Found problems: 649
2018 Hanoi Open Mathematics Competitions, 9
There are three polygons and the area of each one is $3$. They are drawn inside a square of area $6$. Find the greatest value of $m$ such that among those three polygons, we can always find two polygons so that the area of their overlap is not less than $m$.
2004 Oral Moscow Geometry Olympiad, 6
The length of each side and each non-principal diagonal of a convex hexagon does not exceed $1$. Prove that this hexagon contains a principal diagonal whose length does not exceed $\frac{2}{\sqrt3}$.
2002 IMO Shortlist, 6
Let $n\geq3$ be a positive integer. Let $C_1,C_2,C_3,\ldots,C_n$ be unit circles in the plane, with centres $O_1,O_2,O_3,\ldots,O_n$ respectively. If no line meets more than two of the circles, prove that \[ \sum\limits^{}_{1\leq i<j\leq n}{1\over O_iO_j}\leq{(n-1)\pi\over 4}. \]
1989 Greece National Olympiad, 4
A trapezoid with bases $a,b$ and altitude $h$ is circumscribed around a circl.. Prove that $h^2\le ab$.
2002 Federal Math Competition of S&M, Problem 2
Points $A_0,A_1,\ldots,A_{2k}$, in this order, divide a circumference into $2k+1$ equal arcs. Point $A_0$ is connected by chords to all the other points. These $2k$ chords divide the interior of the circle into $2k+1$ parts. These parts are alternately painted red and blue so that there are $k+1$ red and $k$ blue parts. Show that the blue area is larger than the red area.
1966 IMO Longlists, 5
Prove the inequality
\[\tan \frac{\pi \sin x}{4\sin \alpha} + \tan \frac{\pi \cos x}{4\cos \alpha} >1\]
for any $x, \alpha$ with $0 \leq x \leq \frac{\pi }{2}$ and $\frac{\pi}{6} < \alpha < \frac{\pi}{3}.$
1982 IMO Shortlist, 2
Let $K$ be a convex polygon in the plane and suppose that $K$ is positioned in the coordinate system in such a way that
\[\text{area } (K \cap Q_i) =\frac 14 \text{area } K \ (i = 1, 2, 3, 4, ),\]
where the $Q_i$ denote the quadrants of the plane. Prove that if $K$ contains no nonzero lattice point, then the area of $K$ is less than $4.$
2023 Canadian Mathematical Olympiad Qualification, 6
Given triangle $ABC$ with circumcircle $\Gamma$, let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively, and let the lines $AD$, $BE$, and $CF$ intersect $\Gamma$ again at points $J$, $K$, and $L$, respectively. Show that the area of triangle $JKL$ is at least that of triangle $ABC$.
Ukrainian TYM Qualifying - geometry, VII.12
Let $a, b$, and $c$ be the lengths of the sides of an arbitrary triangle, and let $\alpha,\beta$, and $\gamma$ be the radian measures of its corresponding angles. Prove that $$ \frac{\pi}{3}\le \frac{\alpha a +\beta b + \gamma c}{a+b+c} < \frac{\pi}{2}.$$ Suggest spatial analogues of this inequality.
2000 Kazakhstan National Olympiad, 8
Given a triangle $ ABC $ and a point $ M $ inside it. Prove that $$
\min \{MA, MB, MC\} + MA + MB + MC <AB + BC + AC. $$
Ukrainian TYM Qualifying - geometry, XII.17
Given a triangle $ABC$, inside which the point $M$ is marked. On the sides $BC,CA$ and $AB$ the following points $A_1,B_1$ and $C_1$ are chosen, respectively, that $MA_1 \parallel CA$, $MB_1 \parallel AB$, $MC_1 \parallel BC$. Let S be the area of triangle $ABC, Q_M$ be the area of the triangle $A_1 B_1 C_1$.
a) Prove that if the triangle $ABC$ is acute, and M is the point of intersection of its altitudes , then $3Q_M \le S$. Is there such a number $k> 0$ that for any acute-angled triangle $ABC$ and the point $M$ of intersection of its altitudes, such thatthe inequality $Q_M> k S$ holds?
b) For cases where the point $M$ is the point of intersection of the medians, the center of the inscribed circle, the center of the circumcircle, find the largest $k_1> 0$ and the smallest $k_2> 0$ such that for an arbitrary triangle $ABC$, holds the inequality $k_1S \le Q_M\le k_2S$ (for the center of the circumscribed circle, only acute-angled triangles $ABC$ are considered).
1970 IMO Shortlist, 3
In the tetrahedron $ABCD,\angle BDC=90^o$ and the foot of the perpendicular from $D$ to $ABC$ is the intersection of the altitudes of $ABC$. Prove that: \[ (AB+BC+CA)^2\le6(AD^2+BD^2+CD^2). \] When do we have equality?
2008 Balkan MO Shortlist, G1
In acute angled triangle $ABC$ we denote by $a,b,c$ the side lengths, by $m_a,m_b,m_c$ the median lengths and by $r_{b}c,r_{ca},r_{ab}$ the radii of the circles tangents to two sides and to circumscribed circle of the triangle, respectively. Prove that
$$\frac{m_a^2}{r_{bc}}+\frac{m_b^2}{r_{ab}}+\frac{m_c^2}{r_{ab}} \ge \frac{27\sqrt3}{8}\sqrt[3]{abc}$$
1996 All-Russian Olympiad Regional Round, 10.6
Given triangle $A_0B_0C_0$. On the segment $A_0B_0$ points $A_1$, $A_2$, $...$, $A_n$, and on the segment $B_0C_0$ - points $C_1$, $C_2$, $...$, $Cn$ so that all segments $A_iC_{i+1}$ ($i = 0$, $1$, $...$,$n-1$) are parallel to each other and all segments $ C_iA_{i+1}$ ($i = 0$, $1$, $...$,$n-1$) are too. Segments $C_0A_1$, $A_1C_2$, $A_2C_1$ and $C_1A_0$ bound a certain parallelogram, segments $C_1A_2$, $A_2C_3$, $A_3C_2$ and $C_2A_1$ too, etc. Prove that the sum of the areas of all $n -1$ resulting parallelograms less than half the area of triangle $A_0B_0C_0$.
2019 Polish Junior MO Finals, 4.
The point $D$ lies on the side $AB$ of the triangle $ABC$. Assume that there exists such a point $E$ on the side $CD$, that
$$
\sphericalangle EAD = \sphericalangle AED \quad \text{and} \quad \sphericalangle ECB = \sphericalangle CEB.
$$
Show that $AC + BC > AB + CE$.
Estonia Open Senior - geometry, 2005.2.4
Three rays are going out from point $O$ in space, forming pairwise angles $\alpha, \beta$ and $\gamma$ with $0^o<\alpha \le \beta \le \gamma <180^o$. Prove that $\sin \frac{\alpha}{2}+ \sin \frac{\beta}{2} > \sin \frac{\gamma}{2}$.
2012 Sharygin Geometry Olympiad, 7
A convex pentagon $P $ is divided by all its diagonals into ten triangles and one smaller pentagon $P'$. Let $N$ be the sum of areas of five triangles adjacent to the sides of $P$ decreased by the area of $P'$. The same operations are performed with the pentagon $P'$, let $N'$ be the similar difference calculated for this pentagon. Prove that $N > N'$.
(A.Belov)
2015 BMT Spring, 9
Find the side length of the largest square that can be inscribed in the unit cube.
Ukraine Correspondence MO - geometry, 2006.7
Let $D$ and $E$ be the midpoints of the sides $BC$ and $AC$ of a right triangle $ABC$. Prove that if $\angle CAD=\angle ABE$, then $$\frac{5}{6} \le \frac{AD}{AB}\le \frac{\sqrt{73}}{10}.$$
1997 IMO Shortlist, 7
The lengths of the sides of a convex hexagon $ ABCDEF$ satisfy $ AB \equal{} BC$, $ CD \equal{} DE$, $ EF \equal{} FA$. Prove that:
\[ \frac {BC}{BE} \plus{} \frac {DE}{DA} \plus{} \frac {FA}{FC} \geq \frac {3}{2}.
\]
2020 Saint Petersburg Mathematical Olympiad, 3.
On the side $AD$ of the convex quadrilateral $ABCD$ with an acute angle at $B$, a point $E$ is marked.
It is known that $\angle CAD = \angle ADC=\angle ABE =\angle DBE$.
(Grade 9 version) Prove that $BE+CE<AD$.
(Grade 10 version) Prove that $\triangle BCE$ is isosceles.(Here the condition that $\angle B$ is acute is not necessary.)
1984 IMO Shortlist, 4
Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that:
\[ n\minus{}3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n\plus{}1\over 2}\Bigr]\minus{}2,\]
where $ [x]$ denotes the greatest integer not exceeding $ x$.
2015 Oral Moscow Geometry Olympiad, 5
On the $BE$ side of a regular $ABE$ triangle, a $BCDE$ rhombus is built outside it. The segments $AC$ and $BD$ intersect at point $F$. Prove that $AF <BD$.
2001 Mongolian Mathematical Olympiad, Problem 2
In an acute-angled triangle $ABC$, $a,b,c$ are sides, $m_a,m_b,m_c$ the corresponding medians, $R$ the circumradius and $r$ the inradius. Prove the inequality
$$\frac{a^2+b^2}{a+b}\cdot\frac{b^2+c^2}{b+c}\cdot\frac{a^2+c^2}{a+c}\ge16R^2r\frac{m_a}a\cdot\frac{m_b}b\cdot\frac{m_c}c.$$
1987 Greece National Olympiad, 4
Let $A,B$ be two points interior of circle $C(O,R)$ and $M$ a point on the circle. Let $A_1,B_1$ be the intersections of the circle with lines $MA$,$MB$ respectively. Let $G$ be the midpoint of $AB$and $G_1= C\cap MG$. Prove that$$\frac{MA}{AA_1}+ \frac{MB}{BB_1}> 2\frac{MG}{GG_1}$$