Found problems: 333
Consider a square on the hypotenuse of a right triangle $[ABC]$ (right at $B$). Prove that the line segment that joins vertex $B$ with the center of the square makes $45^o$ angles with legs of the triangle.
A ladder is leaning against a house with its lower end $15$ feet from the house. When the lower end is pulled $9$ feet farther from the house, the upper end slides $13$ feet down. How long is the ladder (in feet)?
Consider $\vartriangle ABC$ right at $B, F$ a point such that $B - F - C$ and $AF$ bisects $\angle BAC$, $I$ a point such that $A - I - F$ and CI bisect $\angle ACB$, and $E$ a point such that $A- E - C$ and $AF \perp EI$. If $AB = 4$ and $\frac{AI}{IF}={4}{3}$ , determine $AE$.
Notation: $A-B-C$ means than points $A,B,C$ are collinear in that order i.e. $ B$ lies between $ A$ and $C$.
In triangle $ABC, \angle A = 90^o$. $M$ is the midpoint of $BC$ and $H$ is the foot of the altitude from $A$ to $BC$. The line passing through $M$ and perpendicular to $AC$ meets the circumcircle of triangle $AMC$ again at $P$. If $BP$ intersects $AH$ at $K$, prove that $AK = KH$.
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.
Show that $MK=ML$.
[i]Proposed by Josef Tkadlec, Czech Republic[/i]
In three houses $A,B$ and $C$, forming a right triangle with the legs $AC=30$ and $CB=40$, live three beetles $a,b$ and $c$, capable of moving at speeds of $2, 3$ and $4$, respectively. Suppose that you simultaneously release these bugs from point $M$ and mark the time after which beetles reach their homes. Find on the plane such a point $M$, where is the last time to reach the house a bug would be minimal.
In a triangle $ABC$, right in $A$ and isosceles, let $D$ be a point on the side $AC$ ($A \ne D \ne C$) and $E$ be the point on the extension of $BA$ such that the triangle $ADE$ is isosceles. Let $P$ be the midpoint of segment $BD$, $R$ be the midpoint of the segment $CE$ and $Q$ the intersection point of $ED$ and $BC$. Prove that the quadrilateral $ARQP$ is a square.
Given a circle of radius $r= 1995$. Show that around it you can describe exactly $16$ primitive Pythagorean triangles. The primitive Pythagorean triangle is a right-angled triangle, the lengths of the sides of which are expressed by coprime integers.
In a right triangle with legs $a$, $b$ and hypotenuse $c$, draw semicircles with diameters on the sides of the triangle as indicated in the figure. The purple areas have values $X,Y$ . Calculate $X + Y$.
[img]https://cdn.artofproblemsolving.com/attachments/1/a/5086dc7172516b0a986ef1af192c15eba4d6fc.png[/img]
Let $ABC$ be a right triangle. It is known that there are points $D$ on the side $AC$ and $E$ on the side $BC$ such that $AB = AD = BE$ and $BD$ is perpendicular to $DE$. Calculate the ratios $\frac{AB}{BC}$ and $\frac{BC}{CA}$.
Suppose a triangle $\vartriangle ABC$ with $\angle C = 90^o$ is given. Let $D$ be the midpoint of $AC$, and let $E$ be the foot of the altitude through $C$ on $BD$. Show that the tangent in $C$ of the circumcircle of $\vartriangle AEC$ is perpendicular to $AB$.
Cut a right triangle with an angle of $30^o$ into three isosceles non-acute triangles, among which there are no congruent ones.
(Maria Rozhkova)
$ABC$ is a triangle with $\angle A = 90^o$. Take $E$ such that the triangle $AEC$ is outside $ABC$ and $AE = CE$ and $\angle AEC = 90^o$. Similarly, take $D$ so that $ADB$ is outside $ABC$ and similar to $AEC$. $O$ is the midpoint of $BC$. Let the lines $OD$ and $EC$ meet at $D'$, and the lines $OE$ and $BD$ meet at $E'$. Find area $DED'E'$ in terms of the sides of $ABC$.
Let $D$ and $E$ be the midpoints of the sides $BC$ and $AC$ of a right triangle $ABC$. Prove that if $\angle CAD=\angle ABE$, then $$\frac{5}{6} \le \frac{AD}{AB}\le \frac{\sqrt{73}}{10}.$$
Show that it is possible to color each point of a circle red or blue so that no right-angled triangle inscribed in the circle has its vertices all the same color.
Consider a square. Find the locus of midpoints of the hypothenuses of rightangled triangles with the vertices lying on three different sides of the square and not coinciding with its vertices.
(B.Frenkin)
$ABC$ is a triangle with $\angle A = 90^\circ$, $\angle B = 60^\circ$. The points $A_1$, $B_1$, $C_1$ on $BC$, $CA$, $AB$ respectively are such that $A_1B_1C_1$ is equilateral and the perpendiculars (to $BC$ at $A_1$, to $CA$ at $B_1$ and to $AB$ at $C_1$) meet at a point $P$ inside the triangle. Find the ratios $PA_1:PB_1:PC_1$.
The trapezoid is composed of three conguent right isosceles triangles as shown in the figure. It is necessary to cut it into $4$ equal parts. How to do it?
[img]https://cdn.artofproblemsolving.com/attachments/f/e/87b07ae823190f26b70bfa22824679a829e649.png[/img]
Assume that $C$ is a right angle of triangle $ABC$ and $N$ is a midpoint of the semicircle, constructed on $CB$ as on diameter externally. Prove that $AN$ divides the bisector of angle $C$ in half.
Let $A,B,C,D,E$ be consecutive points on a circle with center $O$ such that $AC=BD=CE=DO$. Let $H_1,H_2,H_3$ be the orthocenters triangles $ACD,BCD,BCE$ respectively. Prove that the triangle $H_1H_2H_3$ is right.
Each point on the boundary of a square has to be colored in one color. Consider all right triangles with the vertices on the boundary of the square. Determine the least number of colors for which there is a coloring such that no such triangle has all its vertices of the same color.
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$.
Show that $MK=ML$.
[i]Proposed by Josef Tkadlec, Czech Republic[/i]
In a right triangle, the point $O$ is the center of the circumcircle. Another circle of smaller radius centered at the point $O$ touches the larger leg and the altitude drawn from the top of the right angle. Find the acute angles of a right triangle and the ratio of the radii of the circumscribed and smaller circles.
The midpoint of the hypotenuse $AB$ of the right triangle $ABC$ is $K$. The point $M$ on the side $BC$ is taken such that $BM = 2 \cdot MC$. Prove that $\angle BAM = \angle CKM$.
Consider $\vartriangle ABC$, right at $B$, let $I$ be its incenter and $F,D,E$ the points where the circle inscribed on sides AB, $BC$ and $AC$, respectively. If $M$ is the intersection point of $CI$ and $EF$, and $N$ is the intersection point of $DM$ and $AB$. Prove that $AN = ID$.