Olympiad problems

2005 iTest, 3

For a convex hexagon $AHSIMC$ whose side lengths are all $1$, let $Z$ and $z$ be the maximum and minimum values, respectively, of the three diagonals $AI$, $HM$, and $SC$. If $\sqrt{x}\le Z \le \sqrt{y} $ and $\sqrt{q}\le z \le \sqrt{r} $ , find the product $qrxy$, if $q$,$ r$, $x$, and $y$ are all integers.

1953 Moscow Mathematical Olympiad, 233

Tags: angle , geometric inequality , trapezoid , geometry

Prove that the sum of angles at the longer base of a trapezoid is less than the sum of angles at the shorter base.

Ukrainian TYM Qualifying - geometry, I.9

Tags: geometric inequality , geometry

Prove that for any interior point of a triangle the sum of squares of distances from it to the sides of a triangle is not less than $\frac{4S^2}{9R^2}$. [hide=about S,R]they are not defined, but I suppose they mean it's area and circumradii respectively[/hide]

Estonia Open Senior - geometry, 2004.1.5

Tags: inequalities , area of triangle , geometric inequality , geometry

Find the smallest real number $x$ for which there exist two non-congruent triangles with integral side lengths having area $x$.

1989 IMO Shortlist, 24

Tags: combinatorics , sphere , 3d geometry , maximization , geometric inequality , geometry

For points $ A_1, \ldots ,A_5$ on the sphere of radius 1, what is the maximum value that $ min_{1 \leq i,j \leq 5} A_iA_j$ can take? Determine all configurations for which this maximum is attained. (Or: determine the diameter of any set $ \{A_1, \ldots ,A_5\}$ for which this maximum is attained.)

Indonesia Regional MO OSP SMA - geometry, 2007.4

Tags: altitude , geometric inequality , geometry

In acute triangles $ABC$, $AD, BE ,CF$ are altitudes, with $D, E, F$ on the sides $BC, CA, AB$, respectively. Prove that $$DE + DF \le BC$$

2017 Moldova Team Selection Test, 7

Tags: inequalities , geometric inequality , incenter , circumcircle , geometry

Let $ABC$ be an acute triangle, and $H$ its orthocenter. The distance from $H$ to rays $BC$, $CA$, and $AB$ is denoted by $d_a$, $d_b$, and $d_c$, respectively. Let $R$ be the radius of circumcenter of $\triangle ABC$ and $r$ be the radius of incenter of $\triangle ABC$. Prove the following inequality: $$d_a+d_b+d_c \le \frac{3R^2}{4r}$$.

2017-IMOC, G6

Tags: geometric inequality , geometry

A point $P$ lies inside $\vartriangle ABC$ such that the values of areas of $\vartriangle PAB, \vartriangle PBC, \vartriangle PCA$ can form a triangle. Let $BC = a,CA = b,AB = c, PA = x,PB = y, PC = z$, prove that $$\frac{(x + y)^2 + (y + z)^2 + (z + x)^2}{x + y + z} \le a + b + c$$

2006 Spain Mathematical Olympiad, 3

Tags: area , square root , diagonal , convex quadrilateral , geometry , geometric inequality

The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ intersect at $E$. Denotes by $S_1,S_2$ and $S$ the areas of the triangles $ABE$, $CDE$ and the quadrilateral $ABCD$ respectively. Prove that $\sqrt{S_1}+\sqrt{S_2}\le \sqrt{S}$ . When equality is reached?

2018 Spain Mathematical Olympiad, 3

Tags: geometry , inequalities , geometric inequality

Let $ABC$ be an acute-angled triangle with circumcenter $O$ and let $M$ be a point on $AB$. The circumcircle of $AMO$ intersects $AC$ a second time on $K$ and the circumcircle of $BOM$ intersects $BC$ a second time on $N$. Prove that $\left[MNK\right] \geq \frac{\left[ABC\right]}{4}$ and determine the equality case.

2000 Romania National Olympiad, 4

Tags: geometric inequality , square , geometry

In the square $ABCD$ we consider $ E \in (AB)$, $ F \in (AD)$ and $EF \cap AC = \{P\}$. Show that: a) $\frac{1}{AE} + \frac{1}{AF} = \frac{\sqrt2}{AP}$ b) $AP^2 \le \frac{AE \cdot AF}{2}$

1998 Tuymaada Olympiad, 3

Tags: geometry , area , inradius , geometric inequality , inequalities , area of a triangle

The segment of length $\ell$ with the ends on the border of a triangle divides the area of that triangle in half. Prove that $\ell >r\sqrt2$, where $r$ is the radius of the inscribed circle of the triangle.

2007 China Northern MO, 2

Tags: inequalities , geometric inequality

Let $ a,\, b,\, c$ be side lengths of a triangle and $ a+b+c = 3$. Find the minimum of \[ a^{2}+b^{2}+c^{2}+\frac{4abc}{3}\]

2014 Iranian Geometry Olympiad (junior), P5

Tags: circumcircle , geometry , arc midpoint , geometric inequality

Two points $X, Y$ lie on the arc $BC$ of the circumcircle of $\triangle ABC$ (this arc does not contain $A$) such that $\angle BAX = \angle CAY$ . Let $M$ denotes the midpoint of the chord $AX$ . Show that $BM +CM > AY$ . by Mahan Tajrobekar

1990 Tournament Of Towns, (275) 3

Tags: geometric inequality , algebra , geometry

There are two identical clocks on the wall, one showing the current Moscow time and the other showing current local time. The minimum distance between the ends of their hour hands equals $m$ and the maximum distance equals $M$. Find the distance between the centres of the clocks. (S Fomin, Leningrad)

1943 Eotvos Mathematical Competition, 2

Tags: geometry , perimeter , geometric inequality

Let $P$ be any point inside an acute triangle. Let $D$ and $d$ be respectively the maximum and minimum distances from $P$ to any point on the perimeter of the triangle. (a) Prove that $D \ge 2d$. (b) Determine when equality holds

2015 Stars Of Mathematics, 4

Tags: geometric inequality , inequalities

Let $S$ be a finite set of points in the plane,situated in general position(any three points in $S$ are not collinear),and let $$D(S,r)=\{\{x,y\}:x,y\in S,\text{dist}(x,y)=r\},$$ where $R$ is a positive real number,and $\text{dist}(x,y)$ is the euclidean distance between points $x$ and $y$.Prove that $$\sum_{r>0}|D(S,r)|^2\le\frac{3|S|^2(|S|-1)}{4}.$$

2022 Sharygin Geometry Olympiad, 8.8

Tags: trapezoid , isosceles , geometry , geometric inequality , inequalities

An isosceles trapezoid $ABCD$ ($AB = CD$) is given. A point $P$ on its circumcircle is such that segments $CP$ and $AD$ meet at point $Q$. Let $L$ be tha midpoint of$ QD$. Prove that the diagonal of the trapezoid is not greater than the sum of distances from the midpoints of the lateral sides to ana arbitrary point of line $PL$.

Kyiv City MO Juniors Round2 2010+ geometry, 2021.7.41

Tags: geometric inequality , right angle , perimeter , geometry

Point $C$ lies inside the right angle $AOB$. Prove that the perimeter of triangle $ABC$ is greater than $2 OC$.

2015 Canada National Olympiad, 2

Tags: geometry , inequalities , geometric inequality

Let $ABC$ be an acute-angled triangle with altitudes $AD,BE,$ and $CF$. Let $H$ be the orthocentre, that is, the point where the altitudes meet. Prove that \[\frac{AB\cdot AC+BC\cdot CA+CA\cdot CB}{AH\cdot AD+BH\cdot BE+CH\cdot CF}\leq 2.\]

1980 Spain Mathematical Olympiad, 1

Tags: geometric inequality , area of a triangle , maximum , geometry

Among the triangles that have a side of length $5$ m and the angle opposite of $30^o$, determine the one with maximum area, calculating the value of the other two angles and area of triangle.

2011 Swedish Mathematical Competition, 2

Tags: acute , geometric inequality , geometry

Given a triangle $ABC$, let $P$ be a point inside the triangle such that $| BP | > | AP |, | BP | > | CP |$. Show that $\angle ABC <90^o$

1989 IMO Shortlist, 13

Tags: geometric inequality , convex quadrilateral , geometry

Let $ ABCD$ be a convex quadrilateral such that the sides $ AB, AD, BC$ satisfy $ AB \equal{} AD \plus{} BC.$ There exists a point $ P$ inside the quadrilateral at a distance $ h$ from the line $ CD$ such that $ AP \equal{} h \plus{} AD$ and $ BP \equal{} h \plus{} BC.$ Show that: \[ \frac {1}{\sqrt {h}} \geq \frac {1}{\sqrt {AD}} \plus{} \frac {1}{\sqrt {BC}} \]

1949-56 Chisinau City MO, 28

Tags: geometric inequality , geometry , inequalities

Prove the inequality $2\sqrt{(p-b)(p-c)}\le a$, where $a, b, c$ are the lengths of the sides, and $p$ is the semiperimeter of some triangle..

2017 Sharygin Geometry Olympiad, 3

Tags: geometric inequality , geometry

The angles $B$ and $C$ of an acute-angled triangle $ABC$ are greater than $60^\circ$. Points $P,Q$ are chosen on the sides $AB,AC$ respectively so that the points $A,P,Q$ are concyclic with the orthocenter $H$ of the triangle $ABC$. Point $K$ is the midpoint of $PQ$. Prove that $\angle BKC > 90^\circ$. [i]Proposed by A. Mudgal[/i]