Olympiad problems

1959 AMC 12/AHSME, 40

In triangle $ABC$, $BD$ is a median. $CF$ intersects $BD$ at $E$ so that $\overline{BE}=\overline{ED}$. Point $F$ is on $AB$. Then, if $\overline{BF}=5$, $\overline{BA}$ equals: $ \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ \text{none of these} $

2022 Portugal MO, 4

Tags: median , geometry , angle

Let $[AD]$ be a median of the triangle $[ABC]$. Knowing that $\angle ADB = 45^o$ and $\angle A CB = 30^o$, prove that $\angle BAD = 30^o$.

2000 AMC 12/AHSME, 14

Tags: arithmetic sequence , list , mean , median , mode

When the mean, median, and mode of the list \[ 10, 2, 5, 2, 4, 2, x\]are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of $ x$? $ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 20$

2011 Sharygin Geometry Olympiad, 14

Tags: geometry , ratio , altitude , median

In triangle $ABC$, the altitude and the median from vertex $A$ form (together with line $BC$) a triangle such that the bisectrix of angle $A$ is the median; the altitude and the median from vertex $B$ form (together with line AC) a triangle such that the bisectrix of angle $B$ is the bisectrix. Find the ratio of sides for triangle $ABC$.

2016 Sharygin Geometry Olympiad, 1

Tags: right triangle , altitude , median , geometry

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

Kyiv City MO Juniors 2003+ geometry, 2021.9.5

Tags: geometry , equal segments , median

Let $BM$ be the median of the triangle $ABC$, in which $AB> BC$. Point $P$ is chosen so that $AB \parallel PC$ and$ PM \perp BM$. The point $Q$ is chosen on the line $BP$ so that $\angle AQC = 90^o$, and the points $B$ and $Q$ lie on opposite sides of the line $AC$. Prove that $AB = BQ$. (Mikhail Standenko)

1968 All Soviet Union Mathematical Olympiad, 106

Tags: geometry , incircle , median

Medians divide the triangle onto $6$ smaller ones. $4$ of the circles inscribed in those small ones are equal. Prove that the triangle is equilateral.

2016 AMC 10, 7

Tags: arithmetic mean , mode , median

The mean, median, and mode of the $7$ data values $60, 100, x, 40, 50, 200, 90$ are all equal to $x$. What is the value of $x$? $\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 75 \qquad\textbf{(D)}\ 90 \qquad\textbf{(E)}\ 100$

Kyiv City MO Juniors 2003+ geometry, 2019.9.2

Tags: median , geometry , perpendicular

In a right triangle $ABC$, the lengths of the legs satisfy the condition: $BC =\sqrt2 AC$. Prove that the medians $AN$ and $CM$ are perpendicular. (Hilko Danilo)

2023-24 IOQM India, 14

Tags: median , geometry , coordinate geometry , centroid

Let $A B C$ be a triangle in the $x y$ plane, where $B$ is at the origin $(0,0)$. Let $B C$ be produced to $D$ such that $B C: C D=1: 1, C A$ be produced to $E$ such that $C A: A E=1: 2$ and $A B$ be produced to $F$ such that $A B: B F=1: 3$. Let $G(32,24)$ be the centroid of the triangle $A B C$ and $K$ be the centroid of the triangle $D E F$. Find the length $G K$.

2018 India PRMO, 10

Tags: median , geometry , perpendicular

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 $\frac{BC^2 + CA^2 + AB^2}{100}$.

Denmark (Mohr) - geometry, 1992.4

Tags: median , geometry , geometric inequality

Let $a, b$ and $c$ denote the side lengths and $m_a, m_b$ and $m_c$ of the median's lengths in an arbitrary triangle. Show that $$\frac34 < \frac{m_a + m_b + m_c}{a + b + c}<1$$ Also show that there is no narrower range that for each triangle that contains the fraction $$\frac{m_a + m_b + m_c}{a + b + c}$$

2010 Belarus Team Selection Test, 5.2

Tags: median , inequalities , geometric inequality , geometry

Numbers $a, b, c$ are the length of the medians of some triangle. If $ab + bc + ac = 1$ prove that a) $a^2b + b^2c + c^2a > \frac13$ b) $a^2b + b^2c + c^2a > \frac12$ (I. Bliznets)

2007 Sharygin Geometry Olympiad, 5

Tags: geometry , centroid , acute , median

Medians $AA'$ and $BB'$ of triangle $ABC$ meet at point $M$, and $\angle AMB = 120^o$. Prove that angles $AB'M$ and $BA'M$ are neither both acute nor both obtuse.

2016 IMO Shortlist, G2

Tags: mixtilinear incircle , projective geometry , median , geometry

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]

Durer Math Competition CD Finals - geometry, 2008.C2

Tags: geometry , geometric inequality , geometric inequalities , median

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

2022 Tuymaada Olympiad, 8

Tags: sine , geometry , median

In an acute triangle $\triangle ABC$ the points $C_m, A_m, B_m$ are the midpoints of $AB, BC, CA$ respectively. Inside the triangle $\triangle ABC$ a point $P$ is chosen so that $\angle PCB = \angle B_mBC$ and $\angle PAB = \angle ABB_m.$ A line passing through $P$ and perpendicular to $AC$ meets the median $BB_m$ at $E.$ Prove that $E$ lies on the circumcircle of the triangle $\triangle A_mB_mC_m.$ [i](K. Ivanov )[/i]

2015 Balkan MO Shortlist, A3

Tags: inequalities , geometric inequality , median

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)

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.

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 $AB \cdot CD = AC \cdot BD$.

1959 IMO Shortlist, 4

Tags: trigonometry , construction , median , geometry

Construct a right triangle with given hypotenuse $c$ such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.

2020 Swedish Mathematical Competition, 2

Tags: median , geometry , geometric inequality

The medians of the sides $AC$ and $BC$ in the triangle $ABC$ are perpendicular to each other. Prove that $\frac12 <\frac{|AC|}{|BC|}<2$.

2014 Belarusian National Olympiad, 6

Tags: equal area , geometry , median , aras

Points $C_1, A_1$ and $B_1$ are marked on the sides $AB, BC$ and $CA$ of a triangle $ABC$ so that the segments $AA_1, BB_1$, and $CC_1$ are concurrent (see the fig.). It is known that the area of the white part of the triangle $ABC$ is equal to the area of its black part. Prove that at least one of the segments $AA_1, BB_1, CC_1$ is a median of the triangle $ABC$. [img]https://1.bp.blogspot.com/-nVVhqdRdf0s/X-WVmt_gyqI/AAAAAAAAM40/943sCRGyCPwT-vqIilTCtXOXHByRLIvPwCLcBGAsYHQ/s0/2014%2Bbelarus%2B11.6.png[/img]

2019 Ramnicean Hope, 2

Tags: median , vector geometry , geometry , circumcircle

Let $ P,Q,R $ be the intersections of the medians $ AD,BE,CF $ of a triangle $ ABC $ with its circumcircle, respectively. Show that $ ABC $ is equilateral if $ \overrightarrow{DP} +\overrightarrow{EQ} +\overrightarrow{FR} =0. $ [i]Dragoș Lăzărescu[/i]

2017 Sharygin Geometry Olympiad, 6

Tags: geometry , covering , median , circumcircle

A median of an acute-angled triangle dissects it into two triangles. Prove that each of them can be covered by a semidisc congruent to a half of the circumdisc of the initial triangle.