Prove that if for angles $A,B,C$ of a triangle holds $$\sin^2 A+\sin^2 B +\sin^2 C=2$$ iff the triangle $ABC$ is right.

Find the area of the figure bounded by two curves $y=x^4,\ y=x^2+2$.

Square pyramid $ABCDE$ has base $ABCD,$ which measures $3$ cm on a side, and altitude $\overline{AE}$ perpendicular to the base$,$ which measures $6$ cm. Point $P$ lies on $\overline{BE},$ one third of the way from $B$ to $E;$ point $Q$ lies on $\overline{DE},$ one third of the way from $D$ to $E;$ and point $R$ lies on $\overline{CE},$ two thirds of the way from $C$ to $E.$ What is the area, in square centimeters, of $\triangle PQR?$ $\textbf{(A) } \frac{3\sqrt2}{2} \qquad\textbf{(B) } \frac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2$

Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 3 by 4 by 5 units. Given that the volume of this set is $(m + n \pi)/p$, where $m$, $n$, and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p$.

Triangles $ ABC$ and $ A_1B_1C_1$ have the same area. Using compass and ruler, can we always construct triangle $ A_2B_2C_2$ equal to triangle $ A_1B_1C_1$ so that the lines $ AA_2$, $ BB_2$, and $ CC_2$ are parallel?

Let $A$ be a point outside the circle $\omega$. Two points $B, C$ lie on $\omega$ such that $AB, AC$ are tangent to $\omega$. Let $D$ be any point on $\omega$ ($D$ is neither $B$ nor $C$) and $M$ the foot of perpendicular from $B$ to $CD$. The line through $D$ and the midpoint of $BM$ meets $\omega$ again at $P$. Prove that $AP \perp CP$

The ratio of the length to the width of a rectangle is $4:3$. If the rectangle has diagonal of length $d$, then the area may be expressed as $kd^2$ for some constant $k$. What is $k$? $\textbf{(A) }\dfrac27\qquad\textbf{(B) }\dfrac37\qquad\textbf{(C) }\dfrac{12}{25}\qquad\textbf{(D) }\dfrac{16}{25}\qquad\textbf{(E) }\dfrac34$

Let $ABC$ be a triangle with $\angle C=90^\circ$. A line joining the midpoint of its altitude $CH$ and the vertex $A$ meets $CB$ at point $K$. Let $L$ be the midpoint of $BC$ ,and $T$ be a point of segment $AB$ such that $\angle ATK=\angle LTB$. It is known that $BC=1$. Find the perimeter of triangle $KTL$.

Consider the parabola consisting of the points $(x, y)$ in the real plane satisfying $$(y + x) = (y - x)^2 + 3(y - x) + 3.$$ Find the minimum possible value of $y$.

Let $M, N$ be the midpoints of sides $AB, AC$ respectively of a triangle $ABC$. The perpendicular bisector to the bisectrix $AL$ meets the bisectrixes of angles $B$ and $C$ at points $P$ and $Q$ respectively. Prove that the common point of lines $PM$ and $QN$ lies on the tangent to the circumcircle of $ABC$ at $A$.

Let $A_1, A_2, \ldots, A_n$ be an $n$ -sided regular polygon. If $\frac{1}{A_1 A_2} = \frac{1}{A_1 A_3} + \frac{1}{A_1A_4}$, find $n$.

Let $ABCD$ be a quadrilateral that has an incircle with centre $O$ and radius $r$. Let $P = AB \cap CD$, $Q = AD \cap BC$, $E = AC \cap BD$. Show that $OE \cdot d = r^2$, where $d$ is the distance of $O$ from $PQ$.

You have a $7\times 7$ board divided into $49$ boxes. Mateo places a coin in a box. a) Prove that Mateo can place the coin so that it is impossible for Emi to completely cover the $48$ remaining squares, without gaps or overlaps, using $15$ $3\times1$ rectangles and a cubit of three squares, like those in the figure. [img]https://cdn.artofproblemsolving.com/attachments/6/9/a467439094376cd95c6dfe3e2ad3e16fe9f124.png[/img] b) Prove that no matter which square Mateo places the coin in, Emi will always be able to cover the 48 remaining squares using $14$ $3\times1$ rectangles and two cubits of three squares.

In triangle $ ABC$, $ M$ is the midpoint of side $ BC$ and $ G$ is the centroid of triangle $ ABC$. A line $ l$ passes through $ G$, intersecting line $ AB$ at $ P$ and line $ AC$ at $ Q$, where $ P\ne B$ and $ Q\ne C$. If $ [XYZ]$ denotes the area of triangle $ XYZ$, show that $ \frac{[BGM]}{[PAG]}\plus{}\frac{[CMG]}{[QGA]}\equal{}\frac32$.

Suppose that a triangle $ABC$ with incenter $I$ satisfies $CA+AI=BC$. Find the ratio between the measures of the angles $\angle BAC$ and $\angle CBA$.

Let $ABCD$ be a cyclic quadrilateral such that $AC =56, BD = 65, BC>DA$ and $AB: BC =CD: DA$. Find the ratio of areas $S (ABC): S (ADC)$.

Let $L$ be an arbitrary point on the minor arc $CD$ of the circumcircle of square $ABCD$. Let $K,M,N$ be the intersection points of $AL,CD$; $CL,AD$; and $MK,BC$ respectively. Prove that $B,M,L,N$ are concyclic.

A line, which passes through the incentre $I$ of the triangle $ABC$, meets its sides $AB$ and $BC$ at the points $M$ and $N$ respectively. The triangle $BMN$ is acute. The points $K,L$ are chosen on the side $AC$ such that $\angle ILA=\angle IMB$ and $\angle KC=\angle INB$. Prove that $AM+KL+CN=AC$. [i]S. Berlov[/i]

An $ 8$-foot by $ 10$-foot floor is tiled with square tiles of size $ 1$ foot by $ 1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $ 1/2$ foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded? [asy]unitsize(2cm); defaultpen(linewidth(.8pt)); fill(unitsquare,gray); filldraw(Arc((0,0),.5,0,90)--(0,0)--cycle,white,black); filldraw(Arc((1,0),.5,90,180)--(1,0)--cycle,white,black); filldraw(Arc((1,1),.5,180,270)--(1,1)--cycle,white,black); filldraw(Arc((0,1),.5,270,360)--(0,1)--cycle,white,black);[/asy]$ \textbf{(A)}\ 80\minus{}20\pi \qquad \textbf{(B)}\ 60\minus{}10\pi \qquad \textbf{(C)}\ 80\minus{}10\pi \qquad \textbf{(D)}\ 60\plus{}10\pi \qquad \textbf{(E)}\ 80\plus{}10\pi$

On the sides of the right triangle, outside are constructed regular nonagons, which are constructed on one of the catheti and on the hypotenuse, with areas equal to $1602$ $cm^2$ and $2019$ $cm^2$, respectively. What is the area of the nonagon that is constructed on the other cathetus of this triangle? (Vladislav Kirilyuk)

Suppose that you are given the foot of the altitude from vertex $A$ of a scalene triangle $ABC$, the midpoint of the arc with endpoints $B$ and $C$, not containing $A$ of the circumscribed circle of $ABC$, and also a third point $P$. Construct the triangle from these three points if $P$ is the a) orthocenter b) centroid c) incenter of the triangle.

Given five segments. It is possible to construct a triangle of every subset of three of them. Prove that at least one of those triangles is acute-angled.

Let $ABC$ be a triangle, let $O$ be its circumcenter, and let $H$ be its orthocenter. Let $P$ be a point on the segment $OH$. Prove that $6r\leq PA+PB+PC\leq 3R$, where $r$ is the inradius and $R$ the circumradius of triangle $ABC$. [b]Moderator edit:[/b] This is true only if the point $P$ lies inside the triangle $ABC$. (Of course, this is always fulfilled if triangle $ABC$ is acute-angled, since in this case the segment $OH$ completely lies inside the triangle $ABC$; but if triangle $ABC$ is obtuse-angled, then the condition about $P$ lying inside the triangle $ABC$ is really necessary.)

Given a triangle $PQX$ in the plane, with $PQ = 3, PX = 2.6$ and $QX = 3.8$. Construct a right-angled triangle $ABC$ such that the incircle of $\vartriangle ABC$ touches $AB$ at $P$ and $BC$ at $Q$, and point $X$ lies on the line $AC$.

Circles $\Omega$ and $\omega$ are internally tangent at point $C$. Chord $AB$ of $\Omega$ is tangent to $\omega$ at $E$, where $E$ is the midpoint of $AB$. Another circle, $\omega_1$ is tangent to $\Omega, \omega,$ and $AB$ at $D,Z,$ and $F$ respectively. Rays $CD$ and $AB$ meet at $P$. If $M$ is the midpoint of major arc $AB$, show that $\tan \angle ZEP = \tfrac{PE}{CM}$. [i]Ray Li.[/i]