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Source: Izho 2020 problem 4

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47 posts

#1 h Jan 11, 2020, 8:42 AM • 3 Y

Y by VicKmath7, Adventure10, Mango247

In a scalene triangle $ABC$ $I$ is the incentr and $CN$ is the bisector of angle $C$ . The line $CN$ meets the circumcircle of $ABC$ again at $M$ . The line $l$ is parallel to $AB$ and touches the incircle of $ABC$ . The point $R$ on $l$ is such. That $CI \bot IR$ . The circumcircle of $MNR$ meets the line $IR$ again at S. Prpve that $AS=BS$ .

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#2 h Jan 11, 2020, 9:09 AM • 3 Y

Y by GeoMetrix, Adventure10, Mango247

IZHO 2020 P4 wrote:

In a scalene triangle $ABC$ $I$ is the incentr and $CN$ is the bisector of angle $C$ . The line $CN$ meets the circumcircle of $ABC$ again at $M$ . The line $l$ is parallel to $AB$ and touches the incircle of $ABC$ . The point $R$ on $l$ is such. That $CI \bot IR$ . The circumcircle of $MNR$ meets the line $IR$ again at S. Prove that $AS=BS$ .

Let $MS\cap AB=K$ . So, it suffices to show that $SK\perp AB$ . Hence, it suffices to show that $INKS$ is a cyclic quadrilateral. So, $\angle RSM=\angle INB=\angle RNM\implies \angle INR=\angle INA$ . This is what we are supposed to show in order to prove the conclusion.

Now we can safely elimate the points $M,K,S$ and $\odot(MNR)$ from the diagram. So now restate the problem as follows.

Restated Problem wrote:

$CAB$ is a triangle with $C$ as its apex. Let the line $\ell$ be tangent to $\odot(I)$ and $\|$ to $AB$ . Let $R\in\ell$ such that $\angle CIR=90^\circ$ . Then show that $\angle ANC=\angle CNR$

For this just note that if $\ell\cap CI=K$ and the tangency point made by $\ell,AB$ with $\odot(I)$ be $X,D$ respectively then $\triangle KXI\cong\triangle NDI\implies \angle KRI=\angle NRI$ . So $\angle CNA=90^\circ-\angle DIN=90^\circ-\angle XIK=90^\circ-\angle XRI=90^\circ-\angle IRN=\angle CNR$ and this is what we were supposed to prove. $\blacksquare$

This post has been edited 15 times. Last edited by amar_04, Dec 25, 2020, 6:50 PM

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UzbekMathematician

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UzbekMathematician

#4 h Jan 11, 2020, 2:07 PM • 4 Y

Y by Mathuzb, MisterJ, Adventure10, Mango247

RnstTrjyn wrote:

In a scalene triangle $ABC$ $I$ is the incentr and $CN$ is the bisector of angle $C$ . The line $CN$ meets the circumcircle of $ABC$ again at $M$ . The line $l$ is parallel to $AB$ and touches the incircle of $ABC$ . The point $R$ on $l$ is such. That $CI \bot IR$ . The circumcircle of $MNR$ meets the line $IR$ again at S. Prpve that $AS=BS$ .

$S'\in IR, MS' \perp AB$ . We know that $MS'$ is a perpendicular bisector of $AB$ . So, it suffices to show that $MNS'R$ is a cyclic.
Let $CN \cap l=T, MS' \cap AB=K , IR \cap AB=L$ . So, $S'INK$ is a cyclic and $LNRT$ is a romb. Since $\angle INR=\angle ANI=\angle IS'M$ . $\implies$ $MNS'R$ is a cyclic quadrilateral. $\blacksquare$

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#5 h Apr 14, 2020, 1:25 PM

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Let $D$ and $D'$ be the tangency point of the incircle with side $BC$ and $l$ respectively and $K=CI\cap l$ .

Lemma 1: $IN=IK$
It is easy to show that $D,I,D'$ are collinear, $DD' \bot BC$ and $DD'\bot l$ . Which implies $\triangle IDN \cong \triangle ID'K$ so, $IN=IK$ .

Lemma 2: $\angle KID'=\angle KRI$
$\angle KID'+ \angle D'IR=\angle D'RI + \angle D'IR=90$
$\Rightarrow \angle KID'=\angle KRI$

Lemma 3: $ID' \parallel SM$
From, Lemma 2 we get $IN=IK$ and $RI\bot NK$ so, $RK=RN$ and $\angle KRI=\angle IRD$ .
Now,
$\angle AID'=\angle KRI=\angle IRD=\angle SRD=\angle SMN=\angle SMA$ So, $ID' \parallel SM$ .

So, $SM$ is perpendicular to $BC$ and $M$ lies on the perpendicular bisector of $BC$ . Which implies $AS=BS$ .

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#6 h Dec 25, 2020, 6:20 PM

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Let $CM\cap \ell=L$ and $SM\cap AB=K$ . Since $AM=BM$ , we need to prove that $SM\perp AB \Leftrightarrow \angle SKA=90^\circ \Leftrightarrow \angle NKM=90^\circ$

$\ell$ is parallel to $AB$ and tangent to the incircle of $ABC$ so we get $LI=IN$ then $LR=RN$ .

$\angle NMK=\angle NMS=\angle NRS=\angle NRI=\angle IRL=90^\circ - \angle ILR=90^\circ - \angle NLR=90^\circ - \angle LNA=90^\circ - \angle MNK$ so $\angle NMK+\angle MNK=90^\circ$ then $\angle NKM=90^\circ$ so we are done.

This post has been edited 1 time. Last edited by BarisKoyuncu, Dec 25, 2020, 6:20 PM
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#7 h Jan 9, 2021, 11:59 PM

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Nice one

$\color{black}\rule{25cm}{1pt}$
First let's add some points.
Let $S'$ be the intersection of the perpendicular bisector of $AB$ with $RI$ , let $N'$ be the intersection of $l$ with $CI$ and let $M'$ be the midpoint of $AB$ .

We start off with this killer lemma.
$\color{red}\rule{25cm}{0.5pt}$
Lemma: $RN$ is tangent to the incircle of $ABC$ .
Proof (Sketch):
Throw the configuration onto the complex plane, with the incircle as the unit circle and let $d=1$ , and add the rest of the touch points.
Then easily get $r$ since $R\in l$ and $CI \perp RI$ . Then calculate $n$ and calculate the touch point (which we call $E$ ) of the other tangent of $R$ onto the incircle, to get that $RN$ contains $E$ . Thus $RN$ is tangent to the incircle of $ABC$ .
$\color{red}\rule{25cm}{0.5pt}$

Now back to the problem.
From our lemma we have that $\angle NRN' = 2 \angle N'RI = 2(90-\angle A- \frac{1}{2}\angle C)$ .
But this implies that $\angle RNA = \angle NRN'$ .
Now we have that $\angle RNM = \angle ANR + \angle ANM = \angle B + \frac{1}{2} \angle C$

But notice how we have that $S'M'NI$ is a cyclic quadrilateral, since we have that $\angle S'IN = S'M'N = 90$ .
This implies that $\angle RS'M = \angle M'NI = \angle B + \frac{1}{2} \angle C$ .
Thus we have that $RS'NM$ is a cyclic quadrilateral, this implies that $S' \equiv S$ .

Thus we have that $SA=SB$ .

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#8 h Jan 12, 2021, 3:35 PM • 1 Y

Y by Mango247

WLOG let $CA > CB.$ First, observe that the reflection of $N$ across $I$ must lie on $\ell$ by homothety. This implies that
$\angle A + \frac{1}{2}\angle B = \angle (\ell, RI) =\angle IRN = \angle SMN,$ so $MS$ must pass through the midpoint of the arc $ACB,$ meaning that $MS$ perpendicularly bisects $AB.$ $\blacksquare$

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#9 h Nov 21, 2021, 2:40 PM

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Let incircle of $ABC$ touches $l$ and $AB$ at $T$ and $D$ , respectively. Consider $P = IR \cap AB$ . Then easy to see the triangles $TIR$ and $PID$ are congruent which follows $PI = IR$ . Therefore segment $NI$ is median and altitude of triangle $PNR$ which implies it's angle bisector, too. So $\angle INR = \angle INP = \angle TIR = \angle ISM$ where the last one is true b/c of $MNSR$ is cyclic. Then $SM \perp AB$ which gives us required result.

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548 posts

#10 h Dec 13, 2021, 7:19 PM • 1 Y

Y by A1inur

Suppose we cut the perpendicular $IR$ from $M$ to $AB$ at $K$ . It is easy to prove that $MNKR$ is cyclic.

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Lemmas

#11 h Jan 9, 2025, 7:58 AM

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Let $L$ be the midpoint of $NR$ . Then $IL\parallel AB$ and after that just angle chasing

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#12 h Mar 3, 2025, 8:13 AM

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RnstTrjyn wrote:

In a scalene triangle $ABC$ $I$ is the incenter and $CN$ is the bisector of angle $C$ . The line $CN$ meets the circumcircle of $ABC$ again at $M$ . The line $l$ is parallel to $AB$ and touches the incircle of $ABC$ . The point $R$ on $l$ is such. That $CI \bot IR$ . The circumcircle of $MNR$ meets the line $IR$ again at S. Prove that $AS=BS$ .

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