This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 131

Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude through $A$. On $AD$, there are distinct points $E$ and $F$ such that $|AE| = |BE|$ and $|AF| =|CF|$. A point$ T \ne D$ satis es $\angle BTE = \angle CTF = 90^o$. Show that $|TA|^2 =|TB| \cdot |TC|$.
In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.
An interior $P$ point to a square $ABCD$ is such that $PA = a, PB = b$ and $PC = b + c$, where the numbers $a, b$ and $c$ satisfy the relationship $a^2 = b^2 + c^2$. Prove that the angle $BPC$ is right.
The center of square $ABCD$ is $K$. The point $P$ is chosen such that $P \ne K$ and the angle $\angle APB$ is right . Prove that the line $PK$ bisects the angle between the lines $AP$ and $BP$.
Let $ABCD$ be a trapezoid with $\angle A = \angle B = 90^o$ and a point $E$ lies on the segment $CD$. Denote $(\omega)$ as incircle of triangle $ABE$ and it is tangent to $AB,AE,BE$ respectively at $P, F,K$. Suppose that $KF$ cuts $BC,AD$ at $M,N$ and $PM,PN$ cut $(\omega)$ at $H, T$. Prove that $PH = PT$.
A parallelogram $ABCD$ is given. On the diagonal BD, a point $P$ is selected such that $AP = BD$ is satisfied. Point $Q$ is the midpoint of segment $CP$. Prove that $\angle BQD = 90^o$. [img]https://cdn.artofproblemsolving.com/attachments/2/0/4bc69ec0330e2afa6b560c56da5dd783b16efb.png[/img] .
Let $ABC$ be a triangle with $\angle A<90^o, AB \ne AC$. Denote $H$ the orthocenter of triangle $ABC$, $N$ the midpoint of segment $[AH]$, $M$ the midpoint of segment $[BC]$ and $D$ the intersection point of the angle bisector of $\angle BAC$ with the segment $[MN]$. Prove that $<ADH=90^o$
Let $X$ be the intersection of the diagonals $AC$ and $BD$ of convex quadrilateral $ABCD$. Let $P$ be the intersection of lines $AB$ and $CD$, and let $Q$ be the intersection of lines $PX$ and $AD$. Suppose that $\angle ABX = \angle XCD = 90^o$. Prove that $QP$ is the angle bisector of $\angle BQC$.
Let $O, M, N$ be the circumcenter, the centroid and the Nagel point of a triangle. Prove that angle $MON$ is right if and only if one of the triangle’s angles is equal to $60^o$.
Circle $C$ center $O$ touches externally circle $C'$ center $O'$. A line touches $C$ at $A$ and $C'$ at $B$. $P$ is the midpoint of $AB$. Show that $\angle OPO' = 90^o$.
Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude through $A$. On $AD$, there are distinct points $E$ and $F$ such that $|AE| = |BE|$ and $|AF| =|CF|$. A point$ T \ne D$ satis es $\angle BTE = \angle CTF = 90^o$. Show that $|TA|^2 =|TB| \cdot |TC|$.
In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.
Let $O$ be the point of intersection of the diagonals of the trapezoid $ABCD$ with the bases $AB$ and $CD$. It is known that $\angle AOB = \angle DAB = 90^o$. On the sides $AD$ and $BC$ take the points $E$ and $F$ so that $EF\parallel AB$ and $EF = AD$. Find the angle $\angle AOE$.
On the sides $AD , BC$ of the square $ABCD$ the points $M, N$ are selected $N$, respectively, such that $AM = BN$. Point $X$ is the foot of the perpendicular from point $D$ on the line $AN$. Prove that the angle $MXC$ is right. (Mirchev Borislav)
Let $ABC$ be a triangle inscribed in circle $(O),$ with its altitudes $BE, CF$ intersect at orthocenter $H$ ($E \in AC, F \in AB$). Let $M$ be the midpoint of $BC, K$ be the orthogonal projection of $H$ on $AM$. $EF$ intersects $BC$ at $P$. Let $Q$ be the intersection of tangent of $(O)$ which passes through $A$ with $BC, T$ be the reflection of $Q$ through $P$. Prove that $\angle OKT = 90^o$.
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]
In an obtuse triangle $ABC$, for which $AC < AB$, the radius of the inscribed circle is $R$, the midpoint of its arc $BC$ (which does not contain $A$) is $S$. A point $T$ is placed on the extension of altitude $AD$ such that $D$ is between $ A$ and $T$ and $AT = 2R$. Prove that $\angle AST = 90^o$.
Let $ABCD$ be a convex quadrilateral such that $AB + BC = 2021$ and $AD = CD$. We are also given that $\angle ABC = \angle CDA = 90^o$. Determine the length of the diagonal $BD$.
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.
Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BH_b, CH_c$ intersect at orthocenter $H$ ($H_b \in AC$, $H_c \in AB$). $H_bH_c$ meets $BC$ at $P$. Let $N$ be the midpoint of $AH, L$ be the orthogonal projection of $O$ on the symmedian with respect to angle $A$ of triangle $ABC$. Prove that $\angle NLP = 90^o$.
You can inscribe a circle in the pentagon $ABCDE$. It is also known that $\angle ABC = \angle BAE = \angle CDE = 90^o$. Find the measure of the angle $ADB$.
Inside the triangle $ABC $, the point $P $ is selected so that $BC = AP $ and $\angle APC = 180 {} ^ \circ - \angle ABC $. On the side $AB $ there is a point $K $, for which $AK = KB + PC $. Prove that $\angle AKC = 90 {} ^ \circ $. (Danilo Hilko)
In a quadrilateral $ABCD$, angle $B$ is equal to $150^o$, angle $C$ is right, and sides $AB$ and $CD$ are equal. Determine the angle between $BC$ and the line connecting the midpoints of sides $BC$ and $AD$.
Let $ABC $be a right triangle with $\angle C = 90^o$ and let $D$ be the foot of the altitude from $C$. Let $E$ be the centroid of triangle $ACD$ and let $F$ be the centroid of triangle $BCD$. The point $P$ satisfies $\angle CEP = 90^o$ and $|CP| = |AP|$, while point $Q$ satisfies $\angle CFQ = 90^o$ and $|CQ| = |BQ|$. Prove that $PQ$ passes through the centroid of triangle $ABC$.
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]