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

A finite number of chords is drawn in a circle $1$ cm in diameter so that any diameter of the circle intersects at most $N$ of these chords. Prove that the sum of the lengths of all chords is less than $3.15 \cdot N$ cm.

Prove theat for an arbitrary triangle holds the inequality $$a \cos A+ b \cos B + c \cos C \le p ,$$ where $a, b, c$ are the sides of the triangle, $A, B, C$ are the angles, $p$ is the semiperimeter.

Prove that the sum of the squares of the distances from a point $P$ to the vertices of a triangle $ABC$ is minimum when $ P$ is the centroid of the triangle.

Let $ABC$ be an equilateral triangle with circumcircle of center $O$ and radius $R$. Point $M$ is exterior to the triangle such that $S_bS_c = S_aS_b+S_aS_c$, where $S_a, S_b, S_c$ are the areas of triangles $MBC,MCA,MAB$ respectively. Prove that $OM \ge R$. Nguyễn Tiến Lâm

A point $X$ is the origin of three rays such that the angle between any two of them equals $120^{\circ}$. Let $\omega$ be an arbitrary circle with radius $R$ such that $X$ lies inside it, and $A$, $B$, $C$ be the common points of the rays with this circle. Find $max(XA+XB+XC)$. Proposed by: F.Nilov

The polygon $P$, cut out of paper, is bent in a straight line and both halves are glued. Can the perimeter of the polygon $Q$ obtained by gluing be larger than the perimeter of the polygon $P$?

Let $P_1, P_2, \dots , P_n$ be distinct points of the plane, $n \geq 2$. Prove that \[ \max_{1\leq i<j\leq n} P_iP_j > \frac{\sqrt 3}{2}(\sqrt n -1) \min_{1\leq i<j\leq n} P_iP_j \]

For a triangle $ ABC,$ let $ k$ be its circumcircle with radius $ r.$ The bisectors of the inner angles $ A, B,$ and $ C$ of the triangle intersect respectively the circle $ k$ again at points $ A', B',$ and $ C'.$ Prove the inequality \[ 16Q^3 \geq 27 r^4 P,\] where $ Q$ and $ P$ are the areas of the triangles $ A'B'C'$ and $ABC$ respectively.

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.)

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

Prove that a tetrahedron with just one edge length greater than $1$ has volume at most $ \frac{1}{8}.$

Three different points $A$, $B$ and $C$ are placed in the plane. Construct a line $m$ through $C$ so that the product of the distances from $A$ and $B$ to $m$ has the maximal value. Is $m$ unique for every triple $A$, $B$ and $C$? (NB Vassiliev)

One convex quadrilateral is inside another. Can it turn out that the sum of the lengths of the diagonals of the outer quadrilateral is less than the sum of the lengths of the diagonals of the inner?

Let triangle $ABC$ be such that its circumradius is $R = 1.$ Let $r$ be the inradius of $ABC$ and let $p$ be the inradius of the orthic triangle $A'B'C'$ of triangle $ABC.$ Prove that \[ p \leq 1 - \frac{1}{3 \cdot (1+r)^2}. \] [hide="Similar Problem posted by Pascual2005"] Let $ABC$ be a triangle with circumradius $R$ and inradius $r$. If $p$ is the inradius of the orthic triangle of triangle $ABC$, show that $\frac{p}{R} \leq 1 - \frac{\left(1+\frac{r}{R}\right)^2}{3}$. [i]Note.[/i] The orthic triangle of triangle $ABC$ is defined as the triangle whose vertices are the feet of the altitudes of triangle $ABC$. [b]SOLUTION 1 by mecrazywong:[/b] $p=2R\cos A\cos B\cos C,1+\frac{r}{R}=1+4\sin A/2\sin B/2\sin C/2=\cos A+\cos B+\cos C$. Thus, the ineqaulity is equivalent to $6\cos A\cos B\cos C+(\cos A+\cos B+\cos C)^2\le3$. But this is easy since $\cos A+\cos B+\cos C\le3/2,\cos A\cos B\cos C\le1/8$. [b]SOLUTION 2 by Virgil Nicula:[/b] I note the inradius $r'$ of a orthic triangle. Must prove the inequality $\frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.$ From the wellknown relations $r'=2R\cos A\cos B\cos C$ and $\cos A\cos B\cos C\le \frac 18$ results $\frac{r'}{R}\le \frac 14.$ But $\frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longleftrightarrow \frac 13\left( 1+\frac rR\right)^2\le \frac 34\Longleftrightarrow$ $\left(1+\frac rR\right)^2\le \left(\frac 32\right)^2\Longleftrightarrow 1+\frac rR\le \frac 32\Longleftrightarrow \frac rR\le \frac 12\Longleftrightarrow 2r\le R$ (true). Therefore, $\frac{r'}{R}\le \frac 14\le 1-\frac 13\left( 1+\frac rR\right)^2\Longrightarrow \frac{r'}{R}\le 1-\frac 13\left( 1+\frac rR\right)^2.$ [b]SOLUTION 3 by darij grinberg:[/b] I know this is not quite an ML reference, but the problem was discussed in Hyacinthos messages #6951, #6978, #6981, #6982, #6985, #6986 (particularly the last message). [/hide]

Let $k_1$ and $k_2$ be two circles with centers $O_1$ and $O_2$ and equal radius $r$ such that $O_1O_2 = r$. Let $A$ and $B$ be two points lying on the circle $k_1$ and being symmetric to each other with respect to the line $O_1O_2$. Let $P$ be an arbitrary point on $k_2$. Prove that \[PA^2 + PB^2 \geq 2r^2.\]

Two right triangles are given, of which the incircle of the first triangle is the circumcircle of the second triangle. Let the areas of the triangles be $S$ and $S'$ respectively. Prove that $\frac{S}{S'} \ge 3 +2\sqrt2$

Let $\omega$ be a circle. For each $n$, let $A_n$ be the area of a regular $n$-sided polygon circumscribed to $\omega$ and $B_n$ the area of a regular $n$-sided polygon inscribed in $\omega$ . Try that $3A_{2015} + B_{2015}> 4A_{4030}$

Let $M$ be the midpoint of segment $BC$ of $\triangle ABC$. Let $D$ be a point such that $AD=AB$, $AD\perp AB$ and points $C$ and $D$ are on different sides of $AB$. Prove that: $$\sqrt{AB\cdot AC+BC\cdot AM}\geq\frac{\sqrt{2}}{2}CD.$$

Let $ABCDEF$ be a hexagon having all interior angles equal to $120^o$ each. Let $P,Q,R, S, T, V$ be the midpoints of the sides of the hexagon $ABCDEF$. Prove the inequality $$p(PQRSTV ) \ge \frac{\sqrt3}{2} p(ABCDEF)$$, where $p(.)$ denotes the perimeter of the polygon. Nguyễn Tiến Lâm

Suppose $n>2$ and let $A_1,\dots,A_n$ be points on the plane such that no three are collinear. [b](a)[/b] Suppose $M_1,\dots,M_n$ be points on segments $A_1A_2,A_2A_3,\dots ,A_nA_1$ respectively. Prove that if $B_1,\dots,B_n$ are points in triangles $M_2A_2M_1,M_3A_3M_2,\dots ,M_1A_1M_n$ respectively then \[|B_1B_2|+|B_2B_3|+\dots+|B_nB_1| \leq |A_1A_2|+|A_2A_3|+\dots+|A_nA_1|\] Where $|XY|$ means the length of line segment between $X$ and $Y$. [b](b)[/b] If $X$, $Y$ and $Z$ are three points on the plane then by $H_{XYZ}$ we mean the half-plane that it's boundary is the exterior angle bisector of angle $\hat{XYZ}$ and doesn't contain $X$ and $Z$ ,having $Y$ crossed out. Prove that if $C_1,\dots ,C_n$ are points in ${H_{A_nA_1A_2},H_{A_1A_2A_3},\dots,H_{A_{n-1}A_nA_1}}$ then \[|A_1A_2|+|A_2A_3|+\dots +|A_nA_1| \leq |C_1C_2|+|C_2C_3|+\dots+|C_nC_1|\] Time allowed for this problem was 2 hours.

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 $PL$.

Let $ABC$ be a triangle with semiperimeter $s$ and inradius $r$. The semicircles with diameters $BC$, $CA$, $AB$ are drawn on the outside of the triangle $ABC$. The circle tangent to all of these three semicircles has radius $t$. Prove that \[\frac{s}{2}<t\le\frac{s}{2}+\left(1-\frac{\sqrt{3}}{2}\right)r. \] [i]Alternative formulation.[/i] In a triangle $ABC$, construct circles with diameters $BC$, $CA$, and $AB$, respectively. Construct a circle $w$ externally tangent to these three circles. Let the radius of this circle $w$ be $t$. Prove: $\frac{s}{2}<t\le\frac{s}{2}+\frac12\left(2-\sqrt3\right)r$, where $r$ is the inradius and $s$ is the semiperimeter of triangle $ABC$. [i]Proposed by Dirk Laurie, South Africa[/i]

Inside the triangle $A_1A_2A_3$ with sides $a_1$, $a_2$, $a_3$, three points are given, which we label $P_1$, $P_2$, $P_3$ so that the product of their distances from the corresponding sides $a_1$, $a_2$, $a_3$ is as large as possible. Prove that the triangles $P_1A_2A_3$, $A_1P_2A_3$, $A_1A_2P_3$ cover the triangle. [hide=original wording]V trojúhelníku A1A2A3 se stranami a1, a2, a3 jsou dány tři body, které označíme Pi, P2, P3 tak, aby součin jejich vzdáleností od odpovídajících stran a1, a2, a3 byl co největší. Dokažte, že trojúhelníky P1A2A3, A1P2A3, A1A2P3 pokrývají trojúhelník.[/quote]

In the triangle $ABC$ with sides $BC> AC> AB$ the angles between altiude and median drawn from one vertex are considered. Find out at which vertex this angle is the largest of the three. (Rozhkova Maria)