Let $ A_1,B_1,C_1 $ be points on the sides (excluding their endpoints) $ BC,CA,AB, $ respectively, of a triangle $ ABC, $ such that $ \angle A_1AB =\angle B_1BC=\angle C_1CA. $ Let $ A^* $ be the intersection of $ BB_1 $ with $ CC_1,B^* $ be the intersection of $ CC_1 $ with $ AA_1, $ and $ C^* $ be the intersection of $ AA_1 $ with $ BB_1. $ Denote with $ r_A,r_B,r_C $ the inradii of $ A^*BC,AB^*C,ABC^*, $ respectively. Prove that $$ \frac{r_A}{BC}=\frac{r_B}{CA}=\frac{r_C}{AB} $$ if and only if $ ABC $ is equilateral. [i]Daniel Văcărețu[/i]

$ ABC$ is a triangle, the bisector of angle $ A$ meets the circumcircle of triangle $ ABC$ in $ A_1$, points $ B_1$ and $ C_1$ are defined similarly. Let $ AA_1$ meet the lines that bisect the two external angles at $ B$ and $ C$ in $ A_0$. Define $ B_0$ and $ C_0$ similarly. Prove that the area of triangle $ A_0B_0C_0 \equal{} 2 \cdot$ area of hexagon $ AC_1BA_1CB_1 \geq 4 \cdot$ area of triangle $ ABC$.

Prove that for a triangle $ ABC $ with $ \angle BAC \ge 90^{\circ } , $ having circumradius $ R $ and inradius $ r, $ the following inequality holds: $$ R\sin A>2r $$ [i]Romeo Ilie[/i]

Let $ABC$ be a triangle with circumradius $R=17$ and inradius $r=7$. Find the maximum possible value of $\sin \frac{A}{2}$.

Let $ABC$ be an equilateral triangle of side $1$, $D$ be a point on $BC$, and $r_1, r_2$ be the inradii of triangles $ABD$ and $ADC$. Express $r_1r_2$ in terms of $p = BD$ and find the maximum of $r_1r_2$.

In the any $[ABCD]$ tetrahedron we denote with $\alpha$, $\beta$, $\gamma$ the measures, in radians, of the angles of the three pairs of opposite edges and with $r$, $R$ the lengths of the rays of the sphere inscribed and respectively circumscribed the tetrahedron. Demonstrate inequality$$\left(\frac{3r}{R}\right)^3\leq \sin \frac{\alpha +\beta +\gamma}{3}$$(A refinement of inequality $R \geq 3r$).

For a given triangle $ ABC$, let $ X$ be a variable point on the line $ BC$ such that $ C$ lies between $ B$ and $ X$ and the incircles of the triangles $ ABX$ and $ ACX$ intersect at two distinct points $ P$ and $ Q.$ Prove that the line $ PQ$ passes through a point independent of $ X$.

Let $O$ and $I$ be the centers of the circumscribed and inscribed circle the acute-angled triangle $ABC$, respectively. It is known that line $OI$ is parallel to the side $BC$ of this triangle. Line $MI$, where $M$ is the midpoint of $BC$, intersects the altitude $AH$ at the point $T$. Find the length of the segment $IT$, if the radius of the circle inscribed in the triangle $ABC$ is equal to $r$. (Grigory Filippovsky)

Points $B = B_1 , B_2, \cdots , B_n , B_{n+1} = C$ are chosen on side $BC$ of a triangle $ABC$ in that order. Let $r_j$ be the inradius of triangle $AB_jB_{j+1}$ for $j = 1, \cdots, n$ , and $r$ be the inradius of $\triangle ABC$. Show that there is a constant $\lambda$ independent of $n$ such that : \[(\lambda -r_1)(\lambda -r_2)\cdots (\lambda -r_n) =\lambda^{n-1}(\lambda -r)\]

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]

A square of side $a$ is inscribed in a triangle so that two of the square’s vertices lie on the base, and the other two lie on the sides of the triangle. Prove that if $r$ is the radius of the circle inscribed in the triangle, then $r\sqrt2 < a < 2r$.

The altitudes of a triangle have integer length and its inradius is a prime number. Find all possible values of the sides of the triangle.

$ \Delta ABC$ is a triangle such that $ AB \neq AC$. The incircle of $ \Delta ABC$ touches $ BC, CA, AB$ at $ D, E, F$ respectively. $ H$ is a point on the segment $ EF$ such that $ DH \bot EF$. Suppose $ AH \bot BC$, prove that $ H$ is the orthocentre of $ \Delta ABC$. Remark: the original question has missed the condition $ AB \neq AC$

The diagonals of a convex quadrilateral divide it into four triangles. The radii of the circles circumscribed around these triangles are equal. Can such a property have a quadrilateral other than: a) parallelogram, b) rhombus? (Sharygin Igor)

The side of an equilateral triangle is $s$. A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is: $ \text{(A)}\ \frac{s^2}{24}\qquad\text{(B)}\ \frac{s^2}{6}\qquad\text{(C)}\ \frac{s^2\sqrt{2}}{6}\qquad\text{(D)}\ \frac{s^2\sqrt{3}}{6}\qquad\text{(E)}\ \frac{s^2}{3} $