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Chile TST (OMCS, OIM, IMO) 23p

geometry problems from Chilean Team Selection Tests  with aops links in the names


collected inside aops here

Cono Sur TST 

There is only one ruler of length $ L $ and constant width $ d $. It is requested to find a point $ C $ on the extension of a line $ \overline {AB} $ in such a way that $ AB = BC $ and it is requested to construct the perpendicular at this point.

Consider an acute triangle with sides $ a, b, $, and $ c $. Let $ m_a, m_b, m_c $ be the altitudes on the sides $ a, b, c $, respectively. If the distances from the respective vertices to the orthocenter are $ d_a, d_b, d_c $, respectively, prove that$$ m_ad_a + m_bd_b + m_cd_c = \frac {a ^ 2 + b ^ 2 + c ^ 2} {2} $$

Let $M$ be the circumcenter of an acute triangle $ABC$ and we assume that the circumscribed circle of $BMA$ intersects segment $BC$ at $P$ and $AC$ at $Q$. Show that the line $CM$ is perpendicular to $PQ$.

Let $ABC$ be a triangle such that $AB = AC$. Let $P$ be a point about $BC$. Let $M, N$ be the feet of the perpendiculars from $P$ to $AB$ and $AC$ respectively. Show that the value of the sum $PM + PN$ it does not depend on the position of the chosen point $P$.



Ibero TST 

They are given in the plane a circle $ \odot (O, r) $, and a line $ L $. The distance from $ O $ to $ L $ is $ d $, with $ d> r $. Points $ M, N \in L $ are chosen such that the circle of diameter $ MN $ is tangent outside $ \odot (O, r) $. Show that there is a point $ A \notin L $ in the plane, such that all segments $MN$ subtend a constant angle in $ A $.

The $ \triangle ABC $ has a circumcenter: $ O $ and an incenter $ I $. Let $ A_1 $ be the point of tangency of the incircle with $BC$.$ AO, AI$ intersect the circumcircle again at $ A', A''$, respectively. Prove that $ A'I \cap A'''A_1 $ is a point on the circumcircle of $ \triangle ABC $

Let $a,b,c$ the sides of a triangle, and $A,B,C$ the respective angles (measured in degrees). Show that:
$$60^o\le \frac{a \cdot A+b\cdot B+c\cdot C}{a+b+c} \le 90^o$$

The circles $ C_1, C_2 $ are tangent internally to the circle $ C $ in the points $ A, B $, respectively. The internal tangent line common to $ C_1 $ and $ C_2 $ touches these circles in $ P $ and $ Q $, respectively. Show that $AP$ and $BQ$ intersect again $ C $ in diametrically opposite points.

Let $ I $ the incenter of the $ \vartriangle ABC$. The incircle of the $ \vartriangle ABC $ is tangent to $BC, CA, AB$ in $ K, L, M $, respectively. The line parallel to $MK$, which passes through $ B $, intersects $ LM, LK $ in $ R, S $, respectively. Prove that $ \angle RIS $ is acute.

An acute triangle $ABC$ is inscribed in a circle $ \Gamma $. Let $D$ be the point of $ \Gamma $ diametrically opposite $C$. Determine all points $X$ of the arc $BC$ (which does not contain the vertex $A$) of $ \Gamma $ such that the quadrilateral $DBXC$ and the triangle $ABC$ have equal areas .

$\vartriangle ABC$ has sidelengths positive integers and $AC = 2007$. The bisector of $\angle BAC$ intersects side $BC$ at point $D$ and $AB = CD$. Find the sides $AB$ and $BC$ of $\vartriangle  ABC$.

Consider $\vartriangle ABC$, $E$ a point on line $AC$ and $F$ a point on side $BC$. Assume that $AE = BF$ and that the circles passing through $A, C, F$ and $B, C, E$ respectively intersect at a point $D\ne C$. Prove that $CD$ is the bisector of the $\angle ACB$.

Let $\Gamma_1$ and $\Gamma_2$ be two circles that intersect at points $P$ and $Q$.
Construct a segment $AB$ that passes through $P$, with $A$ in $\Gamma_1, B$ in $\Gamma_2$ and such that $AP \cdot PB$ be maximum.

In a circle, one of its diameters and an exterior point are drawn on the plane. Construct, using only a ruler (and not a compass), the perpendicular of the point to the extension of the diameter.

Consider $5$ points in the plane, such that there are no $3$ of them collinear. Prove that there is a convex quadrilateral with vertices at $4$ points.

Let $ABC$ a triangle and $l$ is a line where intersects $BC, AC$ and $BA$ in the point(s) $D, E, F$ respectively. Suppose that $l$ don't intersect a vertex in the triangle $ABC$, consider the circle(s) $C, C_b, C_a, C_c$ where are the circumcircles of triangles $ABC, DBF, AEF, DCE$ respectively. Show that this circles $C, C_a, C_b, C_c$ are concurrents.

The incircle triangle $ \vartriangle ABC$ touches $AC$ and $BC$ in $E$ and $D$ respectively. The $A$-excircle touches the extensions of $BC$ in $A_1$, of $CA$ in $B_1$ and $AB$ in $C_1$. Let $ DE \cap  A_1B_1 = L$. Prove that $L$ lies on the circumcircle of the triangle $\vartriangle  A_1BC_1$.

Let $ABC$ be a triangle and points $P, Q, R$ on the sides $AB, BC$ and $CA$ respectively in such a way that $\frac{AP}{AB}= \frac{BQ}{BC}= \frac{CR}{CA}= \frac{1}{n}$ for $n \in  N$. Segments $AQ$ and $CP$ are cut in $D$, segments $BR$ and $AQ$ are cut at $E$ and segments $BR$ and $CP$ are cut at $F$. Calculate the ratio of areas of the triangles $ \frac{(ABC)}{(DEF)}$.

Prove that in a acute scalene triangle, the orthocenter, the incenter and the circumcenter are not collinear.


IMO TST

Let $ A ', B' $ and $ C '$ be the midpoints of the sides $ BC, AC $ and $ AB $ of an acute triangle $ ABC $, respectively. Let $ K $ be the midpoint of the arc $ AB $ of the circle circumscribed to the triangle $ ABC $ that does not contain the point $ C $. Let $ L $ be a point in $ \overrightarrow {KC '} $ such that $ KC' = LC '$. Show that the circle circumscribed to triangle $ A'B'C '$ bisects segment $ \overline {CL} $.

Suppose that $h$ is the length of the maximum height of a triangle, not an obtuse angle. Let $R$ and $r$, respectively, the radii of circumscribed and inscribed circles. Show that $R+r \le h$.

Let $ p $ and $ q $ be the radii of two circles that pass through the vertex $ A $ and tangents at $ B $ and at $ C $ respectively to the side $ BC $ of a triangle $ ABC $. Prove that $ pq = R^2 $, where $ R $ is the radius of the circle circumscribed to the triangle.

Let $ABC$ be an isosceles triangle at $A$. The inscribed circle is tangent to $BC$, $AC$, $AB$ at $D, E, F$, respectively. Let $P$ be on the arc $EF$ that does not contain $D$. Let $Q$ be the point of intersection of $BP$ and the inscribed circle of $ABC$. Lines $EP$ and $EQ$ intersect line $BC$ at $M$ and $N$ respectively. Show that the points $P, F, B$ and $M$ lie on a circle and that $\frac{EM}{EN} =\frac{BF}{BP}$

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