Circle $\omega$ with centre $M$ and diameter $XY$ is given. Point $A$ is chosen on $\omega$ so that $AX<AY$. Points $B, C$ are chosen on segments $XM, YM$, respectively, in a way that $BM=CM$. A parallel line to $AB$ is constructed through $C$; the line intersects $\omega$ at $P$ so that $P$ lies on the smaller arc $\widehat{AY}$. Similarly, a parallel line to $AC$ is constructed through $B$; the line intersects $\omega$ at $Q$ so that $Q$ lies on the smaller arc $\widehat{XA}$. Lines $PQ$ and $XY$ intersect at $S$. Prove that $AS$ is tangent to $\omega$.

Yesterday (=April 22, 2003) was Gittes birthday. She notices that her age equals the sum of the 4 digits of the year she was born in. How old is she?

Let $z_0+z_1+z_2+\cdots$ be an infinite complex geometric series such that $z_0=1$ and $z_{2013}=\dfrac 1{2013^{2013}}$. Find the sum of all possible sums of this series.

JHMT Pizzeria messed up ordering boxes to put their $10$ inch pizza in. ($10$ inch pizza means the diameter of the pizza is $10$ inches) They accidentally ordered an $8\, in \times  8\, in$. box and they immediately need to deliver $10$ inch pizzas to customers, and they decide to cut the pizza minimally so that the most part of the pizza ts in to the $8\, in \times  8\, in$ box. The area they need to cut out from the original $10$ inch pizza can be written in form of $\alpha \arccos( \beta) - \gamma$ . Find the value of $\alpha \beta \gamma$, where $\alpha$ and $\gamma$ are integers and  $\beta$ is a rational number strictly between $\frac12$ and $1$.

Point $P$ is chosen inside triangle $ABC$ so that $\angle APC+\angle ABC=180^o$ and $BC=AP.$ On the side $AB$, a point $K$ is chosen such that $AK = KB + PC$. Prove that $CK \perp AB$.

A subset $\bf{S}$ of the plane is called [i]convex[/i] if given any two points $x$ and $y$ in $\bf{S}$, the line segment joining $x$ and $y$ is contained in $\bf{S}$. A quadrilateral is called [i]convex[/i] if the region enclosed by the edges of the quadrilateral is a convex set. Show that given a convex quadrilateral $Q$ of area $1$, there is a rectangle $R$ of area $2$ such that $Q$ can be drawn inside $R$.

On sides \( AB \) and \( AC \) of an acute-angled, non-isosceles triangle \( ABC \), points \( P \) and \( Q \) are chosen such that the center \( O_9 \) of the nine-point circle of \( \triangle ABC \) is the midpoint of segment \( PQ \). Let \( O \) be the circumcenter of \( \triangle ABC \). On the ray \( OP \) beyond \( P \), segment \( PX \) is marked such that \( PX = AQ \). On the ray \( OQ \) beyond \( Q \), segment \( QY \) is marked such that \( QY = AP \). Prove that the midpoint of side \( BC \), the midpoint of segment \( XY \), and the point \( O_9 \) are collinear. [i]The nine-point circle or the Euler circle[/i] of \( \triangle ABC \) is the circle passing through nine significant points of the triangle — the midpoints of the three sides, the feet of the three altitudes, and the midpoints of the segments connecting the orthocenter with the vertices of \( \triangle ABC \). [i]Proposed by Danylo Khilko[/i]

$$\int_1^2\frac{12x^3+12x+12}{2x^4+3x^2+4x}\,dx$$ [i]Proposed by Connor Gordon[/i]

Areas of four surfaces of a tetrahedron are $S_1,S_2,S_3,S_4$. And the largest one of them is $S$. $\lambda=\frac{S_1+S_2+S_3+S_4}{S}$, then $\lambda$ always satisfies $\text{(A)}2<\lambda\leq4\qquad\text{(B)}3<\lambda<4\qquad\text{(C)}2.5<\lambda\leq4.5\qquad\text{(D)}3.5<\lambda<5.5$

Let $p$ be a prime number and $F=\left \{0,1,2,...,p-1 \right \}$. Let $A$ be a proper subset of $F$ that satisfies the following property: if $a,b \in A$, then $ab+1$ (mod $p$) $ \in A$. How many elements can $A$ have? (Justify your answer.)

Given an acute-angled triangle $ABC$, let points $A' , B' , C'$ be located as follows: $A'$ is the point where altitude from $A$ on $BC$ meets the outwards-facing semicircle on $BC$ as diameter. Points $B', C'$ are located similarly. Prove that $A[BCA']^2 + A[CAB']^2 + A[ABC']^2 = A[ABC]^2$ where $A[ABC]$ is the area of triangle $ABC$.

Let $\theta$ be an obtuse angle with $\sin{\theta}=\frac{3}{5}$. If an ant starts at the origin and repeatedly moves $1$ unit and turns by an angle of $\theta$, there exists a region $R$ in the plane such that for every point $P\in R$ and every constant $c>0$, the ant is within a distance $c$ of $P$ at some point in time (so the ant gets arbitrarily close to every point in the set). What is the largest possible area of $R$? [i]2020 CCA Math Bonanza Individual Round #15[/i]

Consider the sytem of equations \[ a_{11}x_1+a_{12}x_2+a_{13}x_3 = 0 \]\[a_{21}x_1+a_{22}x_2+a_{23}x_3 =0\]\[a_{31}x_1+a_{32}x_2+a_{33}x_3 = 0 \] with unknowns $x_1, x_2, x_3$. The coefficients satisfy the conditions: a) $a_{11}, a_{22}, a_{33}$ are positive numbers; b) the remaining coefficients are negative numbers; c) in each equation, the sum ofthe coefficients is positive. Prove that the given system has only the solution $x_1=x_2=x_3=0$.

In $\vartriangle ABC$, the incircle is tangent to the sides $BC, CA, AB$ at $D, E, F$ respectively. Let $P$ and $Q$ be the midpoints of $DF$ and $DE$ respectively. Lines $P C$ and $DE$ intersect at $R$, and lines $BQ$ and$ DF$ intersect at $S$. Prove that a) Points $B, C, P, Q$ lie on a circle. b) Points $P, Q, R, S$ lie on a circle.

A set of $1985$ points is distributed around the circumference of a circle and each of the points is marked with $1$ or $-1$. A point is called “good” if the partial sums that can be formed by starting at that point and proceeding around the circle for any distance in either direction are all strictly positive. Show that if the number of points marked with $-1$ is less than $662$, there must be at least one good point.

Sofia places the dice on a table as shown in the figure, matching faces that have the same number on each die. She circles the table without touching the dice. What is the sum of the numbers of all the faces that she cannot see? $Note$. In all given the numbers on the opposite faces add up to 7.

Find the greatest real $k$ such that, for every tetrahedron $ABCD$ of volume $V$, the product of areas of faces $ABC,ABD$ and $ACD$ is at least $kV^2$.

The vertices of a regular nonagon are colored such that $1)$ adjacent vertices are different colors and $2)$ if $3$ vertices form an equilateral triangle, they are all different colors. Let $m$ be the minimum number of colors needed for a valid coloring, and n be the total number of colorings using $m$ colors. Determine $mn$. (Assume each vertex is distinguishable.)

Consider the set of all rectangles with a given area $S$. Find the largest value o $ M = \frac{16-p}{p^2+2p}$ where $p$ is the perimeter of the rectangle.

For each positive integer $k$, let $n_k$ be the smallest positive integer such that there exists a finite set $A$ of integers satisfy the following properties: [list] [*]For every $a\in A$, there exists $x,y\in A$ (not necessary distinct) that $$n_k\mid a-x-y$$[/*] [*]There's no subset $B$ of $A$ that $|B|\leq k$ and $$n_k\mid \sum_{b\in B}{b}.$$ [/list] Show that for all positive integers $k\geq 3$, we've $$n_k<\Big( \frac{13}{8}\Big)^{k+2}.$$

Find the number of pairs $(n, q)$, where $n$ is a positive integer and $q$ a non-integer rational number with $0 < q < 2000$, that satisfy $\{q^2\}=\left\{\frac{n!}{2000}\right\}$

Find all sets of real numbers $\{a_1,a_2,\ldots, _{1375}\}$ such that \[2 \left( \sqrt{a_n - (n-1)}\right) \geq a_{n+1} - (n-1), \quad \forall n \in \{1,2,\ldots,1374\},\] and \[2 \left( \sqrt{a_{1375}-1374} \right) \geq a_1 +1.\]

Let $ABC$ be an equilateral triangle. For an arbitrary line $\ell$. through $B$, the orthogonal projections of $A$ and $C$ on $\ell$ are denoted by $D$ and $E$ respectively. If $D\ne E$, equilateral triangles $DEP$ and $DET$ are constructed on different sides of $\ell$. Find the loci of $P$ and $T$.

We consider triangles $ABC$, in which the point $M$ lies on the side $AB$, $AM = a$, $BM = b$, $CM = c$ ($c <a, c <b$). Find the smallest radius of the circumcircle of such triangles.

Bob proposes the following game to Johanna. The board in the figure is an equilateral triangle subdivided in turn into $256$ small equilateral triangles, one of which is painted in black. Bob chooses any point inside the board and places a small token. Johanna can make three types of plays. Each of them consists of choosing any of the $3$ vertices of the board and move the token to the midpoint between the current position of the tile and the chosen vertex. In the second figure we see an example of a move in which Johana chose vertex $A$. Johanna wins if she manages to place her piece inside the triangle black. Prove that Johanna can always win in at most $4$ moves. [asy] unitsize(8 cm); pair A, B, C; int i; A = dir(60); C = (0,0); B = (1,0); fill((6/16*(1,0) + 1/16*dir(60))--(7/16*(1,0) + 1/16*dir(60))--(6/16*(1,0) + 2/16*dir(60))--cycle, gray(0.7)); draw(A--B--C--cycle); for (i = 1; i <= 15; ++i) { draw(interp(A,B,i/16)--interp(A,C,i/16)); draw(interp(B,C,i/16)--interp(B,A,i/16)); draw(interp(C,A,i/16)--interp(C,B,i/16)); } label("$A$", A, N); label("$B$", B, SE); label("$C$", C, SW); [/asy] [asy] unitsize(8 cm); pair A, B, C, X, Y, Z; int i; A = dir(60); C = (0,0); B = (1,0); X = 9.2/16*(1,0) + 3.3/16*dir(60); Y = (A + X)/2; Z = rotate(60,X)*(Y); fill((6/16*(1,0) + 1/16*dir(60))--(7/16*(1,0) + 1/16*dir(60))--(6/16*(1,0) + 2/16*dir(60))--cycle, gray(0.7)); draw(A--B--C--cycle); for (i = 1; i <= 15; ++i) { draw(interp(A,B,i/16)--interp(A,C,i/16)); draw(interp(B,C,i/16)--interp(B,A,i/16)); draw(interp(C,A,i/16)--interp(C,B,i/16)); } draw(A--X, dotted); draw(arc(Z,abs(X - Y),-12,40), Arrow(6)); label("$A$", A, N); label("$B$", B, SE); label("$C$", C, SW); dot(A); dot(X); dot(Y); [/asy]