OIFMAT I-III 2010-13 (Fmat Chile) 7p

 geometry problems from an Internal Olympiad of the Chilean Math Forum FMAT with aops links in the names

aops post collection

geometry collected inside aops here

2010, 2012, 2013

2010 OIFMAT I p2

In an acute angle $\vartriangle ABC$, let $AD, BE, CF$ be their altitudes (with $D, E, F$ lying on $BC, CA, AB$, respectively). Let's call $O, H$ the circumcenter and orthocenter of $\vartriangle ABC$, respectively. Let $P = CF \cap AO$. Suppose the following two conditions are true:

$\bullet$ $FP = EH$

$\bullet$ There is a circle that passes through points $A, O, H, C$

Prove that the $\vartriangle ABC$ is equilateral.

2010 OIFMAT I p6

Let $\vartriangle ABC$ be a triangle with $\angle ACB = 60º$. Let $E$ be a point inside $\overline {AC}$ such that $CE <BC$. Let $D$ over $\overline {BC}$ such that $$\frac {AE} {BD} = \frac {BC} {CE} -1 .$$ Let us call $P$ the intersection of $\overline {AD}$ with $\overline {BE}$ and $Q$ the other point of intersection of the circumcircles of the triangles $AEP$ and $BDP$. Prove that $QE \parallel BC$.

2012 OIFMAT II p3

In the interior of an equilateral triangle $ABC$ a point $P$ is chosen such that $PA ^2 = PB ^2 + PC ^2$. Find the measure of $\angle BPC$.

2012 OIFMAT II p4

Given a $\vartriangle ABC$ with $AB> AC$ and $\angle BAC = 60^o$. Denote the circumcenter and orthocenter as $O$ and $H$ respectively. We also have that $OH$ intersects $AB$ in $P$ and $AC$ in $Q$. Prove that $PO = HQ$.

2013 OIFMAT III p5

In an acute triangle $ABC$ with circumcircle $\Omega$ and circumcenter $O$, the circle $\Gamma$ is drawn, passing through the points $A$, $O$ and $C$ together with its diameter $OQ$, then the points $M$ and $N$ are chosen on the lines $AQ$ and $AC$, respectively, in such a way that the quadrilateral $AMBN$ is a parallelogram.

Prove that the point of intersection of the lines $MN$ and $BQ$ lies on the circle $\Gamma$.

2013 OIFMAT III p6

The acute triangle $ABC$ is inscribed in a circle with center $O$. Let $D$ be the intersection of the bisector of angle $BAC$ with segment $BC$ and $P$ the intersection point of $AB$ with the perpendicular on $OA$ passing through $D$. Show that $AC = AP$.

2013 OIFMAT III p8

$ABCD$ is a trapezoid with $AB$ parallel to $CD$. The external bisectors of the angles at $B$ and $C$ intersect at $P$. The external bisectors of the angles at $A$ and $D$ intersect at $Q$. Show that the length of $PQ$ is equal to half the perimeter of the trapezoid $ABCD$.

source: http://www.fmat.cl/