Found problems: 333
Let $ABCD$ be a convex quadrilateral satisfying $\angle ADB + \angle ACB = \angle CAB + \angle DBA = 30^o$, $AD = BC$. Prove that there exists a right-angled triangle with side lengths $AC$, $BD$, $CD$.
Let $ABC$ be a triangle with right $\angle A$. Consider points $D \in (AC)$ and $E \in (BD)$ such that $\angle ABC = \angle ECD = \angle CED$. Prove that $BE = 2 \cdot AD$
Let us denote successively $r$ and $r_a$ the radii of the inscribed circle and the exscribed circle wrt to side BC of triangle $ABC$. Prove that if it is true that $r+r_a=|BC|$ , then the triangle $ABC$ is a right one
Let $ABC$ be a right angled triangle with $\angle A = 90^o$and $BC = a$, $AC = b$, $AB = c$. Let $d$ be a line passing trough the incenter of triangle and intersecting the sides $AB$ and $AC$ in $P$ and $Q$, respectively.
(a) Prove that $$b \cdot \left( \frac{PB}{PA}\right)+ c \cdot \left( \frac{QC}{QA}\right) =a$$
(b) Find the minimum of $$\left( \frac{PB}{PA}\right)^ 2+\left( \frac{QC}{QA}\right)^ 2$$
Given a right triangle $ABC$ ($\angle A <45^o$,$ \angle C = 90^o$), on the sides $AC$ and $AB$ which are selected points $D,E$ respectively, such that $BD = AD$ and $CB = CE$. Let the segments $BD$ and $CE$ intersect at the point $O$. Prove that $\angle DOE = 90^o$.
Let $r$ be the radius of the inscribed circle of a right triangle $ABC$. Show that $r$ is less than half of either leg and less than one fourth of the hypotenuse.
We are given a right-angled triangle $MNP$ with right angle in $P$. Let $k_M$ be the circle with center $M$ and radius $MP$, and let $k_N$ be the circle with center $N$ and radius $NP$. Let $A$ and $B$ be the common points of $k_M$ and the line $MN$, and let $C$ and $D$ be the common points of $k_N$ and the line $MN$ with with $C$ between $A$ and $B$. Prove that the line $PC$ bisects the angle $\angle APB$.
As shown in figure, given right $\vartriangle ABC$ with $\angle C=90^o$. $I$ is the incenter. The line $BI$ intersects segment $AC$ at the point $D$ . The line passing through $D$ parallel to $AI$ intersects $BC$ at point $E$. The line $EI$ intersects segment $AB$ at point $F$. Prove that $DF \perp AI$.
[img]https://cdn.artofproblemsolving.com/attachments/2/4/6fc94adb4ce12c3bf07948b8c57170ca01b256.png[/img]
Find a right triangle that can be cut into $365$ equal triangles.
Consider the isosceles right triangle $ABC$, with $\angle A = 90^o$ and the point $M \in (BC)$ such that $\angle AMB = 75^o$. On the inner bisector of the angle $MAC$ take a point $F$ such that $BF = AB$. Prove that:
a) the lines $AM$ and $BF$ are perpendicular;
b) the triangle $CFM$ is isosceles.
A right triangle with perpendicular sides $a$ and $b$ and hypotenuse $c$ has the following properties:
$a = p^m$ and $b = q^n$ with $p$ and $q$ prime numbers and $m$ and $n$ positive integers, $c = 2k +1$ with $k$ a positive integer.
Determine all possible values of $c$ and the associated values of $a$ and $b$.
Given is a right triangle $ABC$ with perimeter $2$, with $\angle B=90^o$ . Point $S$ is the center of the excircle to the side $AB$ of the triangle and $H$ is the intersection of the heights of the triangle $ABS$ . Determine the smallest possible length of the segment $HS $.
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 $ABC$.
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 $ABC$ be a triangle and $M$ be the midpoint of $AB$. If it is known that $\angle CAM + \angle MCB = 90^o$, show that triangle $ABC$ is isosceles or right.
Anton has an isosceles right triangle, which he wants to cut into $9$ triangular parts in the way shown in the picture. What is the largest number of the resulting $9$ parts that can be equilateral triangles?
A more formal description of partitioning. Let triangle $ABC$ be given. We choose two points on its sides so that they go in the order $AC_1C_2BA_1A_2CB_1B_2$, and no two coincide. In addition, the segments $C_1A_2$, $A_1B_2$ and $B_1C_2$ must intersect at one point. Then the partition is given by segments $C_1A_2$, $A_1B_2$, $B_1C_2$, $A_1C_2$, $B_1A_2$ and $C_1B_2$.
[img]https://cdn.artofproblemsolving.com/attachments/0/5/5dd914b987983216342e23460954d46755d351.png[/img]
In a right triangle rectangle $ABC$ such that $AB = AC$, $M$ is the midpoint of $BC$. Let $P$ be a point on the perpendicular bisector of $AC$, lying in the semi-plane determined by $BC$ that does not contain $A$. Lines $CP$ and $AM$ intersect at $Q$. Calculate the angles that form the lines $AP$ and $BQ$.
Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$.
Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.
The triangle $ABC$ is right-angled at $A$. Its incentre is $I$, and $H$ is the foot of the perpendicular from $I$ on $AB$. The perpendicular from $H$ on $BC$ meets $BC$ at $E$, and it meets the bisector of $\angle ABC$ at $D$. The perpendicular from $A$ on $BC$ meets $BC$ at $F$. Prove that $\angle EFD = 45^o$
A right triangle is inscribed in parabola $y=x^2$. Prove that it's hypotenuse is not less than $2$.
Let $ABC$ be an equilateral triangle and $\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$, and $CA$ (including $A$, $B$, and $C$). Determine whether, for every partition of $\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle.
Let $ABC$ be a right triangle at $B$. Let $D$ be a point such that $BD\perp AC$ and $DC = AC$. Find the ratio $\frac{AD}{AB}$.
Let $ABCDE$ be a convex pentagon with $\angle AEB=\angle BDC=90^o$ and line $AC$ bisects $\angle BAE$ and $\angle DCB$ internally. The circumcircle of $ABE$ intersects line $AC$ again at $P$.
(a) Show that $P$ is the circumcenter of $BDE$.
(b) Show that $A, C, D, E$ are concyclic.
Consider a parallelogram $ABCD$ such that the midpoint $M$ of the side $CD$ lies on the angle bisector of $\angle BAD$. Show that $\angle AMB$ is a right angle.
In a right triangle $ABC$, M is a point on the hypotenuse $BC$ and $P$ and $Q$ the projections of $M$ on $AB$ and $AC$ respectively. Prove that for no such point $M$ do the triangles $BPM, MQC$ and the rectangle $AQMP$ have the same area.