In the triangle $ABC$ holds $|AB| = |AC|$ and the inscribed circle touches the sides $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$ at the points $D$, $E$ and $F$ respectively . The perpendicular from the point $D$ to the line $EF$ intersects the side $\overline{AB}$ at the point $G$, and the circles circumscribed around the triangles $AEF$ and $ABC$ intersect at the points $A $and $T$. Prove that the lines $T G$ and $T F$ are perpendicular.

It is given function $f(x)=3x-2$ $a)$ Find $g(x)$ if $f(2x-g(x))=-3(1+2m)x+34$ $b)$ Solve the equation: $g(x)=4(m-1)x-4(m+1)$, $m \in \mathbb{R}$

In a given country, all inhabitants are knights or knaves. A knight never lies; a knave always lies. We meet three persons, $A, B$, and $C$. Person $A$ says, “If $C$ is a knight, $B$ is a knave.” Person $C$ says, “$A$ and I are different; one is a knight and the other is a knave.” Who are the knights, and who are the knaves ?

$\triangle ABC$ is a right angle at $C$ and $\angle A = 20^\circ$. If $BD$ is the bisector of $\angle ABC$, then $\angle BDC =$ [asy] size(200); defaultpen(linewidth(0.8)+fontsize(11pt)); pair A= origin, B = 3 * dir(25), C = (B.x,0); pair X = bisectorpoint(A,B,C), D = extension(B,X,A,C); draw(B--A--C--B--D^^rightanglemark(A,C,B,4)); path g = anglemark(A,B,D,14); path h = anglemark(D,B,C,14); draw(g); draw(h); add(pathticks(g,1,0.11,6,6)); add(pathticks(h,1,0.11,6,6)); label("$A$",A,W); label("$B$",B,NE); label("$C$",C,E); label("$D$",D,S); label("$20^\circ$",A,8*dir(12.5)); [/asy] $ \textbf{(A)}\ 40^\circ \qquad \textbf{(B)}\ 45^\circ \qquad \textbf{(C)}\ 50^\circ \qquad \textbf{(D)}\ 55^\circ \qquad \textbf{(E)}\ 60^\circ $

Find all positive integers $k$ for which the equation: $$ \text{lcm}(m,n)-\text{gcd}(m,n)=k(m-n)$$ has no solution in integers positive $(m,n)$ with $m\neq n$.

There are $n$ knights numbered $1$ to $n$ and a round table with $n$ chairs. The first knight chooses his chair, and from him, the knight number $k+1$ sits $ k$ places to the right of knight number $k$ , for all $1 \le k\le n-1$ (occupied and empty seats are counted). In particular, the second knight sits next to the first. Find all values ​​of $n$ such that the $n$ gentlemen occupy the $n$ chairs following the described procedure.

The number $2020$ has three different prime factors. What is their sum?

Let $ABC$ be a right triangle with $\angle ACB = 90^o$ and M the center of $AB$. Let $G$ br any point on the line $MC$ and $P$ a point on the line $AG$, such that $\angle CPA = \angle BAC$ . Further let $Q$ be a point on the straight line $BG$, such that $\angle BQC = \angle CBA$ . Show that the circles of the triangles $AQG$ and $BPG$ intersect on the segment $AB$.

Which statement does not describe benzene, $\ce{C6H6}$? $ \textbf{(A)}\ \text{It is an aromatic hydrocarbon.} \qquad$ $\textbf{(B)}\ \text{It exists in two isomeric forms.} \qquad$ $\textbf{(C)}\ \text{It undergoes substitution reactions.} \qquad$ $\textbf{(D)}\ \text{It can react to form three different products with the formula C}_6\text{H}_4\text{Cl}_2\qquad$

Does there exist a number $n=\overline{a_1a_2a_3a_4a_5a_6}$ such that $\overline{a_1a_2a_3}+4 = \overline{a_4a_5a_6}$ (all bases are $10$) and $n=a^k$ for some positive integers $a,k$ with $k \geq 3 \ ?$

Is it possible to divide all non-empty subsets of a set of 10 elements into triples so that in each triple, two of the subsets do not intersect and in their union give the third? [i]Proposed by Vladislav Frank[/i]

In an arithmetic sequence, the difference of consecutive terms in constant. We consider sequences of integers in which the difference of consecutive terms equals the sum of the differences of all preceding consecutive terms. Which of these sequences with $a_0 = 2012$ and $1\leqslant d = a_1-a_0 \leqslant 43$ contain square numbers?

[b]p1.[/b] Prove that the equation $x^6 - 143x^5 - 917x^4 + 51x^3 + 77x^2 + 291x + 1575 = 0$ has no integer solutions. [b]p2.[/b] There are $81$ wheels in a storage marked by their two types, say first and second type. Wheels of the same type weigh equally. Any wheel of the second type is much lighter than a wheel of the first type. It is known that exactly one wheel is marked incorrectly. Show that it can be detected with certainty after four measurements on a balance scale. [b]p3.[/b] Rob and Ann multiplied the numbers from $1$ to $100$ and calculated the sum of digits of this product. For this sum, Rob calculated the sum of its digits as well. Then Ann kept repeating this operation until he got a one-digit number. What was this number? [b]p4.[/b] Rui and Jui take turns placing bishops on the squares of the $ 8\times 8$ chessboard in such a way that bishops cannot attack one another. (In this game, the color of the rooks is irrelevant.) The player who cannot place a rook loses the game. Rui takes the first turn. Who has a winning strategy, and what is it? [b]p5.[/b] The following figure can be cut along sides of small squares into several (more than one) identical shapes. What is the smallest number of such identical shapes you can get? [img]https://cdn.artofproblemsolving.com/attachments/8/e/9cd09a04209774dab34bc7f989b79573453f35.png[/img] PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

The point $M$ is located inside the triangle $ABC$. The ray $AM$ intersects the circumcircle of the triangle $MBC$ once more at point $D$, the ray $BM$ intersects the circumcircle of the triangle $MCA$ once more at point $E$, and the ray $CM$ intersects the circumcircle of the triangle $MAB$ once more at point $F$. Prove that holds $$\frac{AD}{MD}+\frac{BE}{ME} +\frac{CF}{MF}\ge \frac92 $$

Suppose that $M$ is a set of $2003$ numbers such that, for any distinct $a, b, c \in M$, the number $a^2 + bc$ is rational. Prove that there is a positive integer $n$ such that $a\sqrt n$ is rational for all $a \in M.$

Let $A_1A_2$ and $B_1B_2$ be the diagonals of a rectangle, and let $O$ be its center. Find and construct the set of all points $P$ that satisfy simultaneously the four inequaliies: $$A_1P > OP , \\A_2P > OP, \ \ B_1P > OP , \ \ B_2P > OP.$$

A sequence $a_n$ of real numbers satisfies $a_1=1$, $a_2=0$, and $a_n=(S_{n-1}+1)S_{n-2}$ for all integers $n\geq3$, where $S_k=a_1+a_2+\dots+a_k$ for positive integers $k$. What is the smallest integer $m>2$ such that $127$ divides $a_m$? [i]2020 CCA Math Bonanza Individual Round #9[/i]