This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 649

The $ABCD$ is a cyclic quadrilateral with no parallel sides inscribed in circle $(O, R)$. Let $E$ be the intersection of two diagonals and the angle bisector of $AEB$ cut the lines $AB, BC, CD, DA$ at $M, N, P, Q$ respectively . a) Prove that the circles $(AQM), (BMN), (CNP), (DPQ)$ are passing through a point. Call that point $K$. b) Denote $min \,\{AC, BD\} = m$. Prove that $OK \le \dfrac{2R^2}{\sqrt{4R^2-m^2}}$.
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.
Let $ABCD$ be a convex quadrilateral. Let $E$ be the intersection of its diagonals. Suppose that $CD = BC = BE$. Prove that $AD + DC\ge AB$. [i]Proposed by Dominik Burek - Poland[/i]
Let $ABC$ be a triangle with a circumcircle $(K)$. A circle touching the sides $AB,AC$ is internally tangent to $(K)$ at $K_a$; two other points $K_b,K_c$ are defined in the same manner. Prove that the area of triangle $K_aK_bK_c$ does not exceed that of triangle $ABC$. Nguyen Minh Ha, Hanoi University of Education, Xuan Thuy, Cau Giay, Hanoi.
A tetrahedron $OA_1B_1C_1$ is given. Let $A_2,A_3 \in OA_1, A_2,A_3 \in OA_1, A_2,A_3 \in OA_1$ be points such that the planes $A_1B_1C_1,A_2B_2C_2$ and $A_3B_3C_3$ are parallel and $OA_1 > OA_2 > OA_3 > 0$. Let $V_i$ be the volume of the tetrahedron $OA_iB_iC_i$ ($i = 1,2,3$) and $V$ be the volume of $OA_1B_2C_3$. Prove that $V_1 +V_2 +V_3 \ge 3V$.
Let $a, b, c$ be the side lengths of a triangle. Prove that $2 (a^3 + b^3 + c^3) < (a + b + c) (a^2 + b^2 + c^2) \le 3 (a^3 + b^3 + c^3)$
The height and radius of the base of the cylinder are equal to $1$. What is the smallest number of balls of radius $1$ that can cover the entire cylinder?
A triangle with sides $a$, $b$, and $c$ is given. Denote by $s$ the semiperimeter, that is $s = \frac{a + b + c}{2}$. Construct a triangle with sides $s - a$, $s - b$, and $s - c$. This process is repeated until a triangle can no longer be constructed with the side lengths given. For which original triangles can this process be repeated indefinitely?
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]
In a convex quadrilateral (in the plane) with the area of $32 \text{ cm}^{2}$ the sum of two opposite sides and a diagonal is $16 \text{ cm}$. Determine all the possible values that the other diagonal can have.
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 us consider a triangle $\Delta{PQR}$ in the co-ordinate plane. Show for every function $f: \mathbb{R}^2\to \mathbb{R}\;,f(X)=ax+by+c$ where $X\equiv (x,y) \text{ and } a,b,c\in\mathbb{R}$ and every point $A$ on $\Delta PQR$ or inside the triangle we have the inequality: \begin{align*} & f(A)\le \text{max}\{f(P),f(Q),f(R)\} \end{align*}
Let $ABC$ be such a triangle that $\sphericalangle BAC = 45^{\circ}$ and $ \sphericalangle ACB > 90^{\circ}.$ Show that $BC + (\sqrt{2} - 1)\cdot CA < AB.$
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
$ 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 any noncollinear points $ A,B,C $ and positive real numbers $ x,y, $ the following inequality is true. $$ xAB^2- \frac{xy}{x+y}BC^2 +yCA^2\ge 0 $$ [i]Constantin Rusu[/i]
Find the side length of the largest square that can be inscribed in the unit cube.
In the plane, there are $n \geqslant 6$ pairwise disjoint disks $D_{1}, D_{2}, \ldots, D_{n}$ with radii $R_{1} \geqslant R_{2} \geqslant \ldots \geqslant R_{n}$. For every $i=1,2, \ldots, n$, a point $P_{i}$ is chosen in disk $D_{i}$. Let $O$ be an arbitrary point in the plane. Prove that \[O P_{1}+O P_{2}+\ldots+O P_{n} \geqslant R_{6}+R_{7}+\ldots+R_{n}.\] (A disk is assumed to contain its boundary.)
Let $ABCD$ be a square. Point $P, Q, R, S$ are chosen on the sides $AB$, $BC$, $CD$, $DA$, respectively, such that $AP + CR \ge AB \ge BQ + DS$. Prove that $$area \,\, (PQRS) \le \frac12 \,\, area \,\, (ABCD)$$ and determine all cases when equality holds.
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 median $ BM $ of a triangle $ ABC $ intersects the circumscribed circle at point $ K $. The circumcircle of the triangle $ KMC $ intersects the segment $ BC $ at point $ P $, and the circumcircle of $ AMK $ intersects the extension of $ BA $ at $ Q $. Prove that $ PQ> AC $.
On each side of a regular tetrahedron with edges of length $1$ one constructs exactly such a tetrahedron. This creates a dodecahedron with $8$ vertices and $18$ edges. We imagine that the dodecahedron is hollow. Calculate the length of the largest line segment that fits entirely within this dodecahedron.
In a right-angled triangle let $r$ be the inradius and $s_a,s_b$ be the lengths of the medians of the legs $a,b$. Prove the inequality \[ \frac{r^2}{s_a^2+s_b^2} \leq \frac{3-2 \sqrt2}{5}. \]
Two turtles, the leader and the slave, are crawling along the plane from point $A$ to point $B$. They crawl in turn: first the leader crawls some distance, then the slave crawls some distance in a straight line towards the leading one. Then the leader crawls somewhere again, after which the slave crawls towards the leader, etc. Finally, they both crawl to $B$. Prove that the slave turtle crawled no more than the leading one.