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

Tags were heavily modified to better represent problems.

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Found problems: 65

2012 Junior Balkan Team Selection Tests - Romania, 2

Consider a semicircle of center $O$ and diameter $[AB]$, and let $C$ be an arbitrary point on the segment $(OB)$. The perpendicular to the line $AB$ through $C$ intersects the semicircle in $D$. A circle centered in $P$ is tangent to the arc $BD$ in $F$ and to the segments $[AB]$ and $[CD]$ in $G$ and $E$, respectively. Prove that the triangle $ADG$ is isosceles.

2013 Tournament of Towns, 3

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.

2020 Federal Competition For Advanced Students, P2, 5

Let $h$ be a semicircle with diameter $AB$. Let $P$ be an arbitrary point inside the diameter $AB$. The perpendicular through $P$ on $AB$ intersects $h$ at point $C$. The line $PC$ divides the semicircular area into two parts. A circle will be inscribed in each of them that touches $AB, PC$ and $h$. The points of contact of the two circles with $AB$ are denoted by $D$ and $E$, where $D$ lies between $A$ and $P$. Prove that the size of the angle $DCE$ does not depend on the choice of $P$. (Walther Janous)

2008 Dutch IMO TST, 5

Let $\vartriangle ABC$ be a right triangle with $\angle B = 90^o$ and $|AB| > |BC|$, and let $\Gamma$ be the semicircle with diameter $AB$ that lies on the same side as $C$. Let $P$ be a point on $\Gamma$ such that $|BP| = |BC|$ and let $Q$ be on $AB$ such that $|AP| = |AQ|$. Prove that the midpoint of $CQ$ lies on $\Gamma$.

2017 Estonia Team Selection Test, 10

Let $ABC$ be a triangle with $AB = \frac{AC}{2 }+ BC$. Consider the two semicircles outside the triangle with diameters $AB$ and $BC$. Let $X$ be the orthogonal projection of $A$ onto the common tangent line of those semicircles. Find $\angle CAX$.

2005 Thailand Mathematical Olympiad, 4

Let $O_1$ be the center of a semicircle $\omega_1$ with diameter $AB$ and let $O_2$ be the center of a circle $\omega_2$ inscribed in $\omega_1$ and which is tangent to $AB$ at $O_1$. Let $O_3$ be a point on $AB$ that is the center of a semicircle $\omega_3$ which is tangent to both $\omega_1$ and $\omega_2$. Let $P$ be the intersection of the line through $O_3$ perpendicular to $AB$ and the line through $O_2$ parallel to $AB$. Show that $P$ is the center of a circle $\Gamma$ tangent to all of $\omega_1, \omega_2$ and $\omega_3$.

1974 Bundeswettbewerb Mathematik, 3

A circle $K_1$ of radius $r_1 = 1\slash 2$ is inscribed in a semi-circle $H$ with diameter $AB$ and radius $1.$ A sequence of different circles $K_2, K_3, \ldots$ with radii $r_2, r_3, \ldots$ respectively are drawn so that for each $n\geq 1$, the circle $K_{n+1}$ is tangent to $H$, $K_n$ and $AB.$ Prove that $a_n = 1\slash r_n$ is an integer for each $n$, and that it is a perfect square for $n$ even and twice a perfect square for $n$ odd.

2022 Austrian Junior Regional Competition, 3

A semicircle is erected over the segment $AB$ with center $M$. Let $P$ be one point different from $A$ and $B$ on the semicircle and $Q$ the midpoint of the arc of the circle $AP$. The point of intersection of the straight line $BP$ with the parallel to $P Q$ through $M$ is $S$. Prove that $PM = PS$ holds. [i](Karl Czakler)[/i]

1939 Eotvos Mathematical Competition, 3

$ABC$ is an acute triangle. Three semicircles are constructed outwardly on the sides $BC$, $CA$ and $AB$ respectively. Construct points $A'$ , $B'$ and $C' $ on these semicìrcles respectively so that $AB' = AC'$, $BC' = BA'$ and $CA'= CB'$.

1996 Singapore Senior Math Olympiad, 2

Let $180^o < \theta_1 < \theta_2 <...< \theta_n = 360^o$. For $i = 1,2,..., n$, $P_i = (\cos \theta_i^o, \sin \theta_i^o)$ is a point on the circle $C$ with centre $(0,0)$ and radius $1$. Let $P$ be any point on the upper half of $C$. Find the coordinates of $P$ such that the sum of areas $[PP_1P_2] + [PP_2P_3] + ...+ [PP_{n-1}P_n]$ attains its maximum.

1997 Singapore Senior Math Olympiad, 2

Figure shows a semicircle with diameter $AD$. The chords $AC$ and $BD$ meet at $P$. $Q$ is the foot of the perpendicular from $P$ to $AD$. find $\angle BCQ$ in terms of $\theta$ and $\phi$ . [img]https://cdn.artofproblemsolving.com/attachments/a/2/2781050e842b2dd01b72d246187f4ed434ff69.png[/img]

1995 Chile National Olympiad, 7

In a semicircle of radius $4$ three circles are inscribed, as indicated in the figure. Larger circles have radii $ R_1 $ and $ R_2 $, and the larger circle has radius $ r $. a) Prove that $ \dfrac {1} {\sqrt{r}} = \dfrac {1} {\sqrt{R_1}} + \dfrac {1} {\sqrt{R_2}} $ b) Prove that $ R_1 + R_2 \le 8 (\sqrt{2} -1) $ c) Prove that $ r \le \sqrt{2} -1 $ [img]https://cdn.artofproblemsolving.com/attachments/0/9/aaaa65d1f4da4883973751e1363df804b9944c.jpg[/img]

1989 Chile National Olympiad, 3

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]

Denmark (Mohr) - geometry, 2013.2

The figure shows a rectangle, its circumscribed circle and four semicircles, which have the rectangle’s sides as diameters. Prove that the combined area of the four dark gray crescentshaped regions is equal to the area of the light gray rectangle. [img]https://1.bp.blogspot.com/-gojv6KfBC9I/XzT9ZMKrIeI/AAAAAAAAMVU/NB-vUldjULI7jvqiFWmBC_Sd8QFtwrc7wCLcBGAsYHQ/s0/2013%2BMohr%2Bp3.png[/img]

2023 Bangladesh Mathematical Olympiad, P2

Let the points $A,B,C$ lie on a line in this order. $AB$ is the diameter of semicircle $\omega_1$, $AC$ is the diameter of semicircle $\omega_2$. Assume both $\omega_1$ and $\omega_2$ lie on the same side of $AC$. $D$ is a point on $\omega_2$ such that $BD\perp AC$. A circle centered at $B$ with radius $BD$ intersects $\omega_1$ at $E$. $F$ is on $AC$ such that $EF\perp AC$. Prove that $BC=BF$.