geometry problems from Cordoba of Argeintina, pre TSTs for Cono Sur, IberoAmerican, IMO, organised by La Academia Mate Club from with aops links in the names
collected inside aops here
2014 - 2016
Cono Sur
Let $\vartriangle ABC$ be a triangle. The perpendicular bisector of $AB$ intersects $BC$ at $D$ and the extension of $AC$ at $E$. The bisector of $\angle ABC$ intersects $AC$ at $F$. If $\angle BFD = \angle ABC$ and $\angle FDC = \angle DEB$, find the angles of $\vartriangle ABC$.
Let $\vartriangle ABC$ be a triangle. Let $M$ be the midpoint of side $BC$. Let $P$ be a point on side $AB$. The intersection $X$ between $AM$ and $CP$ is marked. Let $Q$ be the intersection between $BX$ and $AC$. We have that $PQ = 2$, $BC = 4$ and $AX = 3$. Find $AM$.
Given an angle of $13^o$, construct an angle of $1^o$ using only a ruler and compass.
Ibero
Let $ABC$ be a triangle with $AB = 3$ and $BC = 7$. $D$ and $E$ are marked on $BC$ and $AB$ respectively, such that $BD = AE = 2$. Segments $AD$ and $CE$ intersect at $P$. Find $\frac{area\, (APB)}{ area\, (CPB)}$ and $\frac{AP}{PD}$.
Let $ABCD$ be a square. Let $P$ be inside $ABCD$ such that $\angle PAC = \angle PCD = 20^o$. Find $\angle ABP$ .
Let $ABC$ be a triangle such that $AB = BC$. Let $D$ be the foot of the angle bisector from $C$. Let $\Omega$ be the circumscribed circle of triangle $ABC$ and call $P$ the intersection of the tangent to $\Omega$ at $A$ with line $BC$. Show that $CP = AP + AD$.
IMO
Let $ABCD$ be a parallelogram and $O$ an interior point such that $\angle AOB + \angle DOC = 180^o$ .Prove that $\angle OBC = \angle ODC$.
$\bullet$ Prove that using a ruler and a compass it is possible to construct an angle of $3^o$.
$\bullet$ Prove that given any angle whose degrees is a coprime (integer) number with $3$, from it a $19^o$ angle can be constructed using a ruler and a compass.
Clarification: It is not necessary to do all the steps of the constructions to be used as intermediate steps, but it is necessary to indicate what it is possible to construct with a ruler and compass.
Let $\omega$ be a circle and $A$ and $B$ be two points on $\omega$ such that segment $AB$ is not a diameter. If $XY$ is a variable diameter of the same circle, determine the locus of the intersection point of the lines $AX$ and $BY$.
No comments:
Post a Comment