Tuymaada 2010, Junior League, Problem 7

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Tuymaada 2010, Junior League, Problem 7

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#1 h Jul 18, 2010, 9:42 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle, $I$ its incenter, $\omega$ its incircle, $P$ a point such that $PI\perp BC$ and $PA\parallel BC$ , $Q\in (AB), R\in (AC)$ such that $QR\parallel BC$ and $QR$ tangent to $\omega$ .
Show that $\angle QPB = \angle CPR$ .

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#2 h Jul 18, 2010, 6:35 PM • 1 Y

Y by Adventure10

$PR,PQ$ cut $BC$ at $S,T$ and $BR,CQ$ cut $AP$ at $D,E,$ respectively. Line passing through $A \equiv QB \cap RC$ and $P_{\infty} \equiv QR \cap BC$ (lying at infinity) is the polar of $U \equiv BR \cap CQ$ WRT $(I).$ Then if $(I)$ touches $BC$ at $M$ $\Longrightarrow$ $U \in IM.$ We have $P(B,R,U,D)=P(B,S,M,P_{\infty})=-1$ $\Longrightarrow$ $M$ is the midpoint of $BS.$ Similarly, we have $P(C,Q,U,E)=P(C,T,M,P_{\infty})=-1$ $\Longrightarrow$ $M$ is the midpoint of $CT.$ As a result, triangles $\triangle PBS$ and $\triangle PCT$ are both isosceles $\Longrightarrow$ $\angle CPR= \angle QPB.$

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#3 h Jul 18, 2010, 7:25 PM • 3 Y

Y by primesarespecial, Adventure10, Mango247

Let $\{D\}\equiv \omega \cap BC$ , $\{E\}\equiv \omega \cap QR$ , $\{F\}\equiv AD \cap QR$ , $\{S\}\equiv PR \cap AB$ ; it is easy to see that, due to the fact that the triangles are homothetic at $A$ , $F$ is the contact of the incircle of the triangle $AQR$ with $QR$ and $ER = QF$ $(1)$ . From $BC \parallel QR$ we get $\frac{AQ}{AB}=\frac{QF}{BD}=\frac{ER}{DS}$ and, with $(1)$ we get $BD=DS$ , so $BP$ is the reflection of $PR$ in $PI$ .
Similarly $PQ$ and $PC$ , so, by difference, we get the required equality.

Best regards,
sunken rock

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#4 h Jul 21, 2010, 5:00 AM • 2 Y

Y by Adventure10, Mango247

Do an inversion with respect to the incircle. The image of $X$ is denoted as $X'$ , etc. Let $X,Y,Z$ denote the tangency points of the incircle to $BC,CA,AB$ . Let $QR$ touch the incircle at $W$ . It is clear that $W,X,Y,Z$ are not moved and that $Q',R', A',B',C'$ are the midpoints of $Z'W',W'Y',Y'Z',Z'X',X'Y'$ . Also note that $APYIZ$ is cyclic because those points lie on the circle with diameter $AI$ . Consequently, $P'$ lies on the image of that circle which must be a line through $Y'Z'$ . Also $I,P,P'$ are collinear. Further, it is obvious that $I,W,P$ are collinear, so $P'=Y'Z'\cap IW'$ .

Now observe that $\triangle Z'X'P'\sim\triangle W'Y'P'$ and $B'$ , $R'$ are midpoints of $Z'X'$ and $W'Y'$ , so we can easily conclude that $\triangle B'X'P\sim\triangle R'Y'P'$ , so $\angle P'B'X'=\angle Y'R'P'$ .

It is clear that $\angle IB'X'=\angle IR'Y'$ , so subtracting from above, we get $\angle P'B'I=\angle P'R'I$ . $\star$ . By symmetry, we also have $\angle P'C'I=\angle P'Q'I$ . Subtracting these relations gives us

$\angle P'Q'I-\angle P'B'I=\angle P'C'I-\angle P'R'I$

In general for an inversion, $\angle RSI=\angle S'R'I$ where $I$ is the center of inversion and $R',S'$ denote the images of $R,S$ , so using that above, we get

$\angle QPI-\angle BPI=\angle CPI-\angle RPI$
which of course reduces to the desired result of
$\angle QPB=\angle CPR$ .

As a remark, this solution is pretty straightforward application of inversion, you just need to observe in the inverted diagram how to prove the angle equality.

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primesarespecial

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primesarespecial

#5 h Jul 3, 2021, 12:40 PM

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Since $\omega$ is an excircle to $AQR$ , we have $QF=ER$
Then $\frac{QF}{BD}=\frac{AF}{FD}=\frac{PE}{ED}=\frac{ER}{SD}$

Thus $SD=BD$ or $\triangle PBS$ is isosceles .
Now,
$\frac{QJ}{AP}=\frac{AQ}{QB}=\frac{AR}{RC}=\frac{RH}{AP}$

Thus,
$QJ=RH$
And
$\frac{QJ}{GB}=\frac{PJ}{BJ}=\frac{PR}{RC}=\frac{RH}{SC}$

Thus, $\triangle PGC$ is isosceles and we are done.

This post has been edited 1 time. Last edited by primesarespecial, Jul 3, 2021, 12:40 PM

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