Olympiad problems

1994 Canada National Olympiad, 5

Tags: trigonometry , geometry proposed , geometry , Canada , angles , Triangle

Let $ABC$ be an acute triangle. Let $AD$ be the altitude on $BC$, and let $H$ be any interior point on $AD$. Lines $BH,CH$, when extended, intersect $AC,AB$ at $E,F$ respectively. Prove that $\angle EDH=\angle FDH$.

2001 Federal Competition For Advanced Students, Part 2, 3

Tags: geometry , angle bisector , geometry proposed

Let be given a semicircle with the diameter $AB$, and points $C,D$ on it such that $AC = CD$. The tangent at $C$ intersects the line $BD$ at $E$. The line $AE$ intersects the arc of the semicircle at $F$. Prove that $CF < FD$.

2003 Bundeswettbewerb Mathematik, 3

Tags: geometry , cyclic quadrilateral , perpendicular bisector , geometry proposed

Consider a cyclic quadrilateral $ABCD$, and let $S$ be the intersection of $AC$ and $BD$. Let $E$ and $F$ the orthogonal projections of $S$ on $AB$ and $CD$ respectively. Prove that the perpendicular bisector of segment $EF$ meets the segments $AD$ and $BC$ at their midpoints.

2002 France Team Selection Test, 2

Tags: geometry , incenter , circumcircle , trigonometry , geometry proposed

Let $ ABC$ be a non-equilateral triangle. Denote by $ I$ the incenter and by $ O$ the circumcenter of the triangle $ ABC$. Prove that $ \angle AIO\leq\frac{\pi}{2}$ holds if and only if $ 2\cdot BC\leq AB\plus{}AC$.

2013 Sharygin Geometry Olympiad, 18

Tags: Asymptote , geometry , incenter , geometric transformation , reflection , geometry proposed

Let $AD$ be a bisector of triangle $ABC$. Points $M$ and $N$ are projections of $B$ and $C$ respectively to $AD$. The circle with diameter $MN$ intersects $BC$ at points $X$ and $Y$. Prove that $\angle BAX = \angle CAY$.

2009 Kazakhstan National Olympiad, 5

Tags: geometry proposed , geometry

Quadrilateral $ABCD$ inscribed in circle with center $O$. Let lines $AD$ and $BC$ intersects at $M$, lines $AB$ and $CD$- at $N$, lines $AC$ and $BD$ -at $P$, lines $OP$ and $MN$ at $K$. Proved that $ \angle AKP = \angle PKC$. As I know, this problem was very short solution by polars, but in olympiad for this solution gives maximum 4 balls (in marking schemes written, that needs to prove all theorems about properties of polars)

2001 India IMO Training Camp, 3

Tags: geometry , inradius , geometry proposed

Points $B = B_1 , B_2, \cdots , B_n , B_{n+1} = C$ are chosen on side $BC$ of a triangle $ABC$ in that order. Let $r_j$ be the inradius of triangle $AB_jB_{j+1}$ for $j = 1, \cdots, n$ , and $r$ be the inradius of $\triangle ABC$. Show that there is a constant $\lambda$ independent of $n$ such that : \[(\lambda -r_1)(\lambda -r_2)\cdots (\lambda -r_n) =\lambda^{n-1}(\lambda -r)\]

2006 Germany Team Selection Test, 1

Tags: geometry , geometric transformation , reflection , geometry proposed

Let $A$, $B$, $C$, $D$, $E$, $F$ be six points on a circle such that $AE\parallel BD$ and $BC\parallel DF$. Let $X$ be the reflection of the point $D$ in the line $CE$. Prove that the distance from the point $X$ to the line $EF$ equals to the distance from the point $B$ to the line $AC$.

2013 Brazil National Olympiad, 1

Tags: geometry , geometry proposed

Let $\Gamma$ be a circle and $A$ a point outside $\Gamma$. The tangent lines to $\Gamma$ through $A$ touch $\Gamma$ at $B$ and $C$. Let $M$ be the midpoint of $AB$. The segment $MC$ meets $\Gamma$ again at $D$ and the line $AD$ meets $\Gamma$ again at $E$. Given that $AB=a$, $BC=b$, compute $CE$ in terms of $a$ and $b$.

2009 Baltic Way, 12

Tags: geometry , geometric transformation , reflection , geometry proposed

In a quadrilateral $ABCD$ we have $AB||CD$ and $AB=2CD$. A line $\ell$ is perpendicular to $CD$ and contains the point $C$. The circle with centre $D$ and radius $DA$ intersects the line $\ell$ at points $P$ and $Q$. Prove that $AP\perp BQ$.

1997 Baltic Way, 13

Tags: geometry proposed , geometry

Five distinct points $A,B,C,D$ and $E$ lie on a line with $|AB|=|BC|=|CD|=|DE|$. The point $F$ lies outside the line. Let $G$ be the circumcentre of the triangle $ADF$ and $H$ the circumcentre of the triangle $BEF$. Show that the lines $GH$ and $FC$ are perpendicular.

2014 Turkey Junior National Olympiad, 4

Tags: geometry , circumcircle , trapezoid , geometry proposed

$ABC$ is an acute triangle with orthocenter $H$. Points $D$ and $E$ lie on segment $BC$. Circumcircle of $\triangle BHC$ instersects with segments $AD$,$AE$ at $P$ and $Q$, respectively. Prove that if $BD^2+CD^2=2DP\cdot DA$ and $BE^2+CE^2=2EQ\cdot EA$, then $BP=CQ$.

2012 Turkey Team Selection Test, 2

Tags: inequalities , Pythagorean Theorem , geometry , geometry proposed

In a plane, the six different points $A, B, C, A', B', C'$ are given such that triangles $ABC$ and $A'B'C'$ are congruent, i.e. $AB=A'B', BC=B'C', CA=C'A'.$ Let $G$ be the centroid of $ABC$ and $A_1$ be an intersection point of the circle with diameter $AA'$ and the circle with center $A'$ and passing through $G.$ Define $B_1$ and $C_1$ similarly. Prove that \[ AA_1^2+BB_1^2+CC_1^2 \leq AB^2+BC^2+CA^2 \]

2007 Italy TST, 2

Tags: geometry , circumcircle , geometry proposed

Let $ABC$ a acute triangle. (a) Find the locus of all the points $P$ such that, calling $O_{a}, O_{b}, O_{c}$ the circumcenters of $PBC$, $PAC$, $PAB$: \[\frac{ O_{a}O_{b}}{AB}= \frac{ O_{b}O_{c}}{BC}=\frac{ O_{c}O_{a}}{CA}\] (b) For all points $P$ of the locus in (a), show that the lines $AO_{a}$, $BO_{b}$ , $CO_{c}$ are cuncurrent (in $X$); (c) Show that the power of $X$ wrt the circumcircle of $ABC$ is: \[-\frac{ a^{2}+b^{2}+c^{2}-5R^{2}}4\] Where $a=BC$ , $b=AC$ and $c=AB$.

2014 Iran Team Selection Test, 1

Tags: circumcircle , geometry proposed

The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $BC,AC,AB$ at $A_{1},B_{1},C_{1}$ . let $AI,BI,CI$ meets $BC,AC,AB$ at $A_{2},B_{2},C_{2}$. let $A'$ is a point on $AI$ such that $A_{1}A'\perp B_{2}C_{2}$ .$B',C'$ respectively. prove that two triangle $A'B'C',A_{1}B_{1}C_{1}$ are equal.

2012 ELMO Shortlist, 2

Tags: geometry , circumcircle , power of a point , radical axis , cyclic quadrilateral , geometry proposed

In triangle $ABC$, $P$ is a point on altitude $AD$. $Q,R$ are the feet of the perpendiculars from $P$ to $AB,AC$, and $QP,RP$ meet $BC$ at $S$ and $T$ respectively. the circumcircles of $BQS$ and $CRT$ meet $QR$ at $X,Y$. a) Prove $SX,TY, AD$ are concurrent at a point $Z$. b) Prove $Z$ is on $QR$ iff $Z=H$, where $H$ is the orthocenter of $ABC$. [i]Ray Li.[/i]

2003 China Team Selection Test, 1

Tags: geometry , circumcircle , geometry proposed

In triangle $ABC$, $AB > BC > CA$, $AB=6$, $\angle{B}-\angle{C}=90^o$. The incircle touches $BC$ at $E$ and $EF$ is a diameter of the incircle. Radical $AF$ intersect $BC$ at $D$. $DE$ equals to the circumradius of $\triangle{ABC}$. Find $BC$ and $AC$.

2007 Romania Team Selection Test, 1

Tags: geometry , incenter , induction , cyclic quadrilateral , geometry proposed

In a circle with center $O$ is inscribed a polygon, which is triangulated. Show that the sum of the squares of the distances from $O$ to the incenters of the formed triangles is independent of the triangulation.

2015 China Team Selection Test, 3

Tags: geometry proposed , circumcircle , TST , tangent , geometry

Let $ \triangle ABC $ be an acute triangle with circumcenter $ O $ and centroid $ G .$ Let $ D $ be the midpoint of $ BC $ and $ E\in \odot (BC) $ be a point inside $ \triangle ABC $ such that $ AE \perp BC . $ Let $ F=EG \cap OD $ and $ K, L $ be the point lie on $ BC $ such that $ FK \parallel OB, FL \parallel OC . $ Let $ M \in AB $ be a point such that $ MK \perp BC $ and $ N \in AC $ be a point such that $ NL \perp BC . $ Let $ \omega $ be a circle tangent to $ OB, OC $ at $ B, C, $ respectively $ . $ Prove that $ \odot (AMN) $ is tangent to $ \omega $

1998 Baltic Way, 15

Tags: geometry proposed , geometry

Given acute triangle $ABC$. Point $D$ is the foot of the perpendicular from $A$ to $BC$. Point $E$ lies on the segment $AD$ and satisfies the equation \[\frac{AE}{ED}=\frac{CD}{DB}\] Point $F$ is the foot of the perpendicular from $D$ to $BE$. Prove that $\angle AFC=90^{\circ}$.

2001 Irish Math Olympiad, 3

Tags: geometry proposed , geometry

In an acute-angled triangle $ ABC$, $ D$ is the foot of the altitude from $ A$, and $ P$ a point on segment $ AD$. The lines $ BP$ and $ CP$ meet $ AC$ and $ AB$ at $ E$ and $ F$ respectively. Prove that $ AD$ bisects the angle $ EDF$.

2013 China Team Selection Test, 2

Tags: geometry , circumcircle , incenter , trigonometry , geometry proposed

The circumcircle of triangle $ABC$ has centre $O$. $P$ is the midpoint of $\widehat{BAC}$ and $QP$ is the diameter. Let $I$ be the incentre of $\triangle ABC$ and let $D$ be the intersection of $PI$ and $BC$. The circumcircle of $\triangle AID$ and the extension of $PA$ meet at $F$. The point $E$ lies on the line segment $PD$ such that $DE=DQ$. Let $R,r$ be the radius of the inscribed circle and circumcircle of $\triangle ABC$, respectively. Show that if $\angle AEF=\angle APE$, then $\sin^2\angle BAC=\dfrac{2r}R$

2004 Germany Team Selection Test, 1

Tags: vector , geometry , geometric transformation , similar triangles , geometry proposed

Let $ABC$ be an acute triangle, and let $M$ and $N$ be two points on the line $AC$ such that the vectors $MN$ and $AC$ are identical. Let $X$ be the orthogonal projection of $M$ on $BC$, and let $Y$ be the orthogonal projection of $N$ on $AB$. Finally, let $H$ be the orthocenter of triangle $ABC$. Show that the points $B$, $X$, $H$, $Y$ lie on one circle.

2007 Balkan MO Shortlist, G3

Tags: geometry , parallelogram , trapezoid , geometry proposed

Let $ A_{1}A_{2}A_{3}A_{4}A_{5}$ be a convex pentagon, such that \[ [A_{1}A_{2}A_{3}] \equal{} [A_{2}A_{3}A_{4}] \equal{} [A_{3}A_{4}A_{5}] \equal{} [A_{4}A_{5}A_{1}] \equal{} [A_{5}A_{1}A_{2}].\] Prove that there exists a point $ M$ in the plane of the pentagon such that \[ [A_{1}MA_{2}] \equal{} [A_{2}MA_{3}] \equal{} [A_{3}MA_{4}] \equal{} [A_{4}MA_{5}] \equal{} [A_{5}MA_{1}].\] Here $ [XYZ]$ stands for the area of the triangle $ \Delta XYZ$.

2008 China Team Selection Test, 1

Tags: inequalities , geometry , circumcircle , trigonometry , ratio , cyclic quadrilateral , geometry proposed

Let $P$ be an arbitrary point inside triangle $ABC$, denote by $A_{1}$ (different from $P$) the second intersection of line $AP$ with the circumcircle of triangle $PBC$ and define $B_{1},C_{1}$ similarly. Prove that $\left(1 \plus{} 2\cdot\frac {PA}{PA_{1}}\right)\left(1 \plus{} 2\cdot\frac {PB}{PB_{1}}\right)\left(1 \plus{} 2\cdot\frac {PC}{PC_{1}}\right)\geq 8$.