Problem 5 -- Making A Point Here

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Problem 5 -- Making A Point Here

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Source: 46th Austrian Mathematical Olympiad National Competition Part 2 Problem 5

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274 posts

#1 h Jul 14, 2018, 5:59 PM • 4 Y

Y by PavelMath, Adventure10, Mango247, AlexCenteno2007

Let I be the incenter of triangle $ABC$ and let $k$ be a circle through the points $A$ and $B$ . The circle intersects

* the line $AI$ in points $A$ and $P$
* the line $BI$ in points $B$ and $Q$
* the line $AC$ in points $A$ and $R$
* the line $BC$ in points $B$ and $S$

with none of the points $A,B,P,Q,R$ and $S$ coinciding and such that $R$ and $S$ are interior points of the line segments $AC$ and $BC$ , respectively.

Prove that the lines $PS$ , $QR$ , and $CI$ meet in a single point.

(Stephan Wagner)

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185 posts

#2 h Jul 14, 2018, 7:24 PM • 2 Y

Y by Adventure10, Mango247

RockmanEX3 wrote:

Let I be the incenter of triangle $ABC$ and let $k$ be a circle through the points $A$ and $B$ . The circle intersects

* the line $AI$ in points $A$ and $P$
* the line $BI$ in points $B$ and $Q$
* the line $AC$ in points $A$ and $R$
* the line $BC$ in points $B$ and $S$

with none of the points $A,B,P,Q,R$ and $S$ coinciding and such that $R$ and $S$ are interior points of the line segments $AC$ and $BC$ , respectively.

Prove that the lines $PS$ , $QR$ , and $CI$ meet in a single point.

(Stephan Wagner)

Very nice problem!
Denote by $K$ and $L$ the intersection points of the line which passes through the point $I$ parallel to $AB$ with the sides $AC$ and $BC$ respectively. By direct counting of the angles it iseasy to prove that
1) points $K$ , $Q$ , $R$ , $I$ lie on one circle $\omega_1$ ;
2) points $L$ , $P$ , $S$ , $I$ lie on one circle $\omega_2$ ;
3) points $K$ , $R$ , $S$ , $L$ li on one circle $\omega_3$ .
Then point $C$ is the radical center of the circles $\omega_1$ , $\omega_2$ and $\omega_3$ and $(CI)$ is the radical axis of the circles $\omega_1$ and $\omega_2$ .
Denote the circle which passes through the points $A$ , $R$ , $S$ , $B$ by $\Omega$ . Then the radical center of the circles $\omega_1$ , $\omega_2$ and $\Omega$ is the intersection point of the lines $(CI)$ , $(QR)$ and $(PS)$ . Hence these lines are concurrent.

This post has been edited 2 times. Last edited by PavelMath, Jul 14, 2018, 7:26 PM

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#3 h Jul 14, 2018, 8:25 PM • 1 Y

Y by Adventure10

My solution:
It is easy to see by angle chasing that $PQ \perp CI$ . Moreover, $\angle PQR = \angle PAR = \angle BAP = \angle IQP$ , hence the reflection across the line $PQ$ maps $QI$ to $QR$ . Analogously, this reflection maps $PI$ to $PS$ and the line $CI$ to itself. Thus $QR$ , $PS$ and $CI$ concur at the reflection of $I$ across the line $PQ$ .

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9775 posts

jayme

#4 h Jul 16, 2018, 2:08 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
just applie the Pascal's theorem...

Sincerely
Jean-Louis

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108 posts

#5 h Apr 11, 2020, 11:29 AM • 3 Y

Y by Mango247, Mango247, Mango247

Let $QR$ and $PS$ intersect in $X$ . Quadrilateral $ABSR$ is cyclic, so $\angle BSR=180^{\circ}- \alpha$ , so $\angle RSC=\alpha.$
$\angle ARS=180^{\circ}- \beta$ , so $\angle CRS=\beta.$
$\angle RSP=\angle RSX=\angle RAP=\frac{\alpha}{2}$ (angles on the same arc)
$\angle RSX=\angle XSC=\frac{\alpha}{2}$ , so $XS$ is the bisector of $\angle RSC$
Now, $\angle QRA=\angle QBA=\angle CRX=\frac{\beta}{2}$ (angles on the same arc)
$\angle CRX=\angle SRX=\frac{\beta}{2}$ , so $XR$ is the bisector of $\angle SRC$
Obviously $CX$ is the bisector of $\angle SCR \implies CI, QR, PS$ intersect in one point.

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#6 h Apr 11, 2020, 1:17 PM • 1 Y

Y by Mango247

The easy way to solve this problem is used by Pascal's Theorem with hexagon $ARQBSP$ .
We will get that incenter of $\triangle ABC$ ( $I$ ), the intersection point of $PS$ and $QR$ , and point $C$ are collinear.
Thus, $PS$ , $QR$ , and $CI$ are concurrent.

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9775 posts

jayme

#7 h Oct 18, 2020, 9:43 AM • 2 Y

Y by Mango247, Mango247

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/02.%201.%20Points%20sur%20une%20cevienne%20du%20triangle.pdf p.19...

Sincerely

Jean-Louis

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