Found problems: 721
Let $A$ be a point outside the circle $\omega$ . The tangents from $A$ touch the circle at $B$ and $C$. Let $P$ be an arbitrary point on extension of $AC$ towards $C$, $Q$ the projection of $C$ onto $PB$ and $E$ the second intersection point of the circumcircle of $ABP$ with the circle $\omega$ . Prove that $\angle PEQ = 2\angle APB$
On the bisector $AL$ of triangle $ABC$ a point $K$ is chosen such that $\angle BKL=\angle KBL=30^{\circ}$. Lines $AB$ and $CK$ intersect at point $M$, lines $AC$ and $BK$ intersect at point $N$. FInd the measure of angle $\angle AMN$
[I]Proposed by D. Shiryaev, S. Berlov[/i]
You are given a convex quadrilateral $ABCD$. It is known that $\angle CAD = \angle DBA = 40^o$, $\angle CAB = 60^o$, $\angle CBD = 20^o$. Find the angle $\angle CDB $.
Let $ABC$ be a right triangle and isosceles, with $\angle C = 90^o$. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $AC$. Let $ P$ be such that $MNP$ is an equilateral triangle with $ P$ inside the quadrilateral $MBCN$. Calculate the measure of $\angle CAP$
On a plane, a line $\ell$ and two circles $c_1$ and $c_2$ of different radii are given such that $\ell$ touches both circles at point $P$. Point $M \ne P$ on $\ell$ is chosen so that the angle $Q_1MQ_2$ is as large as possible where $Q_1$ and $Q_2$ are the tangency points of the tangent lines drawn from $M$ to $c_i$ and $c_2$, respectively, differing from $\ell$ . Find $\angle PMQ_1 + \angle PMQ_2$·
Let $n \geq 5$ be a given integer. Determine the greatest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-selfintersecting boundary) having $k$ internal right angles.
[i]Proposed by Juozas Juvencijus Macys, Lithuania[/i]
Two segments $A_1B_1$ and $A_2B_2$ are given on the plane, with $\frac{A_2B_2}{A_1B_1} = k < 1$. On segment $A_1A_2$, point $A_3$ is taken, and on the extension of this segment beyond point $A_2$, point $A_4$ is taken, so $\frac{A_3A_2}{A_3A_1} =\frac{A_4A_2}{A_4A_1}= k$. Similarly, point $B_3$ is taken on segment $B_1B_2$ , and on the extension of this the segment beyond point $B_2$ is point $B_4$, so $\frac{B_3B_2}{B_3B_1} =\frac{B_4B_2}{B_4B_1}= k$. Find the angle between lines $A_3B_3$ and $A_4B_4$.
(Netherlands)
A convex quadrilateral $ABCD$ is given in which $\angle DAB = \angle ABC = 45^o$ and $DA = 3$, $AB = 7\sqrt2$, $BC = 4$. Calculate the length of side $CD$.
[img]https://cdn.artofproblemsolving.com/attachments/1/2/046e31a628b3df4d23d3162cb570e1b9cb71e2.png[/img]
$ABC$ is an isosceles triangle such that $AB = AC$ and $\angle BAC = 96^o$. $D$ is the point for which $\angle ACD = 48^o$, $AD = BC$ and triangle $DAC$ is obtuse-angled. Find $\angle DAC$.
The angles of an acute triangle $ABC$ at $\alpha , \beta, \gamma$. Let $AD$ be a height, $CF$ a median, and $BE$ the bisector of $\angle B$. Show that $AD,CF$ and $BE$ are concurrent if and only if $\cos \gamma \tan\beta = \sin \alpha$ .
Let $ABCD$ be a circumscribed quadrilateral and $M$ the centre of the incircle. There are points $P$ and $Q$ on the lines $MA$ and $MC$ such that $\angle CBA= 2\angle QBP.$ Prove that $\angle ADC = 2 \angle PDQ.$
Two equilateral triangles $ABC$ and $CDE$ have a common vertex (see fig). Find the angle between straight lines $AD$ and $BE$.
[img]https://1.bp.blogspot.com/-OWpqpAqR7Zw/Xzj_fyqhbFI/AAAAAAAAMao/5y8vCfC7PegQLIUl9PARquaWypr8_luAgCLcBGAsYHQ/s0/2010%2Boral%2Bmoscow%2Bgeometru%2B8.1.gif[/img]
Consider $D$ the midpoint of the base $[BC]$ of the isosceles triangle ABC in which $\angle BAC < 90^o$. On the perpendicular from $B$ on the line $BC$ consider the point $E$ such that $\angle EAB= \angle BAC$, and on the line passing though $C$ parallel to the line $AB$ we consider the point $F$ such that $F$ and $D$ are on different side of the line $AC$ and $\angle FAC = \angle CAD$. Prove that $AE = CF$ and $BF = EF$
Let $ABC$ be a triangle with $AB=AC$. Let $D$ be a point on $BC$, and $E$ a point on $AD$ such that $\angle BED=\angle BAC=2\angle CED$. Prove that $BD=2CD$.
Given a right triangle $ABC$, the point $M$ is the midpoint of the hypotenuse $AB$. A circle is circumscribed around the triangle $BCM$, which intersects the segment $AC$ at a point $Q$ other than $C$. It turned out that the segment $QA$ is twice as large as the side $BC$. Find the acute angles of triangle $ABC$.
(Mykola Moroz)
Let $a_n$ denote the angle opposite to the side of length $4n^2$ units in an integer right angled triangle with lengths of sides of the triangle being $4n^2, 4n^4+1$ and $4n^4-1$ where $n \in N$. Then find the value of $\lim_{p \to \infty} \sum_{n=1}^p a_n$
(A) $\pi/2$
(B) $\pi/4$
(C) $\pi $
(D) $\pi/3$
The isosceles right triangle $ABC$ has right angle at $C$ and area $12.5$. The rays trisecting $\angle{ACB}$ intersect $AB$ at $D$ and $E$. What is the area of $\triangle{CDE}$?
$\textbf{(A) }\frac{5\sqrt{2}}{3}\qquad\textbf{(B) }\frac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\frac{15\sqrt{3}}{8}\qquad\textbf{(D) }\frac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\frac{25}{6}$
Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. One of the common tangents to the circles touches $C_1$ at $P_1$ and $C_2$ at $P_2$, while the other touches $C_1$ at $Q_1$ and $C_2$ at $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2=\angle M_1AM_2$.
Let $n>1$ be an integer. Suppose we are given $2n$ points in the plane such that no three of them are collinear. The points are to be labelled $A_1, A_2, \dots , A_{2n}$ in some order. We then consider the $2n$ angles $\angle A_1A_2A_3, \angle A_2A_3A_4, \dots , \angle A_{2n-2}A_{2n-1}A_{2n}, \angle A_{2n-1}A_{2n}A_1, \angle A_{2n}A_1A_2$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\circ}$ and $180^{\circ}$). Prove that there exists an ordering of the given points such that the resulting $2n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.
Let $ABCD$ be a quadrilateral with $\angle A + \angle C = 60^o$. If $AB \cdot CD = BC \cdot AD$, prove that $AB \cdot CD = AC \cdot BD$.
Leonard Giugiuc
Three disks of diameter $d$ are touching a sphere in their centers. Besides, every disk touches the other two disks. How to choose the radius $R$ of the sphere in order that axis of the whole figure has an angle of $60^\circ$ with the line connecting the center of the sphere with the point of the disks which is at the largest distance from the axis ? (The axis of the figure is the line having the property that rotation of the figure of $120^\circ$ around that line brings the figure in the initial position. Disks are all on one side of the plane, passing through the center of the sphere and orthogonal to the axis).
Let $ABC$ be an equilateral triangle. On the plane $(ABC)$ rise the perpendiculars $AA'$ and $BB'$ on the same side of the plane, so that $AA' = AB$ and $BB' =\frac12 AB$. Determine the measure the angle between the planes $(ABC)$ and $(A'B'C')$.
In the triangle $ABC, \angle BAC$ is acute, the angle bisector of $\angle BAC$ meets $BC$ at $D, K$ is the foot of the perpendicular from $B$ to $AC$, and $\angle ADB = 45^o$. Point $P$ lies between $K$ and $C$ such that $\angle KDP = 30^o$. Point $Q$ lies on the ray $DP$ such that $DQ = DK$. The perpendicular at $P$ to $AC$ meets $KD$ at $L$. Prove that $PL^2 = DQ \cdot PQ$.
Two straight lines are drawn on a plane, intersecting at an angle of $40^o$. A circle with center at point $O$ touches these lines. Let's consider a line, different from the given ones, tangent to the same circle and intersecting the given lines at points $B$ and $C$. What can the angle $\angle BOC$ be equal to?
The incircle of a scalene triangle $ABC$ touches the sides $BC, CA$, and $AB$ at points $D, E$, and $F$, respectively. Triangles $APE$ and $AQF$ are constructed outside the triangle so that \[AP =PE, AQ=QF, \angle APE=\angle ACB,\text{ and }\angle AQF =\angle ABC.\]Let $M$ be the midpoint of $BC$. Find $\angle QMP$ in terms of the angles of the triangle $ABC$.
[i]Iran, Shayan Talaei[/i]