Olympiad problems

1953 AMC 12/AHSME, 41

A girls' camp is located $ 300$ rods from a straight road. On this road, a boys' camp is located $ 500$ rods from the girls' camp. It is desired to build a canteen on the road which shall be exactly the same distance from each camp. The distance of the canteen from each of the camps is: $ \textbf{(A)}\ 400\text{ rods} \qquad\textbf{(B)}\ 250\text{ rods} \qquad\textbf{(C)}\ 87.5\text{ rods} \qquad\textbf{(D)}\ 200\text{ rods}\\ \textbf{(E)}\ \text{none of these}$

2025 Poland - First Round, 7

Tags: geometry , perpendicular bisector

Circles $o_1, o_2$ with equal radii intersect at points $A, B$. Points $C, D, E, F$ lie in this order on one line, with $C, E$ lying on $o_1$ and $D, F$ on $o_2$. Perpendicular bisectors of $CD$ and $EF$ intersect $AB$ at $X, Y$ respectively. Prove that $AX=BY$.

2001 South africa National Olympiad, 5

Tags: circumcircle , parallelogram , geometry , trapezoid , cyclic quadrilateral , perpendicular bisector , incenter

Starting from a given cyclic quadrilateral $\mathcal{Q}_0$, a sequence of quadrilaterals is constructed so that $\mathcal{Q}_{k + 1}$ is the circumscribed quadrilateral of $\mathcal{Q}_k$ for $k = 0,1,\dots$. The sequence terminates when a quadrilateral is reached that is not cyclic. (The circumscribed quadrilateral of a cylic quadrilateral $ABCD$ has sides that are tangent to the circumcircle of $ABCD$ at $A$, $B$, $C$ and $D$.) Prove that the sequence always terminates, except when $\mathcal{Q}_0$ is a square.

2004 IberoAmerican, 2

Tags: geometry , circumcircle , perpendicular bisector

Given a scalene triangle $ ABC$. Let $ A'$, $ B'$, $ C'$ be the points where the internal bisectors of the angles $ CAB$, $ ABC$, $ BCA$ meet the sides $ BC$, $ CA$, $ AB$, respectively. Let the line $ BC$ meet the perpendicular bisector of $ AA'$ at $ A''$. Let the line $ CA$ meet the perpendicular bisector of $ BB'$ at $ B'$. Let the line $ AB$ meet the perpendicular bisector of $ CC'$ at $ C''$. Prove that $ A''$, $ B''$ and $ C''$ are collinear.

2014 Vietnam National Olympiad, 1

Tags: symmetry , geometry , circumcircle , trigonometry , gauss , radical axis , perpendicular bisector

Given a circle $(O)$ and two fixed points $B,C$ on $(O),$ and an arbitrary point $A$ on $(O)$ such that the triangle $ABC$ is acute. $M$ lies on ray $AB,$ $N$ lies on ray $AC$ such that $MA=MC$ and $NA=NB.$ Let $P$ be the intersection of $(AMN)$ and $(ABC),$ $P\ne A.$ $MN$ intersects $BC$ at $Q.$ a) Prove that $A,P,Q$ are collinear. b) $D$ is the midpoint of $BC.$ Let $K$ be the intersection of $(M,MA)$ and $(N,NA),$ $K\ne A.$ $d$ is the line passing through $A$ and perpendicular to $AK.$ $E$ is the intersection of $d$ and $BC.$ $(ADE)$ intersects $(O)$ at $F,$ $F\ne A.$ Prove that $AF$ passes through a fixed point.

2019 Hong Kong TST, 3

Tags: perpendicular bisector , geometry

Let $\Gamma_1$ and $\Gamma_2$ be two circles with different radii, with $\Gamma_1$ the smaller one. The two circles meet at distinct points $A$ and $B$. $C$ and $D$ are two points on the circles $\Gamma_1$ and $\Gamma_2$, respectively, and such that $A$ is the midpoint of $CD$. $CB$ is extended to meet $\Gamma_2$ at $F$, while $DB$ is extended to meet $\Gamma_1$ at $E$. The perpendicular bisector of $CD$ and the perpendicular bisector of $EF$ meet at $P$. (a) Prove that $\angle{EPF} = 2\angle{CAE}$. (b) Prove that $AP^2 = CA^2 + PE^2$.

2023 Bangladesh Mathematical Olympiad, P7

Tags: reflection , power of a point , perpendicular bisector , geometry

Let $\Delta ABC$ be an acute triangle and $\omega$ be its circumcircle. Perpendicular from $A$ to $BC$ intersects $BC$ at $D$ and $\omega$ at $K$. Circle through $A$, $D$ and tangent to $BC$ at $D$ intersect $\omega$ at $E$. $AE$ intersects $BC$ at $T$. $TK$ intersects $\omega$ at $S$. Assume, $SD$ intersects $\omega$ at $X$. Prove that $X$ is the reflection of $A$ with respect to the perpendicular bisector of $BC$.

2010 Contests, 1

Tags: geometry , circumcircle , perpendicular bisector

Let $ABC$ be a triangle with $\angle BAC \neq 90^{\circ}.$ Let $O$ be the circumcenter of the triangle $ABC$ and $\Gamma$ be the circumcircle of the triangle $BOC.$ Suppose that $\Gamma$ intersects the line segment $AB$ at $P$ different from $B$, and the line segment $AC$ at $Q$ different from $C.$ Let $ON$ be the diameter of the circle $\Gamma.$ Prove that the quadrilateral $APNQ$ is a parallelogram.

2009 Greece Team Selection Test, 2

Tags: circumcircle , geometry , perpendicular bisector

Given is a triangle $ABC$ with barycenter $G$ and circumcenter $O$.The perpendicular bisectors of $GA,GB,GC$ intersect at $A_1,B_1,C_1$.Show that $O$ is the barycenter of $\triangle{A_1B_1C_1}$.

2021 2nd Memorial "Aleksandar Blazhevski-Cane", 5

Tags: homothety , geometry , collinearity , circumcenter , appolonius circle , collinear , perpendicular bisector

Let $\triangle ABC$ be a triangle with circumcenter $O$. The perpendicular bisectors of the segments $OA,OB$ and $OC$ intersect the lines $BC,CA$ and $AB$ at $D,E$ and $F$, respectively. Prove that $D,E,F$ are collinear.

1980 IMO, 1

Tags: geometry , perpendicular bisector , trigonometry , triangle

Let $\alpha, \beta$ and $\gamma$ denote the angles of the triangle $ABC$. The perpendicular bisector of $AB$ intersects $BC$ at the point $X$, the perpendicular bisector of $AC$ intersects it at $Y$. Prove that $\tan(\beta) \cdot \tan(\gamma) = 3$ implies $BC= XY$ (or in other words: Prove that a sufficient condition for $BC = XY$ is $\tan(\beta) \cdot \tan(\gamma) = 3$). Show that this condition is not necessary, and give a necessary and sufficient condition for $BC = XY$.

2012 France Team Selection Test, 3

Tags: geometry , parallelogram , circumcircle , perpendicular bisector , power of a point

Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$. [i]Proposed by Carlos Yuzo Shine, Brazil[/i]

2021 Saudi Arabia Training Tests, 3

Tags: geometry , bisects segment , perpendicular bisector

Let $ABC$ be an acute, non-isosceles triangle inscribed in (O) and $BB'$, $CC'$ are altitudes. Denote $E, F$ as the intersections of $BB'$, $CC'$ with $(O)$ and $D, P, Q$ are projections of $A$ on $BC$, $CE$, $BF$. Prove that the perpendicular bisectors of $PQ$ bisects two segments $AO$, $BC$.

1999 Junior Balkan MO, 4

Tags: geometry , circumcircle , trigonometry , perpendicular bisector

Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$). [i]Greece[/i]

2011 Iran MO (2nd Round), 3

Tags: perpendicular bisector , geometric transformation , incenter , geometry , reflection , circumcircle

The line $l$ intersects the extension of $AB$ in $D$ ($D$ is nearer to $B$ than $A$) and the extension of $AC$ in $E$ ($E$ is nearer to $C$ than $A$) of triangle $ABC$. Suppose that reflection of line $l$ to perpendicular bisector of side $BC$ intersects the mentioned extensions in $D'$ and $E'$ respectively. Prove that if $BD+CE=DE$, then $BD'+CE'=D'E'$.

2011 Turkey MO (2nd round), 2

Tags: geometric transformation , geometry , circumcircle , trigonometry , reflection , cyclic quadrilateral , perpendicular bisector

Let $ABC$ be a triangle $D\in[BC]$ (different than $A$ and $B$).$E$ is the midpoint of $[CD]$. $F\in[AC]$ such that $\widehat{FEC}=90$ and $|AF|.|BC|=|AC|.|EC|.$ Circumcircle of $ADC$ intersect $[AB]$ at $G$ different than $A$.Prove that tangent to circumcircle of $AGF$ at $F$ is touch circumcircle of $BGE$ too.

2006 Peru IMO TST, 4

Tags: circumcircle , parallelogram , geometry , similar triangles , perpendicular bisector

[color=blue][size=150]PERU TST IMO - 2006[/size] Saturday, may 20.[/color] [b]Question 04[/b] In an actue-angled triangle $ABC$ draws up: its circumcircle $w$ with center $O$, the circumcircle $w_1$ of the triangle $AOC$ and the diameter $OQ$ of $w_1$. The points are chosen $M$ and $N$ on the straight lines $AQ$ and $AC$, respectively, in such a way that the quadrilateral $AMBN$ is a parallelogram. Prove that the intersection point of the straight lines $MN$ and $BQ$ belongs the circumference $w_1.$ --- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88513]Spanish version[/url] $\text{\LaTeX}{}$ed by carlosbr

2009 Czech and Slovak Olympiad III A, 6

Tags: circumcircle , geometry , perpendicular bisector

Given two fixed points $O$ and $G$ in the plane. Find the locus of the vertices of triangles whose circumcenters and centroids are $O$ and $G$ respectively.

2019 Middle European Mathematical Olympiad, 3

Tags: circumcircle , moving points , perpendicular bisector , geometry

Let $ABC$ be an acute-angled triangle with $AC>BC$ and circumcircle $\omega$. Suppose that $P$ is a point on $\omega$ such that $AP=AC$ and that $P$ is an interior point on the shorter arc $BC$ of $\omega$. Let $Q$ be the intersection point of the lines $AP$ and $BC$. Furthermore, suppose that $R$ is a point on $\omega$ such that $QA=QR$ and $R$ is an interior point of the shorter arc $AC$ of $\omega$. Finally, let $S$ be the point of intersection of the line $BC$ with the perpendicular bisector of the side $AB$. Prove that the points $P, Q, R$ and $S$ are concyclic. [i]Proposed by Patrik Bak, Slovakia[/i]

2016 Sharygin Geometry Olympiad, P20

Tags: geometry , incircle , perpendicular bisector

The incircle $\omega$ of a triangle $ABC$ touches $BC, AC$ and $AB$ at points $A_0, B_0$ and $C_0$ respectively. The bisectors of angles $B$ and $C$ meet the perpendicular bisector to segment $AA_0$ at points $Q$ and $P$ respectively. Prove that $PC_0$ and $QB_0$ meet on $\omega$ .

2013 Junior Balkan Team Selection Tests - Romania, 3

Tags: geometry , perpendicular , euler circle , perpendicular bisector

Let $D$ be the midpoint of the side $[BC]$ of the triangle $ABC$ with $AB \ne AC$ and $E$ the foot of the altitude from $BC$. If $P$ is the intersection point of the perpendicular bisector of the segment line $[DE]$ with the perpendicular from $D$ onto the the angle bisector of $BAC$, prove that $P$ is on the Euler circle of triangle $ABC$.

2018 lberoAmerican, 6

Tags: geometry , perpendicular bisector

Let $ABC$ be an acute triangle with $AC > AB > BC$. The perpendicular bisectors of $AC$ and $AB$ cut line $BC$ at $D$ and $E$ respectively. Let $P$ and $Q$ be points on lines $AC$ and $AB$ respectively, both different from $A$, such that $AB = BP$ and $AC = CQ$, and let $K$ be the intersection of lines $EP$ and $DQ$. Let $M$ be the midpoint of $BC$. Show that $\angle DKA = \angle EKM$.

1961 IMO Shortlist, 5

Tags: trigonometry , geometry , inequalities , perpendicular bisector

Construct a triangle $ABC$ if $AC=b$, $AB=c$ and $\angle AMB=w$, where $M$ is the midpoint of the segment $BC$ and $w<90$. Prove that a solution exists if and only if \[ b \tan{\dfrac{w}{2}} \leq c <b \] In what case does the equality hold?

2000 APMO, 3

Tags: angle bisector , circumcircle , ratio , trigonometry , geometry , perpendicular bisector

Let $ABC$ be a triangle. Let $M$ and $N$ be the points in which the median and the angle bisector, respectively, at $A$ meet the side $BC$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $NA$ meets $MA$ and $BA$, respectively. And $O$ the point in which the perpendicular at $P$ to $BA$ meets $AN$ produced. Prove that $QO$ is perpendicular to $BC$.

2014 China Team Selection Test, 1

Tags: geometry , perpendicular bisector , cyclic quadrilateral

$ABCD$ is a cyclic quadrilateral, with diagonals $AC,BD$ perpendicular to each other. Let point $F$ be on side $BC$, the parallel line $EF$ to $AC$ intersect $AB$ at point $E$, line $FG$ parallel to $BD$ intersect $CD$ at $G$. Let the projection of $E$ onto $CD$ be $P$, projection of $F$ onto $DA$ be $Q$, projection of $G$ onto $AB$ be $R$. Prove that $QF$ bisects $\angle PQR$.