Prove that the largest area of a triangle with sides $a, b, c$ satisfying the relation $a^2 +b^2 c^2 = 3m^2$, equals to $\frac{\sqrt3}{4}m^2$.

In a convex quadrilateral $ABCD$, the diagonal $AC$ intersects the diagonal $BD$ at its midpoint $S$. The radii of incircles of triangles $ABS,BCS,CDS,DAS$ are $r_1,r_2,r_3,r_4$, respectively. Prove that $$|r_1-r_2+r_3-r_4|\le\frac18|AB-BC+CD-DA|.$$

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

On the sides $AB, BC, CA$ of triangle $ABC$, points $C', A', B'$ are taken. Prove that for the areas of the corresponding triangles, the inequality holds: $$S_{ABC}S^2_{A'B'C'}\ge 4S_{AB'C'}S_{BC'A'}S_{CA'B'}$$ and equality is achieved if and only if the lines $AA', BB', CC'$ intersect at one point.

Let $a,b,c$ be the sides of a triangle, $\alpha ,\beta ,\gamma$ the corresponding angles and $r$ the inradius. Prove that $$a\cdot sin\alpha+b\cdot sin\beta+c\cdot sin\gamma\geq 9r$$

The area $T$ and an angle $\gamma$ of a triangle are given. Determine the lengths of the sides $a$ and $b$ so that the side $c$, opposite the angle $\gamma$, is as short as possible.

In a triangle $ABC$, let $x=\cos\frac{A-B}{2},y=\cos\frac{B-C}{2},z=\cos\frac{C-A}{2}$. Prove that $$x^4+y^4+z^2\leq 1+2x^2y^2z^2.$$

Let $ABCDEF$ be a hexagon with sides $AB,CD,EF$ being equal to $m$ units, sides $BC,DE, FA$ being equal to $n$ units. The diagonals $AD,BE,CF$ have lengths $x, y$, and $z$ units. Prove the inequality $$\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} \ge \frac{3}{(m+ n)^2}$$ Nguyễn Văn Quý

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 $area \, [MNPQ] \le \frac12 area\, [ABC]$.

In a metallic disk, a circular sector is removed, so that with the remaining can form a conical glass of maximum volume. Calculate, in radians, the angle of the sector that is removed. [hide=original wording]En un disco metalico se quita un sector circular, de modo que con la parte restante se pueda formar un vaso c´onico de volumen maximo. Calcular, en radianes, el angulo del sector que se quita.[/hide]

Into an angle $A$ of size $a$, a circle is inscribed tangent to its sides at points $B$ and $C$. A line tangent to this circle at a point M meets the segments $AB$ and $AC$ at points $P$ and $Q$ respectively. What is the minimum $a$ such that the inequality $S_{PAQ}<S_{BMC}$ is possible?

On the base $AB$ of the isosceles triangle $ABC$, lies the point $P$ such that $AP : PB = 1 : 2$. Determine the minimum of $\angle ACP$.

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 $ A_1A_2A_3A_4$ be a tetrahedron, $ G$ its centroid, and $ A'_1, A'_2, A'_3,$ and $ A'_4$ the points where the circumsphere of $ A_1A_2A_3A_4$ intersects $ GA_1,GA_2,GA_3,$ and $ GA_4,$ respectively. Prove that \[ GA_1 \cdot GA_2 \cdot GA_3 \cdot GA_ \cdot4 \leq GA'_1 \cdot GA'_2 \cdot GA'_3 \cdot GA'_4\] and \[ \frac{1}{GA'_1} \plus{} \frac{1}{GA'_2} \plus{} \frac{1}{GA'_3} \plus{} \frac{1}{GA'_4} \leq \frac{1}{GA_1} \plus{} \frac{1}{GA_2} \plus{} \frac{1}{GA_3} \plus{} \frac{1}{GA_4}.\]

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

The altitudes $BB_1$ and $CC_1$ of the acute triangle $\triangle ABC$ intersect at $H$. The circle centered at $O_b$ passes through points $A,C_1$, and the midpoint of $BH$. The circle centered at $O_c$ passes through $A,B_1$ and the midpoint of $CH$. Prove that $B_1 O_b +C_1O_c > \frac{BC}{4}$

Let $I$ be the incenter of triangle $ABC$. The excircle with center $I_A$ touches the side $BC$ at point $A'$. The line $l$ passing through $I$ and perpendicular to $BI$ meets $I_AA'$ at point $K$ lying on the medial line parallel to $BC$. Prove that $\angle B \leq 60^\circ$.

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 $\Delta$.

Show that for any vectors $a, b$ in Euclidean space, \[|a \times b|^3 \leq \frac{3 \sqrt 3}{8} |a|^2 |b|^2 |a-b|^2\] Remark. Here $\times$ denotes the vector product.

In the tetrahedron $ABCD,\angle BDC=90^o$ and the foot of the perpendicular from $D$ to $ABC$ is the intersection of the altitudes of $ABC$. Prove that: \[ (AB+BC+CA)^2\le6(AD^2+BD^2+CD^2). \] When do we have equality?

The lengths, $a$ and $b$, of two sides of a triangle are known. (a) What length should the third side be, in order for the largest angle of the triangle to be of the least possible value? (b) What length should the third side be in order for the smallest angle of the triangle to be of the greatest possible value?

Let $ABC$ be an equilateral triangle, and let $P$ stand for an arbitrary point inside the triangle. Is it true that $| \angle PAB - \angle PAC| \ge | \angle PBC - \angle PCB|$ ?

Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.

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

The insphere of a tetrahedron $ABCD$ touches the faces $ABC$, $BCD$, $CDA$, $DAB$ at $D^{\prime}$, $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ respectively. Denote by $S_{AB}$ the area of the triangle $AC^{\prime}B^{\prime}$. Define similarly $S_{AC}$, $S_{BC},$ $S_{AD}$, $S_{BD}$, $S_{CD}$. Prove that there exists a triangle with sidelengths $\sqrt{S_{AB}S_{CD}}$, $\sqrt{S_{AC}S_{BD}}$ , $\sqrt{S_{AD}S_{BC}}$. Proposed by: S.Arutyunyan