Olympiad problems

1984 IMO Longlists, 33

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$.

2005 iTest, 3

Tags: geometry , geometric inequality

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.

2024 Moldova Team Selection Test, 6

Tags: geometric inequality , barycentric coordinates

Prove that in any triangle the length of the shortest bisector does not exceed three times the radius of the incircle.

1954 Moscow Mathematical Olympiad, 279

Tags: geometric inequality , geometry , concurrency

Given four straight lines, $m_1, m_2, m_3, m_4$, intersecting at $O$ and numbered clockwise with $O$ as the center of the clock, we draw a line through an arbitrary point $A_1$ on $m_1$ parallel to $m_4$ until the line meets $m_2$ at $A_2$. We draw a line through $A_2$ parallel to $m_1$ until it meets $m_3$ at $A_3$. We also draw a line through $A_3$ parallel to $m_2$ until it meets $m_4$ at $A_4$. Now, we draw a line through$ A_4$ parallel to $m_3$ until it meets $m_1$ at $B$. Prove that a) $OB< \frac{OA_1}{2}$ . b) $OB \le \frac{OA_1}{4}$ . [img]https://cdn.artofproblemsolving.com/attachments/5/f/5ea08453605e02e7e1253fd7c74065a9ffbd8e.png[/img]

2006 IMO, 6

Tags: geometry , area , geometric inequality

Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.

1983 All Soviet Union Mathematical Olympiad, 355

Tags: geometry , geometric inequality , inequalities , area

The point $D$ is the midpoint of the side $[AB]$ of the triangle $ABC$ . The points $E$ and $F$ belong to $[AC]$ and $[BC]$ respectively. Prove that the area of triangle $DEF$ area does not exceed the sum of the areas of triangles $ADE$ and $BDF$.

2016 India IMO Training Camp, 1

Tags: geometry , inradius , circumcircle , geometric inequality

Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $A_1,B_1$ and $C_1$ be respectively the midpoints of the arcs $BAC,CBA$ and $ACB$ of $\Gamma$. Show that the inradius of triangle $A_1B_1C_1$ is not less than the inradius of triangle $ABC$.

1966 IMO Longlists, 38

Tags: inequalities , circles , geometry , geometric inequality

Two concentric circles have radii $R$ and $r$ respectively. Determine the greatest possible number of circles that are tangent to both these circles and mutually nonintersecting. Prove that this number lies between $\frac 32 \cdot \frac{\sqrt R +\sqrt r }{\sqrt R -\sqrt r } -1$ and $\frac{63}{20} \cdot \frac{R+r}{R-r}.$

2012 Thailand Mathematical Olympiad, 4

Tags: geometry , right triangle , geometric inequality , incenter , area

Let $ABCD$ be a unit square. Points $E, F, G, H$ are chosen outside $ABCD$ so that $\angle AEB =\angle BF C = \angle CGD = \angle DHA = 90^o$ . Let $O_1, O_2, O_3, O_4$, respectively, be the incenters of $\vartriangle ABE, \vartriangle BCF, \vartriangle CDG, \vartriangle DAH$. Show that the area of $O_1O_2O_3O_4$ is at most $1$.

1980 Spain Mathematical Olympiad, 1

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

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.

2014 Sharygin Geometry Olympiad, 1

Tags: geometry , cyclic quadrilateral , geometric inequality

Let $ABCD$ be a cyclic quadrilateral. Prove that $AC > BD$ if and only if $(AD-BC)(AB- CD) > 0$. (V. Yasinsky)

1988 Tournament Of Towns, (202) 6

Tags: rectangle , geometric inequality , geometry

$M$ is an interior point of a rectangle $ABCD$ and $S$ is its area. Prove that $S \le AM \cdot CM + BM \cdot DM$. (I.J . Goldsheyd)

1989 All Soviet Union Mathematical Olympiad, 497

Tags: geometry , geometric inequality , equal ratio , area

$ABCD$ is a convex quadrilateral. $X$ lies on the segment $AB$ with $\frac{AX}{XB} = \frac{m}{n}$. $Y$ lies on the segment $CD$ with $\frac{CY}{YD} = \frac{m}{n}$. $AY$ and $DX$ intersect at $P$, and $BY$ and $CX$ intersect at $Q$. Show that $\frac{S_{XQYP}}{S_{ABCD}} < \frac{mn}{m^2 + mn + n^2}$.

1992 Austrian-Polish Competition, 7

Tags: 3d geometry , solid geometry , geometric inequality , right angle , geometry

Consider triangles $ABC$ in space. (a) What condition must the angles $\angle A, \angle B , \angle C$ of $\triangle ABC$ fulfill in order that there is a point $P$ in space such that $\angle APB, \angle BPC, \angle CPA$ are right angles? (b) Let $d$ be the longest of the edges $PA,PB,PC$ and let $h$ be the longest altitude of $\triangle ABC$. Show that $\frac{1}{3}\sqrt6 h \le d \le h$.

2012 Mathcenter Contest + Longlist, 4

Tags: geometric inequality , inequalities , algebra , geometry

Let $a,b,c$ be the side lengths of any triangle. Prove that $$\frac{a}{\sqrt{2b^2+2c^2-a^2}}+\frac{b}{\sqrt{2c^2+2a^2-b^2 }}+\frac{c}{\sqrt{2a^2+2b^2-c^2}}\ge \sqrt{3}.$$ [i](Zhuge Liang)[/i]

1992 IMO Longlists, 73

Tags: geometry , geometric inequality , point set

Let $\{A_n | n = 1, 2, \cdots \} $ be a set of points in the plane such that for each $n$, the disk with center $A_n$ and radius $2^n$ contains no other point $A_j$ . For any given positive real numbers $a < b$ and $R$, show that there is a subset $G$ of the plane satisfying: [b](i)[/b] the area of $G$ is greater than or equal to $R$; [b](ii) [/b]for each point $P$ in $G$, $a < \sum_{n=1}^{\infty} \frac{1}{|A_nP|} <b.$

2021 Oral Moscow Geometry Olympiad, 5

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

The trapezoid is inscribed in a circle. Prove that the sum of distances from any point of the circle to the midpoints of the lateral sides are not less than the diagonal of the trapezoid.

2001 All-Russian Olympiad Regional Round, 11.7

Tags: geometry , combinatorics , combinatorial geometry , geometric inequality

There is an infinite set of points $S$ on the plane, and any $1\times 1$ square contains a finite number of points from the set $S$. Prove that there are two different points $A$ and $B$ from $S$ such that for any other point $X$ from $S$ the following inequalities hold: $$|XA|, |XB| \ge 0.999|AB|.$$

1966 IMO Shortlist, 21

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

Prove that the volume $V$ and the lateral area $S$ of a right circular cone satisfy the inequality \[\left( \frac{6V}{\pi}\right)^2 \leq \left( \frac{2S}{\pi \sqrt 3}\right)^3\] When does equality occur?

2019 Jozsef Wildt International Math Competition, W. 69

Tags: geometry , triangle , angle bisector , geometric inequality , inequalities , circumradius , inradius

Denote $\overline{w_a}, \overline{w_b}, \overline{w_c}$ the external angle-bisectors in triangle $ABC$, prove that $$\sum \limits_{cyc} \frac{1}{w_a}\leq \sqrt{\frac{(s^2 - r^2 - 4Rr)(8R^2 - s^2 - r^2 - 2Rr)}{8s^2R^2r}}$$

2018 Kazakhstan National Olympiad, 6

Tags: inequalities , geometric inequality , geometry

Inside of convex quadrilateral $ABCD$ found a point $M$ such that $\angle AMB=\angle ADM+\angle BCM$ and $\angle AMD=\angle ABM+\angle DCM$.Prove that $$AM\cdot CM+BM\cdot DM\ge \sqrt{AB\cdot BC\cdot CD\cdot DA}.$$

2000 Greece JBMO TST, 4

Tags: geometry , geometric inequality , inequalities

Let $a,b,c$ be sidelengths with $a\ge b\ge c$ and $s\ge a+1$ where $s$ be the semiperimeter of the triangle. Prove that $$ \frac{s-c}{\sqrt{a}}+\frac{s-b}{\sqrt{c}}+\frac{s-a}{\sqrt{b}}\ge \frac{s-b}{\sqrt{a}}+\frac{s-c}{\sqrt{b}}+\frac{s-a}{\sqrt{c}}$$

1993 Tournament Of Towns, (393) 1

Tags: geometry , geometric inequality

Two tangents $CA$ and $CB$ are drawn to a circle ($A$ and $B$ being the tangent points). Consider a “triangle” bounded by an arc $AB$ (the smaller one) and segments $CA$ and $CB$. Prove that the length of any segment inside the triangle is not greater than the length of $CA = CB$. (Folklore)

2011 Oral Moscow Geometry Olympiad, 6

Tags: geometry , geometric inequality , inscribed triangle , inscribed , side

One triangle lies inside another. Prove that at least one of the two smallest sides (out of six) is the side of the inner triangle.

VI Soros Olympiad 1999 - 2000 (Russia), 11.3

Tags: geometry , geometric inequality , tangential

A convex quadrilateral $ABCD$ has an inscribed circle touching its sides $AB$, $BC$, $CD$, $DA$ at the points $M$,$N$,$P$,$K$, respectively. Let $O$ be the center of the inscribed circle, the area of the quadrilateral $MNPK$ is equal to $8$. Prove the inequality $$2S \le OA \cdot OC+ OB \cdot OD.$$