227. IMO ShortList 2003 (problem 5).

by Virgil Nicula, Feb 22, 2011, 7:22 PM

Proposed problem. Let $ABC$ be an isosceles triangle with $AC=BC$, whose incentre is $I$. Let $P$ be a point on the circumcircle of the triangle $AIB$

lying inside the triangle $ABC$. The lines through $P$ parallel to $CA$ and $CB$ meet $AB$ at $D$ and $E$, respectively. The line through $P$ parallel to $AB$

meets $CA$ and $CB$ at $F$ and $G$, respectively. Prove that the lines $DF$ and $EG$ intersect on the circumcircle of the triangle $ABC\ .$


Proof. The enunciation of this problem conceals vainly in its debut that the circumcircle of $\triangle AIB$ is the circle with the diameter $II_{c}$ ,

where $I_{c}$ is the $C$- exincenter of $\triangle ABC$ . Denote : the circumcircle $w$ of the triangle $ABC$ ; the circle $\delta$ with the diameter $[II_{c}]$ ;

the circumcircles $w_{a}$ , $w_{b}$ of the isosceles trapezoids $AFPE$ , $BGPD$ respectively ; the second intersection point $R$ between $\overline{FPG}$

and the circle $\delta$ ; the intersection $L\in FD\cap GE\ .$ Prove easily that the lines $CA$ , $CB$ are the tangents from the point $C$ to the circle $\delta$ .

From the relation $FA^{2}=FR\cdot FP$ obtain $\frac{FA}{AD}=\frac{GP}{PE}\ ,$ i.e. the parallelograms $FADP$ and $GPEB$ are similarly as $FADP\sim GPEB$ .

$1\blacktriangleright\ \left\|\begin{array}{ccccc}
\widehat{AEL}\equiv\widehat{GEB}\equiv\widehat{FDP}\equiv\widehat{AFD}\equiv\widehat{AFL} & \Longrightarrow & \widehat{AEL}\equiv\widehat{AFL} & \Longrightarrow & L\in w_{a}\\\\
\widehat{BDL}\equiv\widehat{FDA}\equiv\widehat{GEP}\equiv\widehat{BGE}\equiv\widehat{BGL} & \Longrightarrow & \widehat{BDL}\equiv\widehat{BGL} & \Longrightarrow & L\in w_{b}\end{array}\right\|$ $\implies$ $\widehat{DLE}\equiv\widehat{ABC}$ .

$2\blacktriangleright\ \left\|\begin{array}{c}
\widehat{ALD}\equiv\widehat{ALF}\equiv\widehat{APF}\\\\
\widehat{BLE}\equiv\widehat{BLG}\equiv\widehat{BPG}\end{array}\right\|\implies$ $\widehat{ALD}+\widehat{BLE}\equiv\widehat{APF}+\widehat{BPG}\equiv\widehat{AI_{c}B}\equiv\widehat{ABC}\Longrightarrow m\left(\widehat{ALD}\right)+m\left(\widehat{BLE}\right)=B$ .

$3\blacktriangleright\widehat{ALB}\equiv\widehat{ALD}+\widehat{BLE}+\widehat{DLE}=$ $2\cdot\widehat{ABC}=$ $180^{\circ}-\widehat{ACB}$ $\Longrightarrow$ $\widehat{ALB}+\widehat{ACB}=180^{\circ}$ $\Longrightarrow$ $L\in w$ .

Variation on the same theme. Let $ABC$ be a fixed $A$-isosceles triangle and let mobile $M\in [AB$ , $N\in [AC$ so that $MN\parallel BC$ and exists $P\in (MN)$

for which $PM\cdot PN=BM^{2}$ . Construct $\{S,T\}\subset (BC)$ for which $PS\parallel BM\ ,\ PT\parallel NC\ .$ Ascertain the geometrical locus of $L\in MS\cap NT$ .


Answer.
This post has been edited 11 times. Last edited by Virgil Nicula, Nov 22, 2015, 2:51 PM

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