This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 133

2010 AMC 8, 4

What is the sum of the mean, median, and mode of the numbers, $2,3,0,3,1,4,0,3$? $ \textbf{(A)}\ 6.5 \qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 7.5\qquad\textbf{(D)}\ 8.5\qquad\textbf{(E)}\ 9 $

2019 BAMO, E/3

In triangle $\vartriangle ABC$, we have marked points $A_1$ on side $BC, B_1$ on side $AC$, and $C_1$ on side $AB$ so that $AA_1$ is an altitude, $BB_1$ is a median, and $CC_1$ is an angle bisector. It is known that $\vartriangle A_1B_1C_1$ is equilateral. Prove that $\vartriangle ABC$ is equilateral too. (Note: A median connects a vertex of a triangle with the midpoint of the opposite side. Thus, for median $BB_1$ we know that $B_1$ is the midpoint of side $AC$ in $\vartriangle ABC$.)

2015 European Mathematical Cup, 3

Tags: geometry , circles , median
Circles $k_1$ and $k_2$ intersect in points $A$ and $B$, such that $k_1$ passes through the center $O$ of the circle $k_2$. The line $p$ intersects $k_1$ in points $K$ and $O$ and $k_2$ in points $L$ and $M$, such that the point $L$ is between $K$ and $O$. The point $P$ is orthogonal projection of the point $L$ to the line $AB$. Prove that the line $KP$ is parallel to the $M-$median of the triangle $ABM$. [i]Matko Ljulj[/i]

2000 Argentina National Olympiad, 2

Given a triangle $ABC$ with side $AB$ greater than $BC$, let $M$ be the midpoint of $AC$ and $L$ be the point at which the bisector of angle $\angle B$ intersects side $AC$. The line parallel to $AB$, which intersects the bisector $BL$ at $D$, is drawn by $M$, and the line parallel to the side $BC$ that intersects the median $BM$ at $E$ is drawn by $L$. Show that $ED$ is perpendicular to $BL$.

2024 Yasinsky Geometry Olympiad, 5

Let \( AL \) be the bisector of triangle \( ABC \), \( O \) the center of its circumcircle, and \( D \) and \( E \) the midpoints of \( BL \) and \( CL \), respectively. Points \( P \) and \( Q \) are chosen on segments \( AD \) and \( AE \) such that \( APLQ \) is a parallelogram. Prove that \( PQ \perp AO \). [i]Proposed by Mykhailo Plotnikov[/i]

Durer Math Competition CD Finals - geometry, 2008.C2

Given a triangle with sides $a, b, c$ and medians $s_a, s_b, s_c$ respectively. Prove the following inequality: $$a + b + c> s_a + s_b + s_c> \frac34 (a + b + c) $$

1959 Poland - Second Round, 2

Tags: geometry , similar , median
What relationship between the sides of a triangle makes it similar to the triangle formed by its medians?

2006 Junior Balkan Team Selection Tests - Romania, 1

Tags: geometry , median , ratio , angle
Let $ABC$ be a triangle and $D$ a point inside the triangle, located on the median of $A$. Prove that if $\angle BDC = 180^o - \angle BAC$, then $AB \cdot CD = AC \cdot BD$.

2016 Sharygin Geometry Olympiad, 1

An altitude $AH$ of triangle $ABC$ bisects a median $BM$. Prove that the medians of triangle $ABM$ are sidelengths of a right-angled triangle. by Yu.Blinkov

2017 Brazil Team Selection Test, 3

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$. [i]Proposed by Evan Chen, Taiwan[/i]

1975 Chisinau City MO, 90

Construct a right-angled triangle along its two medians, starting from the acute angles.

1995 Denmark MO - Mohr Contest, 3

Tags: ratio , geometry , median
From the vertex $C$ in triangle $ABC$, draw a straight line that bisects the median from $A$. In what ratio does this line divide the segment $AB$? [img]https://1.bp.blogspot.com/-SxWIQ12DIvs/XzcJv5xoV0I/AAAAAAAAMY4/Ezfe8bd7W-Mfp2Qi4qE_gppbh9Fzvb4XwCLcBGAsYHQ/s0/1995%2BMohr%2Bp3.png[/img]

2020 Yasinsky Geometry Olympiad, 4

The median $AM$ is drawn in the triangle $ABC$ ($AB \ne AC$). The point $P$ is the foot of the perpendicular drawn on the segment $AM$ from the point $B$. On the segment $AM$ we chose such a point $Q$ that $AQ = 2PM$. Prove that $\angle CQM = \angle BAM$.

1998 Italy TST, 2

In a triangle $ABC$, points $H,M,L$ are the feet of the altitude from $C$, the median from $A$, and the angle bisector from $B$, respectively. Show that if triangle $HML$ is equilateral, then so is triangle $ABC$.

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 $$\frac{m_a^2}{r_{bc}}+\frac{m_b^2}{r_{ab}}+\frac{m_c^2}{r_{ab}} \ge \frac{27\sqrt3}{8}\sqrt[3]{abc}$$

1999 Czech And Slovak Olympiad IIIA, 3

Tags: median , sum , ratio , geometry
Show that there exists a triangle $ABC$ such that $a \ne b$ and $a+t_a = b+t_b$, where $t_a,t_b$ are the medians corresponding to $a,b$, respectively. Also prove that there exists a number $k$ such that every such triangle satisfies $a+t_a = b+t_b = k(a+b)$. Finally, find all possible ratios $a : b$ in such triangles.

1995 All-Russian Olympiad Regional Round, 10.3

Tags: median , tangent , geometry
In an acute-angled triangle $ABC$, the circle $S$ with the altitude $BK$ as the diameter intersects $AB$ at $E$ and $BC$ at $F$. Prove that the tangents to $S$ at $E$ and $F$ meet on the median from $B$.

1999 Estonia National Olympiad, 4

For the given triangle $ABC$, prove that a point $X$ on the side $AB$ satisfies the condition $\overrightarrow{XA} \cdot\overrightarrow{XB} +\overrightarrow{XC} \cdot \overrightarrow{XC} = \overrightarrow{CA} \cdot \overrightarrow{CB} $, iff $X$ is the basepoint of the altitude or median of the triangle $ABC$.

1967 All Soviet Union Mathematical Olympiad, 084

a) The maximal height $|AH|$ of the acute-angled triangle $ABC$ equals the median $|BM|$. Prove that the angle $ABC$ isn't greater than $60$ degrees. b) The height $|AH|$ of the acute-angled triangle $ABC$ equals the median $|BM|$ and bisectrix $|CD|$. Prove that the angle $ABC$ is equilateral.

2015 Balkan MO Shortlist, A3

Let a$,b,c$ be sidelengths of a triangle and $m_a,m_b,m_c$ the medians at the corresponding sides. Prove that $$m_a\left(\frac{b}{a}-1\right)\left(\frac{c}{a}-1\right)+ m_b\left(\frac{a}{b}-1\right)\left(\frac{c}{b}-1\right) +m_c\left(\frac{a}{c}-1\right)\left(\frac{b}{c}-1\right)\geq 0.$$ (FYROM)

2018 India PRMO, 10

In a triangle $ABC$, the median from $B$ to $CA$ is perpendicular to the median from $C$ to $AB$. If the median from $A$ to $BC$ is $30$, determine $\frac{BC^2 + CA^2 + AB^2}{100}$.

2000 Belarus Team Selection Test, 5.1

Let $AM$ and $AL$ be the median and bisector of a triangle $ABC$ ($M,L \in BC$). If $BC = a, AM = m_a, AL = l_a$, prove the inequalities: (a) $a\tan \frac{a}{2} \le 2m_a \le a \cot \frac{a}{2} $ if $a < \frac{\pi}{2}$ and $a\tan \frac{a}{2} \ge 2m_a \ge a \cot \frac{a}{2} $ if $a > \frac{\pi}{2}$ (b) $2l_a \le a\cot \frac{a}{2} $.

2004 All-Russian Olympiad Regional Round, 8.3

In an acute triangle, the distance from the midpoint of any side to the opposite vertex is equal to the sum of the distances from it to sides of the triangle. Prove that this triangle is equilateral.

2017 Yasinsky Geometry Olympiad, 2

Medians $AM$ and $BE$ of a triangle $ABC$ intersect at $O$. The points $O, M, E, C$ lie on one circle. Find the length of $AB$ if $BE = AM =3$.

2017 Morocco TST-, 3

Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$. [i]Proposed by Evan Chen, Taiwan[/i]