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Source: 2017 Belarus Team Selection Test 3.1

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184 posts

#1 h Mar 31, 2019, 5:57 PM • 4 Y

Y by GeometryWorld, HWenslawski, Adventure10, lian_the_noob12

Let $I$ be the incenter of a non-isosceles triangle $ABC$ . The line $AI$ intersects the circumcircle of the triangle $ABC$ at $A$ and $D$ . Let $M$ be the middle point of the arc $BAC$ . The line through the point $I$ perpendicular to $AD$ intersects $BC$ at $F$ . The line $MI$ intersects the circle $BIC$ at $N$ .
Prove that the line $FN$ is tangent to the circle $BIC$ .

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2125 posts

#2 h Mar 31, 2019, 6:15 PM • 3 Y

Y by hansu, SenatorPauline, Adventure10

Solution

This post has been edited 7 times. Last edited by AlastorMoody, Aug 24, 2020, 2:11 PM
Reason: nothing is NICEEE in this WARUDOOO, everything SUCKKKKKKKSSSSS

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Synthetic_Potato

114 posts

Synthetic_Potato

#3 h Mar 31, 2019, 6:21 PM • 3 Y

Y by AlastorMoody, Pluto1708, Adventure10

Notice that since $\angle DBM= \angle DCM = 90^\circ$ $MB,MC$ are tangent to $\odot (BIC)$ . Also it is easy to show that $FI$ is tangent to $\odot (BIC)$ . Let $FN'$ be the other tangent from $F$ to $\odot (BIC)$ . From la hire's theorem see that since $F$ lies on polar of $M$ WRT $\odot (BIC)$ , $M\in IN'$ . So $N=N'$ . $\blacksquare$ .

This post has been edited 1 time. Last edited by Synthetic_Potato, Mar 31, 2019, 6:22 PM

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1618 posts

#4 h Apr 1, 2019, 11:06 AM • 2 Y

Y by Adventure10, Mango247

Since: $MB$ , $MC$ tangent ( $BIC$ ) at $B$ , $C$ , we have: $BICN$ is harmonic quadrilateral
So: tangents at $I$ , $N$ of ( $BIC$ ) and $BC$ concurrent or $FN$ tangents ( $BIC$ ) at $N$

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386 posts

#5 h Mar 20, 2020, 3:06 PM

Y by

https://i.imgur.com/s8jk4pc.png

Please note that point $A'$ isn't needed at all.

So let's first of define an inversion , $\psi(\Gamma_{\triangle BIC})$ .
Now we have the following $B \xrightarrow{\psi} B$ , $C \xrightarrow{\psi} C$ , $I \xrightarrow{\psi} I$
$\Gamma_{\triangle ABC} \xrightarrow{\psi} \overline{BC}$ and $M \xrightarrow{\psi} M'$ , $F \xrightarrow{\psi} F'$
The center of $\Gamma_{\triangle BIC}$ is obviously $D$
So let's start with the angle-chase:
We have the following $\angle DF'M = \angle DM'F = 90$ and $\angle DF'I = \angle DIF = 90$ ,
thus the points $M,I,F',N$ are all collinear,thus
$90 = \angle DF'N = \angle DNF$
Thus the line $FN$ is tangent to $\Gamma_{\triangle BIC}$ .....

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110 posts

#6 h Mar 21, 2020, 6:22 AM

Y by

Notice that $M$ is the polar point of $BC$ and $FI$ is tangent to $\odot BIC$

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259 posts

#7 h Mar 26, 2020, 4:42 PM

Y by

Since $M, I, N$ are collinear and $MB$ and $MC$ are tangent to $(BIC), BICN$ is harmonic which implies that $FN$ is tangent to $(BIC$ )

This post has been edited 1 time. Last edited by lahmacun, Apr 3, 2020, 1:01 PM

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1482 posts

#8 h Sep 6, 2021, 12:40 AM

Y by

By incenter-excenter lemma $D$ is center of $(BIC)$ thus $FI$ its tangent to $(BIC)$ and we take polars w.r.t. $(BIC)$
Clearly $\mathcal P_M=BC$ and $\mathcal P_I=FI$ thus by la'hire since $M,I,N$ are colinear we have that $BC, FI, \mathcal P_N$ are concurrent thus $\mathcal P_N=NF$ as desired...

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luosw

#9 h Sep 6, 2021, 1:22 AM

Y by

Oh, how to use tikz in AoPS?

This post has been edited 1 time. Last edited by luosw, Sep 6, 2021, 1:32 AM

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#10 h Sep 6, 2021, 1:23 AM • 1 Y

Y by jhu08

Let the $A$ -mixtilinear incircle touch $(ABC)$ at $T$ . It's well-known that $T \in MI$ and $TI = TN$ .

By the Incenter-Excenter Lemma, $D$ is the center of $(BIC)$ . Because $\angle FID = 90^{\circ}$ , it follows that $FI$ is tangent to $(BIC)$ .

Since $\angle DTI = \angle DTM = 90^{\circ}$ , we know $T$ lies on the circle with diameter $DI$ . Hence, the Radical Axis Theorem on $(DTI), (BIC), (ABC)$ implies $II \equiv IF, DT, BC$ are concurrent at $IF \cap BC = F$ .

Thus, $F$ lies on $DT$ , which is obviously the perpendicular bisector of $IN$ . But this implies $FI = FN$ , which clearly suffices. $\blacksquare$

Remark: The article linked below contains all of the well-known facts I cited in this solution.
https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2020-02/mr_2_2020_mixtilinear.pdf

GeoGebra: https://www.geogebra.org/geometry/nwmgedzz.

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luosw

#11 h Sep 6, 2021, 1:48 AM • 1 Y

Y by jhu08

https://i.loli.net/2021/09/06/bporku3sjqNn87w.png

We can easily find that $MB,MC$ is tangent to the circle $\varGamma$ . So quadrilateral $BICN$ is a harmonic quadrilateral. So $B,C;S,F$ are harmonic points. So $MN$ is the polar of $F$ . $\boxed{}$

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primesarespecial

364 posts

primesarespecial

#12 h Sep 6, 2021, 11:18 AM • 1 Y

Y by jhu08

We have that $MB,MC$ are tangent to $(BIC)$ .
Now, $F$ lies on the polar of $M$ ,so by La Hire's theorem, $M$ lies on the polar of $F$ .
Now,by construction $FI$ is tangent to $(BIC)$ .
So $IM$ is the polar of $F$ and hence $N$ is the other tangency point from $F$ .

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Bonime

#13 h Apr 1, 2025, 1:14 PM

Y by

By IE-lemma, $FI$ is tangent to $(BIC)$ . If $N*$ is the other tangency point of $F$ in $(BIC)$ , since $F \in m$ we get that $M \in f$ (wrt to $(BIC)$ ) so $M - I - N*$ and we´re done. $\blacksquare$

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