geometry problems from an Internal Olympiad of the Chilean Math Forum FMAT with aops links in the names
geometry collected inside aops here
2010, 2012, 2013
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.
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 $.
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 $.
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 $.
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 $.
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 $AC = AP$.
$ABCD$ 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/
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