $BC$ . Thus, $\angle QPA = \angle C$ . Thus we have $\angle QPI = 90^\circ - \frac{\angle C}{2}$ . So $PI$ bisects $\angle BPQ$ and so $I$ is the incenter of $BPQC$ which means $MN$ is tangent to incircle. $\blacksquare$ .

Also, I'm interested to know the source of this problem.

EDIT: @below, a simpler proof of the incircle mapping to mixtilinear is here:
Let $D,E$ be the intouch points on $AB$ and $D',E'$ be the mixtilinear touch point on $AB,AC$ . See that $\angle D'IA = 90^\circ = \angle IDA$ . So, $AI$ is tangent to the circle $IDD'$ . So, $AI^2=AD\cdot AD'$ . So, $D$ maps to $D'$ under this inversion. Also, $AB$ and $AC$ map to themselves. Since incircle is tangent to $AB,AC$ we see that it maps to circle tangent to $AB,AC$ at $D',E'$ which is the mixtilinear incircle.

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#3 h Feb 3, 2019, 4:59 PM • 3 Y

Y by karitoshi, AlastorMoody, Adventure10

Solution

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dchenmathcounts

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dchenmathcounts

#4 h Feb 22, 2019, 4:24 AM • 1 Y

Y by Adventure10

Which year of Peru TST is this?

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9795 posts

jayme

#5 h Feb 22, 2019, 7:35 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers

http://www.artofproblemsolving.com/community/c6t48f6h1096581_mn_is_tangent_to_the_incircle
https://artofproblemsolving.com/community/c6t48f6h1425177_geometry

Sincerely
Jean-Louis

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#6 h Feb 23, 2019, 3:30 PM • 1 Y

Y by Adventure10

Here is my solution for this problem
Solution
Let $K$ , $L$ be intersections of $MN$ with $AC$ , $AB$
It's easy to prove that $BCKL$ is cyclic quadrilateral
But: $\widehat{AML}$ = $\widehat{ABM}$ then $AI^2$ = $AM^2$ = $AL$ . $AB$
So: $\widehat{ALI}$ = $\widehat{AIB}$ = $90^o$ + $\dfrac{\widehat{ACB}}{2}$ or $\widehat{BLI}$ = $90^o$ $-$ $\dfrac{\widehat{ACB}}{2}$
Hence: $LI$ is internal bisector of $\widehat{BLK}$
Similarly: $KI$ is internal bisector of $\widehat{CKL}$
Therefore: $BCKL$ is tangential quadrilateral or $MN$ tangents ( $I$ )

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#7 h Feb 27, 2019, 8:49 PM • 2 Y

Y by Adventure10, Mango247

dchenmathcounts wrote:

Which year of Peru TST is this?

Is (Peru) TST IMO 2015 (Day 2, Problem 8) Link: https://onemperu.files.wordpress.com/2018/12/Selectivos-IMO.pdf (In Spanish)

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#8 h May 16, 2021, 4:12 PM • 1 Y

Y by Bradygho

janssv.200603 wrote:

Let $I$ be the incenter of the $ABC$ triangle. The circumference that passes through $I$ and has center
in $A$ intersects the circumscribed circumference of the $ABC$ triangle at points $M$ and
$N$ . Prove that the line $MN$ is tangent to the inscribed circle of the $ABC$ triangle.

Peru TST for IMO :blush:

:blush:

, lets solve this...
We will use inversion and angle chasing:
Denote $\mathcal I(A, AI)$ the inversion with center $A$ and the radius $AI$ .
Note that $\mathcal I(M)=M$ and $\mathcal I(N)=N$ .
Now $BMNC$ is cyclic and hence $\mathcal I(B)$ and $\mathcal I(C)$ must lie on $MN$ but note that the inversed point is colinear with the original point and the center of invertion and hence:
$\mathcal I(B)=MN \cap AB=B'$ and $\mathcal I(C)=MN \cap AC=C'$ and then $B'C'CB$ is cyclic.
Now the problem is equivalent to prove that the angle bisectors on $B'C'CB$ are concurrent at $I$ .
Since $B'C'CB$ is cyclic $\angle ABC=\angle AC'B'=180º-\angle B'C'C$ and $\angle ACB=\angle AB'C'=180º-\angle BB'C'$ .
By another propety of inversion $AI^2=AC' \cdot AC \implies \triangle AC'I \sim \triangle AIC \implies \angle AC'I=\angle AIC$ and $AI^2=AB' \cdot AB \implies \triangle AB'I \sim \triangle AIB \implies \angle AB'I=\angle AIB$ .
Now by angle chasing:
$\angle AIC=\angle AC'B'+\angle B'C'I=\angle ABC+\angle B'C'I=90º+\frac{\angle ABC}{2} \implies \angle B'C'I=\angle IC'C=90º-\frac{\angle ABC}{2}$
This means $\frac{\angle B'C'C}{2}=90º-\frac{\angle ABC}{2}=\angle B'C'I=\angle IC'C$ and that means $CI$ bisects $\angle B'C'C$ .
By the same way you get $BI$ bisects $\angle BB'C'$ and then $B'C'CB$ has an incircle.
In fact the incircle of $\triangle ABC$ and the incircle of $B'C'CB$ are the same and that means $B'C'$ tangent to the incircle and that means $MN$ tangent to the incircle of $\triangle ABC$ .
Thus we are done :blush:

:blush:

This post has been edited 1 time. Last edited by MathLuis, May 16, 2021, 4:20 PM

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#9 h May 17, 2021, 9:18 AM • 2 Y

Y by Math-48, Mango247

Let $T$ be the midpoint of arc $BC$ not containing $A$
Let the tangents to $\omega$ at $T$ intersect $(ABC)$ at $M'$ and $N'$ . by poncelet porism $M'N'$ tangents $\omega$ and $I$ is the incentre of $\triangle TM'N'$ so $A$ is the midpoint of arc $M'N'$ so by fact5, $AI=AM'=AN'$ so $M=M'$ and $N=N'$ and done!

This post has been edited 1 time. Last edited by Muaaz.SY, May 17, 2021, 9:19 AM

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#10 h Apr 20, 2025, 8:26 AM

Y by

Let $L$ be the midpoint of minor arc $BC$ and let $M'$ and $N'$ be points on $(ABC)$ such that $LM'$ and $LN'$ is tangent to the incircle
Let $N'L$ meet the incircle at $X$
By using sine rule in $\triangle IXL$ --
$\frac{r}{sin \alpha}=IL=2Rsin\angle \frac{A}{2}$ It is well known that -
$r=4Rsin\angle \frac{A}{2}sin\angle \frac{B}{2}sin\angle \frac{C}{2}$ $\therefore sin\alpha =2sin\angle \frac{B}{2}\angle \frac{C}{2}$ $\frac{AN'}{sin \alpha}=2R$ $\therefore AN'=4Rsin\angle \frac{A}{2} sin\angle \frac{B}{2}sin \angle \frac{C}{2}=r$

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