Let $I$ be the incenter of a nonisosceles triangle $ABC$. Prove that there exists a unique pair of points $M$, $N$ lying on the sides $AC$, $BC$ respectively, such that $\angle AIM = \angle BIN$ and $MN|| AB$.
Problem
Source: Sharygin 2018
Tags: geometry
Vrangr
04.04.2018 22:25
I trig bashed this, it got a tad bit bloody but still pretty manageable. @3below, that's what I proved with this long bash.
anantmudgal09
04.04.2018 22:29
One of my favourites from this contest TheDarkPrince wrote: Let $I$ be the incenter of a nonisosceles triangle $ABC$. Prove that there exists a unique pair of points $M$, $N$ lying on the sides $AC$, $BC$ respectively, such that $\angle AIM = \angle BIN$ and $MN|| AB$. Let $P=\overline{AN} \cap \overline{BM}$. Then $\overline{CP}$ is a median in $\triangle CAB$. However, by isogonality lemma, we conclude that $\overline{IP}, \overline{IC}$ are isogonal lines in angle $AIC$. Thus, $\overline{PI} \perp \overline{AC}$ and $\overline{CP}$ is a median, providing a unique characterisation for the pair $(M,N)$.
WizardMath
05.04.2018 04:36
The unique points are where the tangent to the incircle at the antipode of the incircle touch point to $AB$ meets $CB,AC$. Uniqueness is easy, proceeding by contradiction to get $ABC$ isosceles.
TheDarkPrince
05.04.2018 07:34
The unique points are such that $\angle AIM=\angle BIN = 90^{\circ}$.
TheDarkPrince
05.04.2018 09:44
Vrangr wrote: I trig bashed this, it got a tad bit bloody but still pretty manageable.
Let $\angle AIM=\angle BIN=\theta$. We claim that $MN||AB$ if and only if $\theta=90^{\circ}$.
[asy][asy]
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label("$N$", (-0.972,3.566), NE * labelscalefactor);
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/* end of picture */
[/asy][/asy]
Suppose $MN||AB$. From law of sines in $\triangle AIM$ and $\triangle BIN$, we have $\frac{AM}{AI}=\frac{\sin \theta}{\sin\left(\theta+\frac{A}{2}\right)}$ and $\frac{BN}{BI}=\frac{\sin \theta}{\sin \left(\theta +\frac{B}{2}\right)}$. From law of sines in $\triangle AIB$, we have $\frac{AI}{BI}=\frac{\sin \frac{B}{2}}{\sin \frac{A}{2}}$. So, we have
\begin{align*}
\frac{\sin \frac{B}{2}\cos \frac{B}{2}}{\sin \frac{A}{2}\cos \frac{A}{2}}=\frac{\sin B}{\sin A}=\frac{AC}{AB}&=\frac{AM}{BN}\\
&=\frac{AM}{AI}\cdot \frac{BI}{BN}\cdot \frac{AI}{BI}\\
&=\left(\frac{\sin \theta}{\sin \left(\theta + \frac{A}{2}\right)}\right)\cdot \left(\frac{\sin\left(\theta + \frac{B}{2}\right)}{\sin \theta}\right)\cdot \frac{\sin \frac{B}{2}}{\sin \frac{A}{2}}\\
&=\frac{\sin \frac{B}{2}\sin\left(\theta +\frac{B}{2}\right)}{\sin \frac{A}{2}\sin\left(\theta +\frac{A}{2}\right)}.
\end{align*}
So, $\cos \frac{A}{2}\sin\left(\theta + \frac{B}{2}\right)=\cos \frac{B}{2}\sin \left(\theta + \frac{A}{2}\right)$.
\begin{align*}
\Rightarrow \sin \left(\theta + \frac{A}{2}+\frac{B}{2}\right)+\sin \left(\theta + \frac{A}{2}-\frac{B}{2}\right)&=2\cos \frac{B}{2}\sin \left(\theta +\frac{A}{2}\right)\\
&=2\cos \frac{A}{2}\sin\left(\theta +\frac{B}{2}\right)\\
&=\sin \left(\theta + \frac{A}{2}+\frac{B}{2}\right)+\sin \left(\theta + \frac{B}{2}-\frac{A}{2}\right)
\end{align*}
So, $\sin \left(\theta + \frac{B}{2}-\frac{A}{2}\right)=\sin \left(\theta + \frac{A}{2}-\frac{B}{2}\right)$. If $\left(\theta + \frac{B}{2}-\frac{A}{2}\right)=\left(\theta + \frac{A}{2}-\frac{B}{2}\right)$, $A=B$ which is not possible because $\triangle ABC$ is not isosceles.
So, $180^{\circ}=\left(\theta + \frac{A}{2}-\frac{B}{2}\right)+\left(\theta + \frac{B}{2}-\frac{A}{2}\right)=2\theta \Rightarrow \theta = 90^{\circ}$ which proves our claim.
Wizard_32
05.04.2018 12:57
My trig bash is really smooth
(1) By applying the sine rule to $\triangle BIN$, we get $sinx=\frac{BN}{IN}sin \beta$. Similarly by $\triangle AIM$, $sinx=\frac{AM}{IM}sin \alpha$.
(2) Since $MN$ is parallel to $AB$, hence by Thale's theorem, $\frac{BN}{AM}=\frac{CN}{CM}$.
(3) By applying sine rule to $\triangle CIN$, we get $\frac{CN}{IN}=\frac{sin \angle CIN}{sin \gamma}$. By $\triangle BIC$, we get $\angle CIN=180^{o}-\gamma-\beta-x=90^{o}-x+\alpha$, since $\alpha+\beta+\gamma=90^{o}$.
Thus, $\frac{CN}{IN}=\frac{cos(x-\alpha)}{sin \gamma}$, as $sin(90^{o}+\theta)=cos\theta$.
Similarly, by $\triangle CMI$, $\frac{CM}{IM}=\frac{cos(x-\beta)}{sin \gamma}$.
Now, by (1), we get $$\frac{BN}{IN}sin \beta=sinx=\frac{AM}{IM}sin \alpha \Rightarrow \frac{BN}{AM} \cdot \frac{IM}{IN}=\frac{sin \alpha}{sin \beta}$$Now using (2) and (3), we further get
$$\frac{sin \alpha}{sin \beta}=\frac{BN}{AM} \cdot \frac{IM}{IN}=\frac{CN}{CM} \cdot \frac{IM}{IN}=\frac{CN}{IN} \cdot \frac{IM}{CM}=\frac{cos(x-\alpha)}{sin \gamma} \cdot \frac{sin \gamma}{cos(x-\beta)}=\frac{cos(x-\alpha)}{cos(x-\beta)}$$Hence, $cos(x-\alpha)sin\beta=cos(x-\beta)sin\alpha$. Using the sum formula, this is equivalent to
$$(cosx \cdot cos\alpha+sinx \cdot sin\alpha)sin\beta=(cosx \cdot cos\beta+sinx \cdot sin\beta)sin\alpha \Rightarrow (cosx)(cos\alpha)(cos\beta)(tan\alpha-tan\beta)=0$$Clearly, $0 < \alpha,\beta < 90^{o}$ and $\alpha \ne \beta$ (since the triangle is non-isosceles) and so we must have $cosx=0$ or $x=90^{o}$, proving the claim.
USER378398
06.04.2018 09:37
WizardMath wrote: The unique points are where the tangent to the incircle at the antipode of the incircle touch point to $AB$ meets $CB,AC$. Uniqueness is easy, proceeding by contradiction to get $ABC$ isosceles. I did the same.
Synthetic_Potato
29.05.2018 17:17
Not a really bad Trig Bash My Solution: Let us assume that we found $M,N$ satisfying the conditions. Let $\angle AIM=\angle BIN = x$. So, from sine rule in $BIN$ and $AIM$, we get ${\frac{AI}{AM}=\frac{\sin{A/2}}{\sin{(x+A/2)}}}$ and ${\frac{BI}{BN}=\frac{\sin{B/2}}{\sin{(x+B/2)}}}$. Dividing, we get $\frac{BN}{AM}\cdot \frac{AI}{IB}=\frac{\sin{(x+A/2)}}{\sin{(x+B/2)}}$. We will use the fact that $\sin{(a+b)}=\sin a \cos b +\cos a \sin b$. Also, from sine rule in $AIB$, $\frac{AI}{IB}=\frac{\sin B/2}{\sin A/2}$. Also, $\frac{BN}{AM}=\frac{BC}{AC}=\frac{\sin A}{\sin B}$. So, $\frac{BN}{AM}\cdot \frac{AI}{IB}=\frac{\cos A/2}{\cos B/2}$ from the fact that $\sin A = 2\sin A/2 \cos A/2$. Using the identity and using these result in the previously derived equation, $$\frac{\sin x \cos A/2 + \cos x \sin A/2}{\sin x \cos B/2 + \cos x \sin B/2}=\frac{\cos A/2}{\cos B/2}$$ Note that $\cos A/2, \cos B/2 $ are non zero. So, cancelling out $\cos A/2, \cos B/2$ from the numerator and denominator of both fractions, we get $\cos x (\tan A/2 - \tan B/2)=0$. As $ABC$ is non isosceles, we get $\tan A/2 \not = \tan B/2$. So, $\cos x = 0$. Hence $x=90^\circ$. So there is only one $x$ for which the conditions are satisfied. Note that $x=90^\circ$ indeed works. That happens when $MN$ is the tangent to the incircle parallel to $AB$. Then $\angle MIA=\angle NIB=90^\circ$. Also, Trigonometric proof is also possible, to prove $MN\parallel BC$ in this case. From sine rule in triangles $AIM,BIN$ we get $\frac{\sin (90^\circ + B/2)}{\sin 90^\circ}= \sin (90^\circ+B/2) =\frac{AI}{AM}$. Similarly, $\sin (90^\circ +B/2) = \frac{BI}{BN}$. Dividing and noting that $\frac{AI}{BI}=\frac{\sin B/2}{\sin A/2}$ and $\sin (90^\circ + A/2) = \cos A/2$, and $\sin (90^\circ + B/2) = \cos B/2$. So, we get $\frac{AM}{BN}= \frac{\sin A}{\sin B}$, because $2\sin A/2 \cos A/2 = \sin A$. As $\frac{\sin A}{\sin B}=\frac{BC}{CA}$, we get $MN\parallel BC$. $\blacksquare$.
PROF65
30.05.2018 00:05
Let $D=(I) \cap AB$ .First remak that the parallel tangent to $(I)$ touch the incircle at $T$ and cuts $AC,BC$ at $M,N$ respectively.$\angle AIM=\frac 1 2 \angle DIT=90^\circ$ idem $\angle BIN=90^\circ $ suppose that we have an other $M',N'$ verifying the conditions and consider $K=AN\cap BM,K'=AN'\cap AM'$ then : $IM,IN$ are isogonal wrt $\angle AIB$ so by isogonality lemma $IK,IC $ are also isogonal similarly $IK',IC$ are isogonal wrt $\angle AIB$ hence $I,K,K'$ are collinear but $MN \parallel BC $ so $ AK$ is the median idem for $AK'$ whence $I$ is on the median which means $C AB$ is isoceles therefore there is unique a pair of $M,N$