Olympiad problems

1931 Eotvos Mathematical Competition, 3

Let $A$ and $B$ be two given points, distance $1 $ apart. Determine a point $P$ on the line $AB$ such that $$\frac{1}{1 + AP}+\frac{1}{1 + BP}$$ is a maximum.

2010 Saudi Arabia BMO TST, 2

Tags: geometry , geometric inequality , inscribed , square , area

Let $ABC$ be an acute triangle and let $MNPQ$ be a square inscribed in the triangle such that $M ,N \in BC$, $P \in AC$, $Q \in AB$. Prove that $area \, [MNPQ] \le \frac12 area\, [ABC]$.

2024 South Africa National Olympiad, 5

Tags: geometric inequality , geometry , inequalities

Consider three circles $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$, with centres $O_1$, $O_2$ and $O_3$, respectively, such that each pair of circles is externally tangent. Suppose we have another circle $\Gamma$ with centre $O$ on the line segment $O_1O_3$ such that $\Gamma_1$, $\Gamma_2$ and $\Gamma_3$ are each internally tangent to $\Gamma$. Show that $\angle O_1O_2O_3$ measures less than $90^\circ$.

2009 Sharygin Geometry Olympiad, 1

Tags: geometric inequality , geometry

Let $a, b, c$ be the lengths of some triangle's sides, $p, r$ be the semiperimeter and the inradius of triangle. Prove an inequality $\sqrt{\frac{ab(p- c)}{p}} +\sqrt{\frac{ca(p- b)}{p}} +\sqrt{\frac{bc(p-a)}{p}} \ge 6r$ (D.Shvetsov)

2009 Bundeswettbewerb Mathematik, 3

Tags: geometry , angle bisector , circumcircle , geometric inequality

Given a triangle $ABC$ and a point $P$ on the side $AB$ . Let $Q$ be the intersection of the straight line $CP$ (different from $C$) with the circumcicle of the triangle. Prove the inequality $$\frac{\overline{PQ}}{\overline{CQ}} \le \left(\frac{\overline{AB}}{\overline{AC}+\overline{CB}}\right)^2$$ and that equality holds if and only if the $CP$ is bisector of the angle $ACB$. [img]https://cdn.artofproblemsolving.com/attachments/b/1/068fafd5564e77930160115a1cd409c4fdbf61.png[/img]

1989 Poland - Second Round, 3

Tags: 3d geometry , tetrahedron , volume , geometric inequality , geometry

Given is a trihedral angle $ OABC $ with a vertex $ O $ and a point $ P $ in its interior. Let $ V $ be the volume of a parallelepiped with two vertices at points $ O $ and $ P $, whose three edges are contained in the rays $ \overrightarrow{OA} $, $ \overrightarrow{OB} $, $ \overrightarrow{OC} $. Calculate the minimum volume of a tetrahedron whose three faces are contained in the faces of the trihedral angle $OABC$ and the fourth face contains the point $P$.

2019 Yasinsky Geometry Olympiad, p5

Tags: inequalities , geometry , inradius , geometric inequality , radius

In the triangle $ABC$ it is known that $BC = 5, AC - AB = 3$. Prove that $r <2 <r_a$ . (here $r$ is the radius of the circle inscribed in the triangle $ABC$, $r_a$ is the radius of an exscribed circle that touches the sides of $BC$). (Mykola Moroz)

Ukrainian TYM Qualifying - geometry, XII.15

Tags: geometry , trigonometry , 3d geometry , geometric inequality

Given a triangular pyramid $SABC$, in which $\angle BSC = \alpha$, $\angle CSA =\beta$, $\angle ASB = \gamma$, and the dihedral angles at the edges $SA$ and $SB$ have the value of $\phi$ and $\delta$, respectively. Prove that $\gamma > \alpha \cdot \cos \delta +\beta \cdot \cos \phi.$$

1999 Brazil Team Selection Test, Problem 2

Tags: geometric inequality , inequalities , geometry , ratio

In a triangle $ABC$, the bisector of the angle at $A$ of a triangle $ABC$ intersects the segment $BC$ and the circumcircle of $ABC$ at points $A_1$ and $A_2$, respectively. Points $B_1,B_2,C_1,C_2$ are analogously defined. Prove that $$\frac{A_1A_2}{BA_2+CA_2}+\frac{B_1B_2}{CB_2+AB_2}+\frac{C_1C_2}{AC_2+BC_2}\ge\frac34.$$

1989 IMO Longlists, 40

Tags: geometry , geometric inequality , convex quadrilateral

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: \[ \frac {1}{\sqrt {h}} \geq \frac {1}{\sqrt {AD}} \plus{} \frac {1}{\sqrt {BC}} \]

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)

2002 Regional Competition For Advanced Students, 3

Tags: geometry , perimeter , inequalities , convex , hexagon , geometric inequality

In the convex $ABCDEF$ (has all interior angles less than $180^o$) with the perimeter $s$ the triangles $ACE$ and $BDF$ have perimeters $u$ and $v$ respectively. a) Show the inequalities $\frac{1}{2} \le \frac{s}{u+v}\le 1$ b) Check whether $1$ is replaced by a smaller number or $1/2$ by a larger number can the inequality remains valid for all convex hexagons.

Ukrainian TYM Qualifying - geometry, I.9

Tags: geometry , geometric inequality

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]

1990 ITAMO, 4

Tags: inequalities , algebra , geometric inequality

Let $a,b,c$ be side lengths of a triangle with $a+b+c = 1$. Prove that $a^2 +b^2 +c^2 +4abc \le \frac12$ .

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: $$a + b + c> s_a + s_b + s_c> \frac34 (a + b + c) $$

Ukraine Correspondence MO - geometry, 2006.7

Tags: geometry , geometric inequality , right triangle

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 $$\frac{5}{6} \le \frac{AD}{AB}\le \frac{\sqrt{73}}{10}.$$

2022 Taiwan TST Round 1, G

Tags: geometric inequality , geometry , inequalities

Two triangles $ABC$ and $A'B'C'$ are on the plane. It is known that each side length of triangle $ABC$ is not less than $a$, and each side length of triangle $A'B'C'$ is not less than $a'$. Prove that we can always choose two points in the two triangles respectively such that the distance between them is not less than $\sqrt{\dfrac{a^2+a'^2}{3}}$.

1998 Belarus Team Selection Test, 2

Tags: geometry , perimeter , geometric inequality , incircle , inradius

The incircle of the triangle $ABC$ touches its sides $AB,BC,CA$ at points $C_1,A_1,B_1$ respectively. If $r$ is the inradius of $\vartriangle ABC, P,P_1$ are the perimeters of $\vartriangle ABC, \vartriangle A_1B_1C_1$ respectively, prove that $P+P_1 \ge 9 \sqrt3 r$. I. Voronovich

1972 Poland - Second Round, 3

Tags: geometry , circumcircle , analytic geometry , lattice points , geometric inequality

The coordinates of the triangle's vertices in the Cartesian system $XOY$ are integers. Prove that the diameter of the circle circumscribed by this triangle is not greater than the product of the lengths of the triangle's sides.

2016 Korea National Olympiad, 3

Tags: geometry , geometric inequality

Acute triangle $\triangle ABC$ has area $S$ and perimeter $L$. A point $P$ inside $\triangle ABC$ has $dist(P,BC)=1, dist(P,CA)=1.5, dist(P,AB)=2$. Let $BC \cap AP = D$, $CA \cap BP = E$, $AB \cap CP= F$. Let $T$ be the area of $\triangle DEF$. Prove the following inequality. $$ \left( \frac{AD \cdot BE \cdot CF}{T} \right)^2 > 4L^2 + \left( \frac{AB \cdot BC \cdot CA}{24S} \right)^2 $$

2018 Irish Math Olympiad, 7

Tags: geometric inequality , inequalities , algebra

Let $a, b, c$ be the side lengths of a triangle. Prove that $2 (a^3 + b^3 + c^3) < (a + b + c) (a^2 + b^2 + c^2) \le 3 (a^3 + b^3 + c^3)$

2020 Estonia Team Selection Test, 2

Tags: geometry , geometric inequality , area

The radius of the circumcircle of triangle $\Delta$ is $R$ and the radius of the inscribed circle is $r$. Prove that a circle of radius $R + r$ has an area more than $5$ times the area of triangle $\Delta$.

2015 Czech and Slovak Olympiad III A, 5

Tags: geometry , geometric inequality , incenter

In given triangle $\triangle ABC$, difference between sizes of each pair of sides is at least $d>0$. Let $G$ and $I$ be the centroid and incenter of $\triangle ABC$ and $r$ be its inradius. Show that $$[AIG]+[BIG]+[CIG]\ge\frac{2}{3}dr,$$ where $[XYZ]$ is (nonnegative) area of triangle $\triangle XYZ$.

Kyiv City MO 1984-93 - geometry, 1985.8.3

Tags: geometry , geometric inequality

The longest diagonal of a convex hexagon is $2$. Is there necessarily a side or diagonal in this hexagon whose length does not exceed $1$?

1987 IMO Shortlist, 4

Tags: geometry , geometric inequality , 3d geometry , polyhedron

Let $ABCDEFGH$ be a parallelepiped with $AE \parallel BF \parallel CG \parallel DH$. Prove the inequality \[AF + AH + AC \leq AB + AD + AE + AG.\] In what cases does equality hold? [i]Proposed by France.[/i]