[asy] draw(unitsquare);draw((0,0)--(.4,1)^^(0,.6)--(1,.2)); label("D",(0,1),NW);label("E",(.4,1),N);label("C",(1,1),NE); label("P",(0,.6),W);label("M",(.25,.55),E);label("Q",(1,.2),E); label("A",(0,0),SW);label("B",(1,0),SE); //Credit to Zimbalono for the diagram[/asy] Inside square $ABCD$ (See figure) with sides of length $12$ inches, segment $AE$ is drawn where $E$ is the point on $DC$ which is $5$ inches from $D$. The perpendicular bisector of $AE$ is drawn and intersects $AE$, $AD$, and $BC$ at points $M$, $P$, and $Q$ respectively. The ratio of segment $PM$ to $MQ$ is $\textbf{(A) }5:12\qquad\textbf{(B) }5:13\qquad\textbf{(C) }5:19\qquad\textbf{(D) }1:4\qquad \textbf{(E) }5:21$

Does there exist a cube in space such that the perpendiculars dropped from its eight vertices to a given plane are of length $0, 1, 2, 3, 4, 5, 6$ and $7$? (V Proizvolov)

Let the Nagel point of triangle $ABC$ be $N$. We draw lines from $B$ and $C$ to $N$ so that these lines intersect sides $AC$ and $AB$ in $D$ and $E$ respectively. $M$ and $T$ are midpoints of segments $BE$ and $CD$ respectively. $P$ is the second intersection point of circumcircles of triangles $BEN$ and $CDN$. $l_1$ and $l_2$ are perpendicular lines to $PM$ and $PT$ in points $M$ and $T$ respectively. Prove that lines $l_1$ and $l_2$ intersect on the circumcircle of triangle $ABC$. [i]Proposed by Nima Hamidi[/i]

Find, with proof, the point $P$ in the interior of an acute-angled triangle $ABC$ for which $BL^2+CM^2+AN^2$ is a minimum, where $L,M,N$ are the feet of the perpendiculars from $P$ to $BC,CA,AB$ respectively. [i]Proposed by United Kingdom.[/i]

Let $P$ be an arbitrary interior point of the equilateral triangle $ABC$. From $P$ draw parallel to the sides: $A'_1A_1 \parallel AB$, $B' _1B_1 \parallel BC$ and $C'_1C_1 \parallel CA$. Prove that the sum of legths $| AC_1 | + | BA_1 | + | CB_1 |$ is independent of the choice of point $P$. [img]https://cdn.artofproblemsolving.com/attachments/5/a/15b06706c09e2458fb5938807b9f3833ffb62e.png[/img]

A triangle $ABC$ is given. Let $D$ be a point on the extension of the segment $AB$ beyond $A$ such that $AD=BC,$ and $E$ be a point on the extension of the segment $BC$ beyond $B$ such that $BE=AC.$ Prove that the circumcircle of the triangle $DEB$ passes through the incenter of the triangle $ABC.$

Prove that there exists a positive real number $N$ such that for arbitrary real numbers $a,b>N$ it is possible to cover the perimeter of a rectangle with side lengths $a$ and $b$ using non-overlapping unit disks (the unit disks can be tangent to each other). [i]Submitted by Benedek Váli, Budapest[/i]

The points $S(i, j)$ with integer Cartesian coordinates $0 < i \leq n, 0 < j \leq m, m \leq n$, form a lattice. Find the number of: [b](a)[/b] rectangles with vertices on the lattice and sides parallel to the coordinate axes; [b](b)[/b] squares with vertices on the lattice and sides parallel to the coordinate axes; [b](c)[/b] squares in total, with vertices on the lattice.

Given a non-degenerate triangle $ABC$, let $A_{1}, B_{1}, C_{1}$ be the point of tangency of the incircle with the sides $BC, AC, AB$. Let $Q$ and $L$ be the intersection of the segment $AA_{1}$ with the incircle and the segment $B_{1}C_{1}$ respectively. Let $M$ be the midpoint of $B_{1}C_{1}$. Let $T$ be the point of intersection of $BC$ and $B_{1}C_{1}$. Let $P$ be the foot of the perpendicular from the point $L$ on the line $AT$. Prove that the points $A_{1}, M, Q, P$ lie on a circle.

Let $M$ be the midpoint of $AC$ in an acute triangle $ABC$. Let $K$ be a point on the minor arc $AC$, such that $\angle AKM=90^{o}$. Let $BK \cap AM=X$ and the $A$-altitude meets $BM$ at $Y$. Show that $XY \parallel AB$.

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is $ \textbf{(A)}\ \frac{\sqrt{3}}{12} \qquad\textbf{(B)}\ \frac{\sqrt{6}}{16} \qquad\textbf{(C)}\ \frac{1}{6} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{8} \qquad\textbf{(E)}\ \frac{1}{4} $

In an right isosceles triangle $ABC$, there are two points on the hypotenuse $AB, K$ and $M$, respectively, such that $KCM$ angle is $45^o$ (point $K$ lies between $A$ and $M$). Prove that $AK^2 + MB^2 = KM^2$ [img]https://cdn.artofproblemsolving.com/attachments/2/c/e7c57e0651e5a4c492cc4ae4b115bf68a7a833.png[/img]

Let ­ $\Omega$ be a circle of radius $8$ centered at point $O$, and let $M$ be a point on ­$\Omega$. Let $S$ be the set of points $P$ such that $P$ is contained within $\Omega$ ­, or such that there exists some rectangle $ABCD$ containing $P$ whose center is on ­ $\Omega$ with$ AB = 4$, $BC = 5$, and $BC \parallel OM$. Find the area of $S$.

Let $ABC$ be a triangle and $D$ a point on its side $BC.$ Points $E, F$ lie on the lines $AB, AC$ beyond vertices $B, C,$ respectively, such that $BE = BD$ and $CF = CD.$ Let $P$ be a point such that $D$ is the incenter of triangle $P EF.$ Prove that $P$ lies inside the circumcircle $\Omega$ of triangle $ABC$ or on it.

In triangle $ABC$, the angle bisectors $AD$, $BE$ and $CF$ are drawn, intersecting at point $I$. The perpendicular bisector of the segment $AD$ intersects lines $BE$ and $CF$ at points $M$ and $N$, respectively. Prove that points $A$, $I$, $M$ and $ N$ lie on the same circle.

Let $ ABC$ be a non-obtuse triangle with $ CH$ and $ CM$ are the altitude and median, respectively. The angle bisector of $ \angle BAC$ intersects $ CH$ and $ CM$ at $ P$ and $ Q$, respectively. Assume that \[ \angle ABP\equal{}\angle PBQ\equal{}\angle QBC,\] (a) prove that $ ABC$ is a right-angled triangle, and (b) calculate $ \dfrac{BP}{CH}$. [i]Soewono, Bandung[/i]

Given an acute isosceles triangle $ABC, AK$ and $CN$ are its angle bisectors, $I$ is their intersection point . Let point $X$ be the other intersection point of the circles circumscribed around $\vartriangle ABC$ and $\vartriangle KBN$. Let $M$ be the midpoint of $AC$. Prove that the Euler line of $\vartriangle ABC$ is perpendicular to the line $BI$ if and only if the points $X, I$ and $M$ lie on the same line. (Kivva Bogdan)

We consider an equilateral triangle with its circumscribed circle, of center $O$, and radius $4$cm. We rotate the triangle $90º$ around $O$. Compute the common area that was covered by the previous position of the triangle and is also covered by the new one.

Suppose $\triangle ABC$ has $D$ as the midpoint of $BC$ and orthocenter $H$. Let $P$ be an arbitrary point on the nine point circle of $ABC$. The line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. The line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. If $M$ is the midpoint of $AQ$, show that $HX \perp DM$.

$ABC$ is an acute triangle with orthocenter $H$. Point $P$ is in triangle $BHC$ that $\angle HPC = 3 \angle HBC $ and $\angle HPB =3 \angle HCB $. Reflection of point $P$ through $BH,CH$ is $X,Y$. if $S$ is the center of circumcircle of $AXY$ , Prove that: $$\angle BAS = \angle CAP$$ [i]Proposed by Pouria Mahmoudkhan Shirazi [/i]

Let $ABC$ be an acute triangle with orthocenter $H$. Let $A_1$, $B_1$ and $C_1$ be the feet of the altitudes of triangle $ABC$ opposite to vertices $A$, $B$, and $C$ respectively. Let $B_2$ and $C_2$ be the midpoints of $BB_1$ and $CC_1$, respectively. Let $O$ be the intersection of lines $BC_2$ and $CB_2$. Prove that $O$ is the circumcenter of triangle $ABC$ if and only if $H$ is the midpoint of $AA_1$. [i]Proposed by Dorlir Ahmeti[/i]

Given two circles $\omega_1$ and $\omega_2$ where $\omega_2$ is inside $\omega_1$. Show that there exists a point $P$ such that for any line $\ell$ not passing through $P$, if $\ell$ intersects circle $\omega_1$ at $A,B$ and $\ell$ intersects circle $\omega_2$ at $C,D$, where $A,C,D,B$ lie on $\ell$ in this order, then $\angle APC=\angle BPD$.

Spot spotlight located at vertex $B$ of an equilateral triangle $ABC$, illuminates angle $\alpha$. Find all such values of $\alpha$, not exceeding $60^o$, which at any position of the spotlight, when the illuminated corner is entirely located inside the angle $ABC$, from the illuminated and two unlit segments of side $AC$ can be formed into a triangle.

Points $K$ and $L$ on the sides $AB$ and $BC$ of parallelogram $ABCD$ are such that $\angle AKD = \angle CLD$. Prove that the circumcenter of triangle $BKL$ is equidistant from $A$ and $C$. [i]Proposed by I.I.Bogdanov[/i]

Let $ABCDEF$ be a regular hexagon, in the sides $AB$, $CD$, $DE$ and $FA$ we choose four points $P,Q,R$ and $S$ respectively, such that $PQRS$ is a square. Prove that $PQ$ and $BC$ are parallel.