Found problems: 333
Prove that if a triangle has integral side lengths and its circumradius is a prime number then the triangle is right-angled.
We say that a triangle $ABC$ is great if the following holds: for any point $D$ on the side $BC$, if $P$ and $Q$ are the feet of the perpendiculars from $D$ to the lines $AB$ and $AC$, respectively, then the reflection of $D$ in the line $PQ$ lies on the circumcircle of the triangle $ABC$. Prove that triangle $ABC$ is great if and only if $\angle A = 90^{\circ}$ and $AB = AC$.
[i]Senior Problems Committee of the Australian Mathematical Olympiad Committee[/i]
In the right-angled triangle $ABC$ with a right angle at $C$, the side $BC$ is longer than the side $AC$. The perpendicular bisector of $AB$ intersects the line $BC$ at point $D$ and the line $AC$ at point $E$. The segments $DE$ has the same length as the side $AB$. Find the measures of the angles of the triangle $ABC$.
(R. Henner, Vienna)
Let $AB = 10$ be a diameter of circle $P$. Pick point $C$ on the circle such that $AC = 8$. Let the circle with center $O$ be the incircle of $\vartriangle ABC$. Extend line $AO$ to intersect circle $P$ again at $D$. Find the length of $BD$.
Let $XYZ$ be a right triangle of area $1$ m$^2$ . Consider the triangle $X'Y'Z'$ such that $X'$ is the symmetric of X wrt side $YZ$, $Y'$ is the symmetric of $Y$ wrt side $XZ$ and $Z' $ is the symmetric of $Z$ wrt side $XY$. Calculate the area of the triangle $X'Y'Z'$.
In triangle $ABC$ we have $\angle ACB = 90^o$. The point $M$ is the midpoint of $AB$. The line through $M$ parallel to $BC$ intersects $AC$ in $D$. The midpoint of line segment $CD$ is $E$. The lines $BD$ and $CM$ are perpendicular.
(a) Prove that triangles $CME$ and $ABD$ are similar.
(b) Prove that $EM$ and $AB$ are perpendicular.
[asy]
unitsize(1 cm);
pair A, B, C, D, E, M;
A = (0,0);
B = (4,0);
C = (2.6,2);
M = (A + B)/2;
D = (A + C)/2;
E = (C + D)/2;
draw(A--B--C--cycle);
draw(C--M--D--B);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, N);
dot("$D$", D, NW);
dot("$E$", E, NW);
dot("$M$", M, S);
[/asy]
[i]Be aware: the figure is not drawn to scale.[/i]
The point $P$ lies inside $\vartriangle ABC$ so that $\vartriangle BPC$ is isosceles, and angle $P$ is a right angle. Furthermore both $\vartriangle BAN$ and $\vartriangle CAM$ are isosceles with a right angle at $A$, and both are outside $\vartriangle ABC$. Show that $\vartriangle MNP$ is isosceles and right-angled.
[img]https://1.bp.blogspot.com/-i9twOChu774/XzcBLP-RIXI/AAAAAAAAMXA/n5TJCOJypeMVW28-9GDG4st5C47yhvTCgCLcBGAsYHQ/s0/2005%2BMohr%2Bp3.png[/img]
An arbitrary point $D$ is marked on the hypotenuse $AB$ of a right triangle $ABC$. The circle circumscribed around the triangle $ACD$ intersects the line $BC$ at the point $E$ for the second time, and the circle circumscribed around the triangle $BCD$ intersects the line $AC$ for the second time at the point $F$. Prove that the line $EF$ passes through the point $D$.
The sides of the triangle $ABC$ satisfy the relation $c^2 = a^2 + b^2$. Show that angle $C$ is right.
In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$.
[img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]
On the sides of triangle $ABC$, isosceles right-angled triangles $AUB, CVB$, and $AWC$ are placed. These three triangles have their right angles at vertices $U, V$ , and $W$, respectively. Triangle $AUB$ lies completely inside triangle $ABC$ and triangles $CVB$ and $AWC$ lie completely outside $ABC$. See the figure. Prove that quadrilateral $UVCW$ is a parallelogram.
[asy]
import markers;
unitsize(1.5 cm);
pair A, B, C, U, V, W;
A = (0,0);
B = (2,0);
C = (1.7,2.5);
U = (B + rotate(90,A)*(B))/2;
V = (B + rotate(90,C)*(B))/2;
W = (C + rotate(90,A)*(C))/2;
draw(A--B--C--cycle);
draw(A--W, StickIntervalMarker(1,1,size=2mm));
draw(C--W, StickIntervalMarker(1,1,size=2mm));
draw(B--V, StickIntervalMarker(1,2,size=2mm));
draw(C--V, StickIntervalMarker(1,2,size=2mm));
draw(A--U, StickIntervalMarker(1,3,size=2mm));
draw(B--U, StickIntervalMarker(1,3,size=2mm));
draw(rightanglemark(A,U,B,5));
draw(rightanglemark(B,V,C,5));
draw(rightanglemark(A,W,C,5));
dot("$A$", A, S);
dot("$B$", B, S);
dot("$C$", C, N);
dot("$U$", U, NE);
dot("$V$", V, NE);
dot("$W$", W, NW);
[/asy]
The point $P$ lies inside $\vartriangle ABC$ so that $\vartriangle BPC$ is isosceles, and angle $P$ is a right angle. Furthermore both $\vartriangle BAN$ and $\vartriangle CAM$ are isosceles with a right angle at $A$, and both are outside $\vartriangle ABC$. Show that $\vartriangle MNP$ is isosceles and right-angled.
[img]https://1.bp.blogspot.com/-i9twOChu774/XzcBLP-RIXI/AAAAAAAAMXA/n5TJCOJypeMVW28-9GDG4st5C47yhvTCgCLcBGAsYHQ/s0/2005%2BMohr%2Bp3.png[/img]
Let $CD$ be an altitude of right-angled triangle $ABC$ with $\angle C = 90$. Regular triangles$ AED$ and $CFD$ are such that $E$ lies on the same side from $AB$ as $C$, and $F$ lies on the same side from $CD$ as $B$. The line $EF$ meets $AC$ at $L$. Prove that $FL = CL + LD$
Let $\gamma$ be the circumscribed circle about a right-angled triangle $ABC$ with right angle $C$. Let $\delta$ be the circle tangent to the sides $AC$ and $BC$ and tangent to the circle $\gamma$ internally.
(a) Find the radius $i$ of $\delta$ in terms of $a$ when $AC$ and $BC$ both have length $a$.
(b) Show that the radius $i$ is twice the radius of the inscribed circle of $ABC$.
Let $P$ be a point inside a triangle $ABC$ with $\angle C = 90^o$ such that $AP = AC$, and let $M$ be the midpoint of $AB$ and $CH$ be the altitude. Prove that $PM$ bisects $\angle BPH$ if and only if $\angle A = 60^o$.
Consider the isosceles right triangle $ABC$ ($AB = AC$) and the points $M, P \in [AB]$ so that $AM = BP$. Let $D$ be the midpoint of the side $BC$ and $R, Q$ the intersections of the perpendicular from $A$ on$ CM$ with $CM$ and $BC$ respectively. Prove that
a) $\angle AQC = \angle PQB$
b) $\angle DRQ = 45^o$
Let $\vartriangle ABC$ be a right-angled triangle with $\angle A = 90^o$ and circumcircle $\Gamma$. The inscribed circle is tangent to $BC$ in point $D$. Let $E$ be the midpoint of the arc $AB$ of $\Gamma$ not containing $C$ and let $F$ be the midpoint of the arc $AC$ of $\Gamma$ not containing $B$.
(a) Prove that $\vartriangle ABC \sim \vartriangle DEF$.
(b) Prove that $EF$ goes through the points of tangency of the incircle to $AB$ and $AC$.
Let $ABC$ be a right triangle with $\angle A= 90^{\circ}$. A circle $\omega$ centered on $BC$ is tangent to $AB$ at $D$ and $AC$ at $E$. Let $F$ and $G$ be the intersections of $\omega$ and $BC$ so that $F$ lies between $B$ and $G$. If lines $DG$ and $EF$ intersect at $X$, show that $AX=AD.$
Let $ABC$ be a triangle with $\angle A = 90^o, \angle B = 75^o$ and $AB = 2$. The points $P$ and $Q$ on the sides $AC$ and $BC$ respectively are such that $\angle APB = \angle CPQ$ and $\angle BQA = \angle CQP$ . Calculate the measurement of the segment $QA $.
The areas of two right-angled triangles have ratio equal to the squares of their hypotenuses. Show that these triangles are similar.
Four right triangles, each with the sides $1$ and $2$, are assembled to a figure as shown.
How large a fraction does the area of the small circle make up of that of the big one?
[img]https://1.bp.blogspot.com/-XODK1XKCS0Q/XzXDtcA-xAI/AAAAAAAAMWA/zSLPpf3IcX0rgaRtOxm_F2begnVdUargACLcBGAsYHQ/s0/2010%2BMohr%2Bp1.png[/img]
Triangle $ABC$ is right-angled at $B$ and has incentre $I$. Points $D$, $E$ and $F$ are the points where the incircle of the triangle touches the sides $BC$, $AC$ and AB respectively. Lines $CI$ and $EF$ intersect at point $P$. Lines $DP$ and $AB$ intersect at point $Q$. Prove that $AQ = BF$.
Let $ABC$ be a triangle with $\angle A = 90^o$ and let $D$ be an orthogonal projection of $A$ onto $BC$. The midpoints of $AD$ and $AC$ are called $E$ and $F$, respectively. Let $M$ be the circumcentre of $\vartriangle BEF$. Prove that $AC\parallel BM$.
Let $ACE$ be a triangle with $\angle ECA=60^{\circ}, \angle AEC=90^{\circ}$. Let $B$ and $D$ be points on the sides $AC$ and $CE$ respectively such that the $\triangle BCD$ is equilateral. Now suppose $BD \cap AE=F$. Find $\angle EAC+\angle EFD$.
In a right triangle $ABC$ with a right angle $C$, let $AK$ and $BN$ be the angle bisectors. Let $D,E$ be the projections of $C$ on $AK, BN$ respectively. Prove that the length of the segment $DE$ is equal to the radius of the inscribed circle.