Three faces $X , Y, Z$ of a unit cube share a common vertex. Suppose the projections of $X , Y, Z$ onto a fixed plane $P$ have areas $x, y, z$, respectively. If $x : y : z = 6 : 10 : 15$, then $x + y + z$ can be written as $m/n$ , where $m, n$ are positive integers and $gcd(m, n) = 1$. Find $100m + n$.

Inside the segment $AC$, an arbitrary point $B$ is selected and circles with diameters $AB$ and $BC$ are constructed. Points $M$ and $L$ are chosen on the circles (in one half-plane with respect to $AC$), respectively, so that $\angle MBA = \angle LBC$. Points $K$ and $F$ are marked, respectively, on rays $BM$ and $BL$ so that $BK = BC$ and $BF = AB$. Prove that points $M, K, F$ and $L$ lie on the same circle.

Two circles $w_1$ and $w_2$ intersect at points $A$ and $B$. Tangents $\ell_1$ and $\ell_2$ respectively are drawn to them through point $A$. The perpendiculars dropped from point $B$ to $\ell_2$ and $\ell_1$ intersects the circles $w_1$ and $w_2$, respectively, at points $K$ and $N$. Prove that points $K, A$ and $N$ lie on one straight line.

The following is known about the quadrilateral $ABCD$: triangles $ABC$ and $CDA$ are equal in area, the area of triangle $BCD$ is $k$ times greater than the area of triangle $DAB$, the bisectors of angles $ABC$ and $CDA$ intersect on the diagonal $AC$, straight lines $AC$ and $BD$ are not perpendicular. Find the ratio $AC/BD$.

Let triangle ABC has semiperimeter $ p$. E,F are located on AB such that $ CE\equal{}CF\equal{}p$. Prove that the C-excircle of triangle ABC touches the circumcircle (EFC).

Let $ ABC$ be an acute-angled triangle. Let $ L$ be any line in the plane of the triangle $ ABC$. Denote by $ u$, $ v$, $ w$ the lengths of the perpendiculars to $ L$ from $ A$, $ B$, $ C$ respectively. Prove the inequality $ u^2\cdot\tan A \plus{} v^2\cdot\tan B \plus{} w^2\cdot\tan C\geq 2\cdot S$, where $ S$ is the area of the triangle $ ABC$. Determine the lines $ L$ for which equality holds.

Points $P,Q,R,S$ lie on a circle and $\angle PSR$ is right. $H,K$ are the projections of $Q$ on lines $PR,PS$. Prove that $HK$ bisects segment $ QS$.

We inscribed in triangle $ABC$ the rectangle $DEFG$ such that $D$ and $E$ fall on side $AB$, $F$ on side $BC$, and $G$ on side $AC$. We know that $AF$ bisects angle $\angle BAC$, and that $\frac{AD}{DE} = \frac12$. What is the measure of angle $\angle CAB$?

et $ABCDEF$ be a convex hexagon which has an inscribed circle and a circumcribed. Denote by $\omega_{A}, \omega_{B},\omega_{C},\omega_{D},\omega_{E}$ and $\omega_{F}$ the inscribed circles of the triangles $FAB, ABC, BCD, CDE, DEF$ and $EFA$, respecitively. Let $l_{AB}$, be the external of $\omega_{A}$ and $\omega_{B}$; lines $l_{BC}$, $l_{CD}$, $l_{DE}$, $l_{EF}$, $l_{FA}$ are analoguosly defined. Let $A_1$ be the intersection point of the lines $l_{FA}$ and $l_{AB}$, $B_1, C_1, D_1, E_1, F_1$ are analogously defined. Prove that $A_1D_1, B_1E_1, C_1F_1$ are concurrent.

Squares $ABCD$ and $AEFG$ each with side length $12$ overlap so that $\vartriangle AED$ is an equilateral triangle as shown. The area of the region that is in the interior of both squares which is shaded in the diagram is $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m + n$. [img]https://cdn.artofproblemsolving.com/attachments/c/2/a2f8d2a090a6342610c43b3fed8a87fa5d7f03.png[/img]

Let $O$ be the circumcenter of an acute-angled triangle $ABC$ with ${\angle B<\angle C}$. The line $AO$ meets the side $BC$ at $D$. The circumcenters of the triangles $ABD$ and $ACD$ are $E$ and $F$, respectively. Extend the sides $BA$ and $CA$ beyond $A$, and choose on the respective extensions points $G$ and $H$ such that ${AG=AC}$ and ${AH=AB}$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if ${\angle ACB-\angle ABC=60^{\circ }}$. [i]Proposed by Hojoo Lee, Korea[/i]

The minimal value of $ f(x) \equal{} \sqrt{a^2 \plus{} x^2} \plus{} \sqrt{(x\minus{}b)^2 \plus{} c^2}$ is A. $ a\plus{}b\plus{}c$ B. $ \sqrt{a^2 \plus{} (b \plus{} c)^2}$ C. $ \sqrt{b^2 \plus{} (a\plus{}c)^2}$ D. $ \sqrt{(a\plus{}b)^2 \plus{} c^2}$ E. None of these

A triangle of area $1$ is cut out of paper. Prove that it can be bent along a straight segment so that the area of the resulting figure is less than $s_0$, where $s_0=\frac{\sqrt5-1}{2}$. Note. The value $s_0$ specified in the condition can be reduced (the smallest value of$s_0$ is unknown to the authors of the problem). If you manage to do this (and justify it), write.

Point $E$ is on side $AB$ of square $ABCD$. If $EB$ has length one and $EC$ has length two, then the area of the square is $\text{(A)}\ \sqrt{3} \qquad \text{(B)}\ \sqrt{5} \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 2\sqrt{3} \qquad \text{(E)}\ 5$

A regular tetrahedron of volume $1$ is filled with water of total volume $\frac{7}{16}$. Is it possible that the center of the tetrahedron lies on the surface of the water? How about in a cube of volume $1$?

A circle with area $40$ is tangent to a circle with area $10$. Let R be the smallest rectangle containing both circles. The area of $R$ is $\frac{n}{\pi}$. Find $n$. [asy] defaultpen(linewidth(0.7)); size(120); real R = sqrt(40/pi), r = sqrt(10/pi); draw(circle((0,0), R)); draw(circle((R+r,0), r)); draw((-R,-R)--(-R,R)--(R+2*r,R)--(R+2*r,-R)--cycle);[/asy]

Let $\omega$ be a circle with center $O$, and let $l$ be a line not intersecting $\omega$. $E$ is a point on $l$ such that $OE$ is perpendicular with $l$. Let $M$ be an arbitrary point on $M$ different from $E$. Let $A$ and $B$ be distinct points on the circle $\omega$ such that $MA$ and $MB$ are tangents to $\omega$. Let $C$ and $D$ be the foot of perpendiculars from $E$ to $MA$ and $MB$ respectively. Let $F$ be the intersection of $CD$ and $OE$. As $M$ moves, determine the locus of $F$.

$\vartriangle ABC$ is an isosceles triangle with $AB = AC$. Let $M$ be the midpoint of $BC$ and $E$ be the point on AC such that $AE :CE = 5 : 3$. Let $X$ be the intersection of $BE$ and $AM$. Given that the area of $\vartriangle CM X$ is $15$, find the area of $\vartriangle ABC$.

Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$. Rays $AO$, $CO$ intersect sides $BC, BA$ in points $A_1, C_1$ respectively, $K$ is the projection of $O$ onto the segment $A_1C_1$, $M$ is the midpoint of $AC$. Prove that $\angle HMA = \angle BKC_1$. [i]Proposed by Anton Trygub[/i]

Let $ ABC$ be an equilateral triangle, and $ P,Q,R$ three points in its interior satisfying \[ \measuredangle PCA \equal{} \measuredangle CAR \equal{} 15^{\circ},\ \measuredangle RBC \equal{} \measuredangle BCQ \equal{} 20^{\circ},\ \measuredangle QAB \equal{} \measuredangle ABP \equal{} 25^{\circ}.\] Compute the angles of triangle $ PQR$.

Let $\triangle ABC$ be a triangle and points $P, Q, R$ on the sides $AB, BC,$ and $CA$ respectively, such that: \[ \frac{AP}{AB} = \frac{BQ}{BC} = \frac{CR}{CA} = \frac{1}{n} \] for $n \in \mathbb{N}$. The segments $AQ$ and $CP$ intersect at $D$, the segments $BR$ and $AQ$ intersect at $E$, and the segments $BR$ and $CP$ intersect at $F$. Compute the ratio: \[ \frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle DEF)}. \]

Through the point of intersection of the altitudes of an acute triangle $ABC$ three circles pass through, each of which touches one of the sides triangle at the foot of the altitude . Prove that the second intersection points of the circles are the vertices of a triangle similar to the original one.

A rectangular piece of paper whose length is $\sqrt3$ times the width has area $A$. The paper is divided into equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area $B$. What is the ratio $B:A$? [asy] import graph; size(6cm); real L = 0.05; pair A = (0,0); pair B = (sqrt(3),0); pair C = (sqrt(3),1); pair D = (0,1); pair X1 = (sqrt(3)/3,0); pair X2= (2*sqrt(3)/3,0); pair Y1 = (2*sqrt(3)/3,1); pair Y2 = (sqrt(3)/3,1); dot(X1); dot(Y1); draw(A--B--C--D--cycle, linewidth(2)); draw(X1--Y1,dashed); draw(X2--(2*sqrt(3)/3,L)); draw(Y2--(sqrt(3)/3,1-L)); [/asy] $ \textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 3:5\qquad\textbf{(C)}\ 2:3\qquad\textbf{(D)}\ 3:4\qquad\textbf{(E)}\ 4:5 $

On the side $BC$ of the triangle $ABC$ there are two points $D$ and $E$ such that $BD < BE$. Denote by $p_1$ and $p_2$ the perimeters of triangles $ABC$ and $ADE$ respectively. Prove that \[p_1 > p_2 + 2\cdot \min\{BD, EC\}.\]

Angle $C$ of an isosceles triangle $ABC$ equals $120^o$. Each of two rays emitting from vertex $C$ (inwards the triangle) meets $AB$ at some point ($P_i$) reflects according to the rule the angle of incidence equals the angle of reflection" and meets lateral side of triangle $ABC$ at point $Q_i$ ($i = 1,2$). Given that angle between the rays equals $60^o$, prove that area of triangle $P_1CP_2$ equals the sum of areas of triangles $AQ_1P_1$ and $BQ_2P_2$ ($AP_1 < AP_2$).