Let $ABC$ be an isosceles triangle ($AB = AC$), $D$ be the midpoint of $BC$, and $M$ be the midpoint of $AD$. On the segment $BM$ take a point $N$ such that $\angle BND = 90^o$. Find the angle $ANC$.

The height drawn on the hypotenuse of a right triangle divides the hypotenuse into two sections with a length ratio of $9: 1$ and two triangles of the starting triangle with a difference of areas of $48$ cm$^2$. Find the original triangle sidelengths.

he center of the circle passing through the midpoints of all sides of triangle $ABC$ lies on the bisector of its angle $C$. Find the side $AB$ if $BC = a$, $AC = b$ ($a$ is not equal to $b$).

For an isosceles triangle $ABC$ where $AB=AC$, it is possible to construct, using only compass and straightedge, an isosceles triangle $PQR$ where $PQ=PR$ such that triangle $PQR$ is similar to triangle $ABC$, point $P$ is in the interior of line segment $AC$, point $Q$ is in the interior of line segment $AB$, and point $R$ is in the interior of line segment $BC$. Describe one method of performing such a construction. Your method should work on every isosceles triangle $ABC$, except that you may choose an upper limit or lower limit on the size of angle $BAC$. [asy] defaultpen(linewidth(0.7)); pair a= (79,164),b=(19,22),c=(138,22),p=(109,91),q=(38,67),r=(78,22); pair point = ((p.x+q.x+r.x)/3,(p.y+q.y+r.y)/3); draw(a--b--c--cycle); draw(p--q--r--cycle); label("$A$",a,dir(point--a)); label("$B$",b,dir(point--b)); label("$C$",c,dir(point--c)); label("$P$",p,dir(point--p)); label("$Q$",q,dir(point--q)); label("$R$",r,dir(point--r));[/asy]

[b]p1.[/b] Find any integer solution of the puzzle: $WE+ST+RO+NG=128$ (different letters mean different digits between $1$ and $9$). [b]p2.[/b] A $5\times 6$ rectangle is drawn on the piece of graph paper (see the figure below). The side of each square on the graph paper is $1$ cm long. Cut the rectangle along the sides of the graph squares in two parts whose areas are equal but perimeters are different by $2$ cm. $\begin{tabular}{|l|l|l|l|l|l|} \hline & & & & & \\ \hline & & & & & \\ \hline & & & & & \\ \hline & & & & & \\ \hline \end{tabular}$ [b]p3.[/b] Three runners started simultaneously on a $1$ km long track. Each of them runs the whole distance at a constant speed. Runner $A$ is the fastest. When he runs $400$ meters then the total distance run by runners $B$ and $C$ together is $680$ meters. What is the total combined distance remaining for runners $B$ and $C$ when runner $A$ has $100$ meters left? [b]p4.[/b] There are three people in a room. Each person is either a knight who always tells the truth or a liar who always tells lies. The first person said «We are all liars». The second replied «Only you are a liar». Is the third person a liar or a knight? [b]p5.[/b] A $5\times 8$ rectangle is divided into forty $1\times 1$ square boxes (see the figure below). Choose 24 such boxes and one diagonal in each chosen box so that these diagonals don't have common points. $\begin{tabular}{|l|l|l|l|l|l|l|l|} \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline & & & & & & & \\ \hline \end{tabular}$ PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

Let $r_1, \ldots , r_n$ be the radii of $n$ spheres. Call $S_1, S_2, \ldots , S_n$ the areas of the set of points of each sphere from which one cannot see any point of any other sphere. Prove that \[\frac{S_1}{r_1^2} + \frac{S_2}{r_2^2}+\cdots+\frac{S_n}{r_n^2} = 4 \pi.\]

In a trapezoid $ABCD$ such that the internal angles are not equal to $90^{\circ}$, the diagonals $AC$ and $BD$ intersect at the point $E$. Let $P$ and $Q$ be the feet of the altitudes from $A$ and $B$ to the sides $BC$ and $AD$ respectively. Circumscribed circles of the triangles $CEQ$ and $DEP$ intersect at the point $F\neq E$. Prove that the lines $AP$, $BQ$ and $EF$ are either parallel to each other, or they meet at exactly one point.

Let $O$ be the circumcenter of $\triangle ABC.$ The circle $k$ passing through $A$ and $B$ cuts $AC$ and $BC$ at $P$ and $Q,$ respectively. Prove that $PQ$ and $OC$ are perpendicular.

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP \perp BC$. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ \perp HO$. [asy] import olympiad; unitsize(30); pair A,B,C,H,O,P,Q,R,Y,Z,Q2,R2,P2; A = (-14.8, -6.6); B = (-10.9, 0.3); C = (-3.1, -7.1); O = circumcenter(A,B,C); H = orthocenter(A,B,C); P = 1.2 * H - 0.2 * A; Q = reflect(A, C) * P; R = reflect(A, B) * P; Y = foot(R, C, A); Z = foot(Q, A, B); P2 = foot(A, B, C); Q2 = foot(P, C, A); R2 = foot(P, A, B); draw(B--(1.6*A-0.6*B)); draw(B--C--A); draw(P--R, blue); draw(R--Y, red); draw(P--Q, blue); draw(Q--Z, red); draw(A--P2, blue); draw(O--H, darkgreen+linewidth(1.2)); draw((1.4*Z-0.4*Y)--(4.6*Y-3.6*Z), red+linewidth(1.2)); draw(rightanglemark(R,Y,A,10), red); draw(rightanglemark(Q,Z,B,10), red); draw(rightanglemark(C,Q2,P,10), blue); draw(rightanglemark(A,R2,P,10), blue); draw(rightanglemark(B,P2,H,10), blue); label("$\textcolor{blue}{H}$",H,NW); label("$\textcolor{blue}{P}$",P,N); label("$A$",A,W); label("$B$",B,N); label("$C$",C,S); label("$O$",O,S); label("$\textcolor{blue}{Q}$",Q,E); label("$\textcolor{blue}{R}$",R,W); label("$\textcolor{red}{Y}$",Y,S); label("$\textcolor{red}{Z}$",Z,NW); dot(A, filltype=FillDraw(black)); dot(B, filltype=FillDraw(black)); dot(C, filltype=FillDraw(black)); dot(H, filltype=FillDraw(blue)); dot(P, filltype=FillDraw(blue)); dot(Q, filltype=FillDraw(blue)); dot(R, filltype=FillDraw(blue)); dot(Y, filltype=FillDraw(red)); dot(Z, filltype=FillDraw(red)); dot(O, filltype=FillDraw(black)); [/asy]

The sides of a convex $200$-gon $A_1 A_2 \dots A_{200}$ are colored red and blue in an alternating fashion. Suppose the extensions of the red sides determine a regular $100$-gon, as do the extensions of the blue sides. Prove that the $50$ diagonals $\overline{A_1A_{101}},\ \overline{A_3A_{103}},\ \dots, \ \overline{A_{99}A_{199}}$ are concurrent. [i]Proposed by: [b]Ankan Bhattacharya[/b][/i]

Let $H$ be the orthocenter of an acute-angled triangle $ABC$ and $P$ a point on the circumcenter of triangle $ABC$. Prove that the Simson line of $P$ bisects the segment $[P H]$.

The bisectors of the angles $B,C$ of a triangle $ABC$ intersect the opposite sides in $B', C'$ respectively. Prove that the straight line $B'C'$ intersects the inscribed circle in two different points.

Rhombus $ABCD$ has side length $2$ and $\angle B = 120 ^\circ$. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$? $ \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad \textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad \textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad \textbf{(D)}\ 1+\frac{\sqrt{3}}{3} \qquad \textbf{(E)}\ 2 $

Let $AA_1,BB_1,CC_1$ be the altitudes of a triangle $ABC$. If $\overrightarrow{AA_1}+\overrightarrow{BB_1}+\overrightarrow{CC_1}=0$ prove that the triangle $ABC$ is equilateral.

$\triangle ABC$, $\angle A=23^{\circ},\angle B=46^{\circ}$. Let $\Gamma$ be a circle with center $C$, radius $AC$. Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$. Find $\angle MAN$.

In trapezoid $ABCD$, $AD$ is parallel to $BC$. $\angle A = \angle D = 45^o$, while $\angle B = \angle C = 135^o$. If $AB = 6$ and the area of $ABCD$ is $30$, find $BC$. [img]https://cdn.artofproblemsolving.com/attachments/0/8/d667522259c773435bc53f5988831aceaef7b7.png[/img]

Moe uses a mower to cut his rectangular $ 90$-foot by $ 150$-foot lawn. The swath he cuts is $ 28$ inches wide, but he overlaps each cut by $ 4$ inches to make sure that no grass is missed. He walks at the rate of $ 5000$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn? $ \textbf{(A)}\ 0.75 \qquad \textbf{(B)}\ 0.8 \qquad \textbf{(C)}\ 1.35 \qquad \textbf{(D)}\ 1.5 \qquad \textbf{(E)}\ 3$

Baron Munchausen plays billiards on a table with the shape of an equilateral triangle. He claims to have shot a ball from one of the sides of this table so that it passed through a certain point three times in three different directions and then returned to the original point on the side. Can that be true, assuming that the usual law of reflection holds? (Μ Evdokimov)

The orthogonal projections of the vertices $A, B, C$ of the tetrahedron $ABCD$ on the opposite faces are denoted by $A', B', C'$ respectively. Suppose that point $A'$ is the circumcenter of the triangle $BCD$, point $B'$ is the incenter of the triangle $ACD$ and $C'$ is the centroid of the triangle $ABD$. Prove that tetrahedron $ABCD$ is regular.

The points $E$ and $F$ divide the diagonal $BD$ of the convex quadrilateral $ABCD$ into three equal parts, i.e. $| BE | = | EF | = | F D |$. Line $AE$ interects side $BC$ at $X$ and line $AF$ intersects $DC$ at $Y$. Prove that: a) if $ABCD$ is parallelogram then $X ,Y$ are the midpoints of $BC, DC$, respectively, b) if the points $X , Y$ are the midpoints of $BC, DC$, respectively , then $ABCD$ is parallelogram

Let the medians of the triangle $ABC$ intersect at point $M$. A line $t$ through $M$ intersects the circumcircle of $ABC$ at $X$ and $Y$ so that $A$ and $C$ lie on the same side of $t$. Prove that $BX\cdot BY=AX\cdot AY+CX\cdot CY$.

Prove that any triangle can be divided into $22$ triangles, each of which has an angle of $22^o$, and another $23$ triangles, each of which has an angle of $23^o$.

A circle which is tangent to the sides $ [AB]$ and $ [BC]$ of $ \triangle ABC$ is also tangent to its circumcircle at the point $ T$. If $ I$ is the incenter of $ \triangle ABC$ , show that $ \widehat{ATI}\equal{}\widehat{CTI}$

In the convex quadrilateral $ABCD$, the bisectors of angles $A$ and $C$ intersect in $I$. Prove that $ABCD$ is circumscriptible if and only if $$S[AIB] + S[CID] =S[AID]+S[BIC]$$ ( $S[XYZ]$ denotes the area of the triangle $XYZ$)

A polygon with sides running along the sides of the squares was cut out of an endless chessboard. A segment of the perimeter of a polygon is called black if the polygon adjacent to it from the inside is which cell is black, respectively white if the cell is white. Let $A$ be the number of black segments on the perimeter, and $B$ be the number of white ones, Let the polygon consist of $a$ black and $b$ white cells. Prove that $A-B = 4(a -b)$.