Found problems: 25757
1998 Argentina National Olympiad, 5
Let $ABC$ a right isosceles triangle with hypotenuse $AB=\sqrt2$ . Determine the positions of the points $X,Y,Z$ on the sides $BC,CA,AB$ respectively so that the triangle $XYZ$ is isosceles, right, and with minimum area.
May Olympiad L2 - geometry, 1996.4
Let $ABCD$ be a square and let point $F$ be any point on side $BC$. Let the line perpendicular to $DF$, that passes through $B$, intersect line $DC$ at $Q$. What is value of $\angle FQC$?
2000 Czech and Slovak Match, 1
$a,b,c$ are positive real numbers which satisfy $5abc>a^3+b^3+c^3$. Prove that $a,b,c$ can form a triangle.
2014 All-Russian Olympiad, 2
Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $ \Omega $ is a circle passing through $A,B,C,D$. Let $ \omega $ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic.
[i]I. Bogdanov[/i]
2018 Turkey EGMO TST, 1
Let $ABCD$ be a cyclic quadrilateral and $w$ be its circumcircle. For a given point $E$ inside $w$, $DE$ intersects $AB$ at $F$ inside $w$. Let $l$ be a line passes through $E$ and tangent to circle $AEF$. Let $G$ be any point on $l$ and inside the quadrilateral $ABCD$. Show that if $\angle GAD =\angle BAE$ and $\angle GCB + \angle GAB = \angle EAD + \angle AGD + \angle ABE$ then $BC$, $AD$ and $EG$ are concurrent.
2017 Ukrainian Geometry Olympiad, 3
Circles ${w}_{1},{w}_{2}$ intersect at points ${{A}_{1}} $ and ${{A}_{2}} $. Let $B$ be an arbitrary point on the circle ${{w}_{1}}$, and line $B{{A}_{2}}$ intersects circle ${{w}_{2}}$ at point $C$. Let $H$ be the orthocenter of $\Delta B{{A}_{1}}C$. Prove that for arbitrary choice of point $B$, the point $H$ lies on a certain fixed circle.
2006 Hanoi Open Mathematics Competitions, 6
The figure $ABCDEF$ is a regular hexagon. Find all points $M$ belonging to the hexagon such that
Area of triangle $MAC =$ Area of triangle $MCD$.
2013 Abels Math Contest (Norwegian MO) Final, 2
In a triangle $T$, all the angles are less than $90^o$, and the longest side has length $s$. Show that for every point $p$ in $T$ we can pick a corner $h$ in $T$ such that the distance from $p$ to $h$ is less than or equal to $s/\sqrt3$.
2004 National High School Mathematics League, 12
In rectangular coordinate system, give two points $M(-1,2),N(1,4)$, $P$ is a moving point on $x$-axis, when $\angle MPN$ takes its maximum value, the $x$-axis of $P$ is________.
1969 IMO Longlists, 10
$(BUL 4)$ Let $M$ be the point inside the right-angled triangle $ABC (\angle C = 90^{\circ})$ such that $\angle MAB = \angle MBC = \angle MCA =\phi.$ Let $\Psi$ be the acute angle between the medians of $AC$ and $BC.$ Prove that $\frac{\sin(\phi+\Psi)}{\sin(\phi-\Psi)}= 5.$
2007 Peru MO (ONEM), 4
Let $ABCD$ be rhombus $ABCD$ where the triangles $ABD$ and $BCD$ are equilateral. Let $M$ and $N$ be points on the sides $BC$ and $CD$, respectively, such that $\angle MAN = 30^o$. Let $X$ be the intersection point of the diagonals $AC$ and $BD$. Prove that $\angle XMN = \angle\ DAM$ and $\angle XNM = \angle BAN$.
1991 IMO, 1
Given a triangle $ \,ABC,\,$ let $ \,I\,$ be the center of its inscribed circle. The internal bisectors of the angles $ \,A,B,C\,$ meet the opposite sides in $ \,A^{\prime },B^{\prime },C^{\prime }\,$ respectively. Prove that
\[ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}.
\]
2019 HMNT, 1
Dylan has a $100\times 100$ square, and wants to cut it into pieces of area at least $1$. Each cut must be a straight line (not a line segment) and must intersect the interior of the square. What is the largest number of cuts he can make?
Russian TST 2017, P1
The diagonals of a convex quadrilateral divide it into four triangles. Prove that the nine point centers of these four triangles either lie on one straight line, or are the vertices of a parallelogram.
2001 AMC 10, 17
Which of the cones listed below can be formed from a $ 252^\circ$ sector of a circle of radius $ 10$ by aligning the two straight sides?
[asy]import graph;unitsize(1.5cm);defaultpen(fontsize(8pt));draw(Arc((0,0),1,-72,180),linewidth(.8pt));draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt));label("$10$",(-0.5,0),S);draw(Arc((0,0),0.1,-72,180));label("$252^{\circ}$",(0.05,0.05),NE);[/asy]
[asy]
import three;
picture mainframe;
defaultpen(fontsize(11pt));
picture conePic(picture pic, real r, real h, real sh)
{
size(pic, 3cm);
triple eye = (11, 0, 5);
currentprojection = perspective(eye);
real R = 1, y = 2;
triple center = (0, 0, 0);
triple radPt = (0, R, 0);
triple negRadPt = (0, -R, 0);
triple heightPt = (0, 0, y);
draw(pic, arc(center, radPt, negRadPt, heightPt, CW));
draw(pic, arc(center, radPt, negRadPt, heightPt, CCW), linetype("8 8"));
draw(pic, center--radPt, linetype("8 8"));
draw(pic, center--heightPt, linetype("8 8"));
draw(pic, negRadPt--heightPt--radPt);
label(pic, (string) r, center--radPt, dir(270));
if (h != 0)
{
label(pic, (string) h, heightPt--center, dir(0));
}
if (sh != 0)
{
label(pic, (string) sh, heightPt--radPt, dir(0));
}
return pic;
}
picture pic1;
pic1 = conePic(pic1, 6, 0, 10);
picture pic2;
pic2 = conePic(pic2, 6, 10, 0);
picture pic3;
pic3 = conePic(pic3, 7, 0, 10);
picture pic4;
pic4 = conePic(pic4, 7, 10, 0);
picture pic5;
pic5 = conePic(pic5, 8, 0, 10);
picture aux1; picture aux2; picture aux3;
add(aux1, pic1.fit(), (0,0), W);
label(aux1, "$\textbf{(A)}$", (0,0), 22W, linewidth(4));
label(aux1, "$\textbf{(B)}$", (0,0), 3E);
add(aux1, pic2.fit(), (0,0), 35E);
add(aux2, aux1.fit(), (0,0), W);
label(aux2, "$\textbf{(C)}$", (0,0), 3E);
add(aux2, pic3.fit(), (0,0), 35E);
add(aux3, aux2.fit(), (0,0), W);
label(aux3, "$\textbf{(D)}$", (0,0), 3E);
add(aux3, pic4.fit(), (0,0), 35E);
add(mainframe, aux3.fit(), (0,0), W);
label(mainframe, "$\textbf{(E)}$", (0,0), 3E);
add(mainframe, pic5.fit(), (0,0), 35E);
add(mainframe.fit(), (0,0), N);
[/asy]
2012 Online Math Open Problems, 37
In triangle $ABC$, $AB = 1$ and $AC = 2$. Suppose there exists a point $P$ in the interior of triangle $ABC$ such that $\angle PBC = 70^{\circ}$, and that there are points $E$ and $D$ on segments $AB$ and $AC$, such that $\angle BPE = \angle EPA = 75^{\circ}$ and $\angle APD = \angle DPC = 60^{\circ}$. Let $BD$ meet $CE$ at $Q,$ and let $AQ$ meet $BC$ at $F.$ If $M$ is the midpoint of $BC$, compute the degree measure of $\angle MPF.$
[i]Authors: Alex Zhu and Ray Li[/i]
2013 Kurschak Competition, 2
Consider the closed polygonal discs $P_1$, $P_2$, $P_3$ with the property that for any three points $A\in P_1$, $B\in P_2$, $C\in P_3$, we have $[\triangle ABC]\le 1$. (Here $[X]$ denotes the area of polygon $X$.)
(a) Prove that $\min\{[P_1],[P_2],[P_3]\}<4$.
(b) Give an example of polygons $P_1,P_2,P_3$ with the above property such that $[P_1]>4$ and $[P_2]>4$.
2009 Sharygin Geometry Olympiad, 21
The opposite sidelines of quadrilateral $ ABCD$ intersect at points $ P$ and $ Q$. Two lines passing through these points meet the side of $ ABCD$ in four points which are the vertices of a parallelogram. Prove that the center of this parallelogram lies on the line passing through the midpoints of diagonals of $ ABCD$.
Kvant 2024, M2817
We are given fixed circles $\Omega$ and $\omega$ such that there exists a hexagon $ABCDEF$ inscribed in $\Omega$ and circumscribed around $\omega$. (Note that then, by virtue of Poncelet's theorem, there is an infinite family of such hexagons.) Prove that the value of $\dfrac{S_{ABCDEF}}{AD+BE+CF}$ it does not depend on the choice of the hexagon $ABCDEF$.
[i]A. Zaslavsky and Tran Quang Hung[/i]
2015 Iran Team Selection Test, 6
$ABCD$ is a circumscribed and inscribed quadrilateral. $O$ is the circumcenter of the quadrilateral. $E,F$ and $S$ are the intersections of $AB,CD$ , $AD,BC$ and $AC,BD$ respectively. $E'$ and $F'$ are points on $AD$ and $AB$ such that $A\hat{E}E'=E'\hat{E}D$ and $A\hat{F}F'=F'\hat{F}B$. $X$ and $Y$ are points on $OE'$ and $OF'$ such that $\frac{XA}{XD}=\frac{EA}{ED}$ and $\frac{YA}{YB}=\frac{FA}{FB}$. $M$ is the midpoint of arc $BD$ of $(O)$ which contains $A$.
Prove that the circumcircles of triangles $OXY$ and $OAM$ are coaxal with the circle with diameter $OS$.
1962 Miklós Schweitzer, 3
Let $ A$ and $ B$ be two Abelian groups, and define the sum of two homomorphisms $ \eta$ and $ \chi$ from $ A$ to $ B$ by \[ a( \eta\plus{}\chi)\equal{}a\eta\plus{}a\chi \;\textrm{for all}\ \;a \in A\ .\] With this addition, the set of homomorphisms from $ A$ to $ B$ forms an Abelian group $ H$. Suppose now that $ A$ is a $ p$-group ( $ p$ a prime number). Prove that in this case $ H$ becomes a topological group under the topology defined by taking the subgroups $ p^kH \;(k\equal{}1,2,...)$ as a neighborhood base of $ 0$. Prove that $ H$ is complete in this topology and that every connected component of $ H$ consists of a single element. When is $ H$ compact in this topology? [L. Fuchs]
2011 Iran Team Selection Test, 6
The circle $\omega$ with center $O$ has given. From an arbitrary point $T$ outside of $\omega$ draw tangents $TB$ and $TC$ to it. $K$ and $H$ are on $TB$ and $TC$ respectively.
[b]a)[/b] $B'$ and $C'$ are the second intersection point of $OB$ and $OC$ with $\omega$ respectively. $K'$ and $H'$ are on angle bisectors of $\angle BCO$ and $\angle CBO$ respectively such that $KK' \bot BC$ and $HH'\bot BC$. Prove that $K,H',B'$ are collinear if and only if $H,K',C'$ are collinear.
[b]b)[/b] Consider there exist two circle in $TBC$ such that they are tangent two each other at $J$ and both of them are tangent to $\omega$.and one of them is tangent to $TB$ at $K$ and other one is tangent to $TC$ at $H$. Prove that two quadrilateral $BKJI$ and $CHJI$ are cyclic ($I$ is incenter of triangle $OBC$).
2014 Junior Regional Olympiad - FBH, 3
Let $ABC$ be a right angled triangle. Prove that angle bisector of right angle is simultaneously an angle bisector of angle between median and altitude to hypotenuse.
2006 Taiwan TST Round 1, 2
Let $p,q$ be two distinct odd primes. Calculate
$\displaystyle \sum_{j=1}^{\frac{p-1}{2}}\left \lfloor \frac{qj}{p}\right \rfloor +\sum_{j=1}^{\frac{q-1}{2}}\left \lfloor \frac{pj}{q}\right\rfloor$.
2007 ISI B.Stat Entrance Exam, 4
Show that it is not possible to have a triangle with sides $a,b,$ and $c$ whose medians have length $\frac{2}{3}a, \frac{2}{3}b$ and $\frac{4}{5}c$.