Is it possible that two cross-sections of a tetrahedron by two different cutting planes are two squares, one with a side of length no greater than $1$ and another with a side of length at least $100$? Mikhail Evdokimov

In a convex hexagon $ABCDEF$ equalities $\angle ABC= \angle CDE= \angle EFA$ hold, and the angle bisectors of angles $ABC$, $CDE$ and $EFA$ intersect in one point. Rays $AB$ and $DC$ intersect at $P$, rays $BC$ and $ED$ - at $Q$, rays $CD$ and $FE$ - at $R$, rays $DE$ and $AF$ - at $S$. Prove that $PR=QS$ [i]M. Zorka[/i]

Let $ABC$ be a triangle with circumcircle $\omega$, incenter $I$, and $A$-excenter $I_A$. Let the incircle and the $A$-excircle hit $BC$ at $D$ and $E$, respectively, and let $M$ be the midpoint of arc $BC$ without $A$. Consider the circle tangent to $BC$ at $D$ and arc $BAC$ at $T$. If $TI$ intersects $\omega$ again at $S$, prove that $SI_A$ and $ME$ meet on $\omega$. [i]Amol Aggarwal.[/i]

Let $ABC$ be an acute triangle. The tangent at $A,B$ of the circumcircle of $ABC$ intersect at $T$. Line $CT$ meets side $AB$ at $D$. Denote by $\Gamma_1,\Gamma_2$ the circumcircle of triangle $CAD$, and the circumcircle of triangle $CBD$, respectively. Let line $TA$ meet $\Gamma_1$ again at $E$ and line $TB$ meet $\Gamma_2$ again at $F$. Line $EF$ intersects sides $AC,BC$ at $P,Q$, respectively. Prove that $EF=PQ+AB$.

Let $ ABCD$ be a convex quadrilateral. The diagonals $ AC$ and $ BD$ intersect at $ K$. Show that $ ABCD$ is cyclic if and only if $ AK \sin A \plus{} CK \sin C \equal{} BK \sin B \plus{} DK \sin D$.

The altitudes through the vertices $ A,B,C$ of an acute-angled triangle $ ABC$ meet the opposite sides at $ D,E, F,$ respectively. The line through $ D$ parallel to $ EF$ meets the lines $ AC$ and $ AB$ at $ Q$ and $ R,$ respectively. The line $ EF$ meets $ BC$ at $ P.$ Prove that the circumcircle of the triangle $ PQR$ passes through the midpoint of $ BC.$

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 $ABCD$ be a convex quadrilateral. Let $E,F,G,H$ be points on segments $AB$, $BC$, $CD$, $DA$, respectively, and let $P$ be the intersection of $EG$ and $FH$. Given that quadrilaterals $HAEP$, $EBFP$, $FCGP$, $GDHP$ all have inscribed circles, prove that $ABCD$ also has an inscribed circle. [i]Evan O'Dorney.[/i]

1.) In the plane let us consider a set $S$ consisting of $n \geq 3$ distinct points satisfying the following three conditions: [b]I.[/b] The distance between any two points $\in S$ is not greater than 1. [b]II.[/b] For every point $A \in S$, there are exactly two “neighbor” points, i.e. two points $X, Y \in S$ for which $AX = AY = 1$. [b]III. [/b] For arbitrary two points $A, B \in S$, let $A', A''$ be the two neighbors of $A, B', B''$ the two neighbors of $B$, then $A'AA'' = B'BB''$. Is there such a set $S$ if $n = 1991$? If $n = 2000$ ? Explain your answer.

Determine all integer numbers $n\ge 3$ such that the regular $n$-gon can be decomposed into isosceles triangles by non-intersecting diagonals.

Six segments are such that any three can form a triangle. Is it true that these segments can be used to form a tetrahedron? (S. Markelov)

Given triangle ABC with incenter I. Let P,Q be point on circumcircle such that $\angle API=\angle CPI$ and $\angle BQI=\angle CQI$.Prove that $BP,AQ$ and $OI$ are concurrent.

The pentagon $ABCDE$ which is inscribed in a circle with $AB < DE$ is the base of a pyramid with apex $S$. If the longest side from $S$ is $SA$, prove that $BS > CS$.

[b]8.1 [/b] Construct a quadrilateral using side lengths and distances between the midpoints of the diagonals. [b]8.2[/b] It is known that $a,b$ and $\sqrt{a}+\sqrt{b} $ are rational numbers. Prove that then $\sqrt{a}$, $\sqrt{b} $ are rational. [b]8.3 / 9.2[/b] Solve equation $x^3 - [x]=3$ [b]8.4[/b] Prove that if in a triangle the angle bisector of the vertex, bisects the angle between the median and the altitude, then the triangle either isosceles or right. . [b]8.5[/b] Given $n$ numbers $x_1, x_2, . . . , x_n$, each of which is equal to $+1$ or $-1$. At the same time $$x_1x_2 + x_2x_3 + . . . + x_{n-1}x_n + x_nx_1 = 0 .$$ Prove that $n$ is divisible by $4$. [b]8.6[/b] There are $n$ points marked on the circle, and it is known that for of any two, one of the arcs connecting them has a measure less than $120^0$.Prove that all points lie on an arc of size $120^0$. PS. You should use hide for answers.Collected [url=https://artofproblemsolving.com/community/c3983442_1961_leningrad_math_olympiad]here[/url].

Let $ABCS$ be an isosceles trapezoid. Denote $A',B',C',D'$ the incenters of triangles $BCD,CDA,$ $DAB,ABC,$ respectively. Show that $A',B',C',D'$ are vertices of a rectangle.

Find the equation of the smallest sphere which is tangent to both of the lines $$\begin{pmatrix} x\\y\\z \end{pmatrix} =\begin{pmatrix} t+1\\ 2t+4\\ -3t +5 \end{pmatrix},\;\;\;\begin{pmatrix} x\\y\\z \end{pmatrix} =\begin{pmatrix} 4t-12\\ -t+8\\ t+17 \end{pmatrix}.$$

A semicircle of radius $5$ and a quarter of a circle of radius $8$ touch each other and are located inside the square as shown in the figure. Find the length of the part of the common tangent, enclosed in the same square. [img]https://cdn.artofproblemsolving.com/attachments/f/2/010f501a7bc1d34561f2fe585773816f168e93.png[/img]

Consider an arbitrary equilateral triangle $KLM$, whose vertices $K, L$ and $M$ lie on the sides $AB, BC$ and $CD$, respectively, of a given square $ABCD$. Find the locus of the midpoints of the sides $KL$ of all such triangles $KLM$.

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 circle centered at a point $F$ and a parabola with focus $F$ have two common points. Prove that there exist four points $A, B, C, D$ on the circle such that the lines $AB, BC, CD$ and $DA$ touch the parabola.

The circle $ \omega $ inscribed in triangle $ABC$ touches its sides $AB, BC, CA$ at points $K, L, M$ respectively. In the arc $KL$ of the circle $ \omega $ that does not contain the point $M$, we select point $S$. Denote by $P, Q, R, T$ the intersection points of straight $AS$ and $KM, ML$ and $SC, LP$ and $KQ, AQ$ and $PC$ respectively. It turned out that the points $R, S$ and $M$ are collinear. Prove that the point $T$ also lies on the line $SM$.

Given an acute triangle $ABC$, let $P$ be an arbitrary point on segment $BC$. A line passing through $P$ and perpendicular to $AC$ intersects $AB$ at $P_b$. A line passing through $P$ and perpendicular to $AB$ intersects $AC$ at $P_c$. Prove that the circumcircle of triangle $AP_bP_c$ passes through a fixed point other than $A$ when $P$ varies on segment $BC$. [i]Proposed by ltf0501[/i]

Let us have a line $\ell$ in the space and a point $A$ not lying on $\ell.$ For an arbitrary line $\ell'$ passing through $A$, $XY$ ($Y$ is on $\ell'$) is a common perpendicular to the lines $\ell$ and $\ell'.$ Find the locus of points $Y.$

In the plane a point $O$ is and a sequence of points $P_1, P_2, P_3, \ldots$ are given. The distances $OP_1, OP_2, OP_3, \ldots$ are $r_1, r_2, r_3, \ldots$ Let $\alpha$ satisfies $0 < \alpha < 1.$ Suppose that for every $n$ the distance from the point $P_n$ to any other point of the sequence is $\geq r^{\alpha}_n.$ Determine the exponent $\beta$, as large as possible such that for some $C$ independent of $n$ \[r_n \geq Cn^{\beta}, n = 1,2, \ldots\]

Quadrilateral $ABCD$ has the property that $AD = BD = CD$ and $\angle ADB < \angle CDB$. Let the circumcircle of $ABD$ be $O$. $O$ intersects $BC$ at $E$ and $CD$ at $F$. Next, extend $AB$ and $CD$ to intersect at a point $G$. Suppose that $BE = 1$, $EF = 3$, and $F D = 4$. Compute the perimeter of $\vartriangle ADG$.