problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Given an acute triangle $ABC$ . with $H$ as its orthocenter, lines $\ell_1$ and $\ell_2$ go through $H$ and are perpendicular to each other. Line $\ell_1$ cuts $BC$ and the extension of $AB$ on $D$ and $Z$ respectively. Whereas line $\ell_2$ cuts $BC$ and the extension of $AC$ on $E$ and $X$ respectively. If the line through $D$ and parallel to $AC$ and the line through $E$ parallel to $AB$ intersects at $Y$ , prove that $X,Y,Z$ are collinear. | [
"Same problem as [Iranian Geometry Olympiad 2015 Advanced P3](https://artofproblemsolving.com/community/c6h1137975p5325937)"
] | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735507.json"
} | |
Prove that there exists a set $X \subseteq \mathbb{N}$ which contains exactly 2022 elements such that for every distinct $a, b, c \in X$ the following equality:
\[ \gcd(a^n+b^n, c) = 1 \] is satisfied for every positive integer $n$ . | Let $a_1 = 21$ and inductively add new elements by the following procedure, suppose we have until $a_k$ . We're going to always have $a_i$ to have two new $3 \pmod 4$ primes dividing it. So let the primes so far be $p_1, p_2, \cdots, p_{2k}$ . Construct two new primes $p,q$ such that both satisfy $\left(\frac{p_i}{p} \right) = 1$ as well as $3 \pmod 4$ for all $1 \le i \le 2k$ which is possible by Dirichlet. Note that by quadratic reciprocity, we have $\left(\frac{p}{p_i} \right) = -1$ too. I claim this works.
Suppose that for some prime $p_k$ , we have $p_k | a^n + b^n \implies \left(\frac{a}{b} \right)^n \equiv -1 \pmod {p_k}$ . But by the way we have constructed the numbers, both $a,b$ are QR's $\pmod {p_k}$ (since each has $2$ primes dividing them which are both QR's or NQR's). Therefore, $\frac{a}{b}$ is a QR as well, so we have $z^2 \equiv -1 \pmod {p_k}$ . But $p_k \equiv 3 \pmod 4$ so this is impossible. So this indeed works. $\blacksquare$ | [
"<details><summary>Solution</summary><details><summary>FLT On Steroids</summary>Let $p_1, p_2, \\cdots, p_{2022}$ be the smallest odd primes and let \\[ N = \\prod_{i=1}^{2022} (p_i - 1). \\] Then, the set $X = \\{ p_1^N, p_2^N, \\ldots, p_{2022}^N\\}$ works. Since $p_i - 1 \\vert N$ for every such $i$ , by ... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 138,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735508.json"
} |
Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that $$ (a + b + c)(ab + bc + ca) + 3\ge 4(a + b + c). $$ | <blockquote>Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that $$ (a + b + c)(ab + bc + ca) + 3\ge 4(a + b + c). $$ </blockquote>
pqr method: $p = a + b + c, q = ab + bc + ca, r = abc = 1$ .
We need to prove that $pq + 3 \ge 4p$ .
Using $q^2\ge 3pr$ , we have $q\ge \sqrt{3pr} = \sqrt{3p}$ , and it suffices to prove that $p\sqrt{3p} + 3 \ge 4p$ .
Since $p\ge 3$ , letting $p = 3u^2$ ( $u\ge 1$ ), it suffices to prove that $9u^3 - 12u^2 + 3 \ge 0$ or $3(u - 1)(3u^2 - u - 1) \ge 0$ which is true. We are done. | [
"deleted ....",
"<blockquote>**Rewrite as** $(a+b+c)(ab+bc+ca-4)>=-3$ AM-GM $(a+b+c)>=3$ AM-HM $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}>=3$ $(ab+bc+ca)>=3$ </blockquote>\nPlease, stop writing nonsense. You can not use $a+b+c\\ge 3$ when $ab+bc+ca-4$ is negative. \n",
"<blockquote><blockquote>**Rewrite as** ... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2742895.json"
} |
Distinct pebbles are placed on a $1001 \times 1001$ board consisting of $1001^2$ unit tiles, such that every unit tile consists of at most one pebble. The *pebble set* of a unit tile is the set of all pebbles situated in the same row or column with said unit tile. Determine the minimum amount of pebbles that must be placed on the board so that no two distinct tiles have the same *pebble set*.
<details><summary>Where's the Algebra Problem?</summary>It's already posted [here](https://artofproblemsolving.com/community/c6h2742895_simple_inequality).</details> | [
"silimar with [2021Benelux](https://artofproblemsolving.com/community/c6h2549153p21775820)"
] | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2743542.json"
} | |
Given that $ABC$ is a triangle, points $A_i, B_i, C_i \hspace{0.15cm} (i \in \{1,2,3\})$ and $O_A, O_B, O_C$ satisfy the following criteria:
a) $ABB_1A_2, BCC_1B_2, CAA_1C_2$ are rectangles not containing any interior points of the triangle $ABC$ ,
b) $\displaystyle \frac{AB}{BB_1} = \frac{BC}{CC_1} = \frac{CA}{AA_1}$ ,
c) $AA_1A_3A_2, BB_1B_3B_2, CC_1C_3C_2$ are parallelograms, and
d) $O_A$ is the centroid of rectangle $BCC_1B_2$ , $O_B$ is the centroid of rectangle $CAA_1C_2$ , and $O_C$ is the centroid of rectangle $ABB_1A_2$ .
Prove that $A_3O_A, B_3O_B,$ and $C_3O_C$ concur at a point.
*Proposed by Farras Mohammad Hibban Faddila* | The number of participants that got this one correct during the test is lower than expected...
I think this is because of the format of the test, which consists of all the four topics ACGN, and is not ordered by difficulty.
Anyway here are some solutions.
<details><summary>Solution 1 (Finding the concurrency point explicitly)</summary>[asy]
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(130), B = dir(210), C = dir(-30);
pair A2 = rotate(-90, A) * ((B + 2 * A)/3);
pair B1 = rotate(90, B) * ((A + 2 * B)/3);
pair B2 = rotate(-90, B) * ((C + 2 * B)/3);
pair C1 = rotate(90, C) * ((B + 2 * C)/3);
pair C2 = rotate(-90, C) * ((A + 2 * C)/3);
pair A1 = rotate(90, A) * ((C + 2 * A)/3);
pair A3 = A1 + A2 - A, B3 = B1 + B2 - B, C3 = C1 + C2 - C;
pair Oa = (B + C1)/2, Ob = (C + A1)/2, Oc = (A + B1)/2;
pair H = orthocenter(A, B, C);
pair O = circumcenter(A, B, C);
pair G = (A+B+C)/3;
pair Ma = (B + C)/2;
draw(A--A2--B1--B--cycle^^A2--A3^^B1--B3, blue+linewidth(1));
draw(B--B2--C1--C--cycle^^B2--B3^^C1--C3, deepcyan+linewidth(1));
draw(C--C2--A1--A--cycle^^C2--C3^^A1--A3, heavymagenta+linewidth(1));
draw(H--O, orange+linewidth(1));
draw(A3--Oa, black+dashed);
draw(O--Oa^^A3--H, orange+linewidth(1));
string[] names = {" $A$ ", " $B$ ", " $C$ ", " $A_1$ ", " $A_2$ ", " $B_1$ ", " $B_2$ ", " $C_1$ ", " $C_2$ ",
" $A_3$ ", " $B_3$ ", " $C_3$ ", " $O_A$ ", " $O_B$ ", " $O_C$ ", " $H$ ", " $O$ ", " $G$ ", " $M_A$ "
};
pair[] points = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc, H, O, G, Ma};
pair[] ll = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc, H, O, G, Ma};
for (int i=0; i<names.length; ++i)
dot(names[i], points[i], dir(ll[i]));
[/asy]
We claim the desired concurrency point is point $G$ , the centroid of $\bigtriangleup ABC$ . It suffices to show that $A_3O_A$ passes through $G$ , and the rest will be similar.
Let $M_A$ be the midpoint of $BC$ . It is well-known that $AH=2OM_A$ .
Since $\angle AA_2A_3 = 180^{\circ} - \angle A_2AA_1 = 180^{\circ} - (360^{\circ} - 90^{\circ} - 90^{\circ} - \angle A) = \angle A$ , and $\frac{AA_2}{A_2A_3} = \frac{AA_2}{AA_1} = \frac{AB}{AC}$ , we have $\bigtriangleup AA_2A_3 \sim \bigtriangleup BAC$ (and denote $k$ as their similarity ratio, i.e. $k = \frac{AA_2}{AB}$ ). Therefore, $$ \angle A_3AA_2 + \angle A_2AB + \angle BAH = \angle B + 90^{\circ} + (90^{\circ} - \angle B) = 180^{\circ}, $$ $\rightarrow A_3, A, H$ is collinear, and $AA_3 = k \cdot BC = BB_2 = 2 * M_AO_A$ . Therefore, $A_3H = 2 * OO_A$ , so $A_3OA$ meets $OH$ at $G$ , as desired. $\blacksquare$</details>
<details><summary>Solution 2 (Some collinearity observation)</summary>[asy]
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(130), B = dir(210), C = dir(-30);
pair A2 = rotate(-90, A) * ((B + 2 * A)/3);
pair B1 = rotate(90, B) * ((A + 2 * B)/3);
pair B2 = rotate(-90, B) * ((C + 2 * B)/3);
pair C1 = rotate(90, C) * ((B + 2 * C)/3);
pair C2 = rotate(-90, C) * ((A + 2 * C)/3);
pair A1 = rotate(90, A) * ((C + 2 * A)/3);
pair A3 = A1 + A2 - A, B3 = B1 + B2 - B, C3 = C1 + C2 - C;
pair Oa = (B + C1)/2, Ob = (C + A1)/2, Oc = (A + B1)/2;
draw(A--A2--B1--B--cycle^^A2--A3^^B1--B3, blue+linewidth(1));
draw(B--B2--C1--C--cycle^^B2--B3^^C1--C3, deepcyan+linewidth(1));
draw(C--C2--A1--A--cycle^^C2--C3^^A1--A3, heavymagenta+linewidth(1));
draw(A3--Oa^^B3--Ob^^C3--Oc, grey+dashed);
draw(A3--B3--C3--cycle, black+dashed);
string[] names = {" $A$ ", " $B$ ", " $C$ ", " $A_1$ ", " $A_2$ ", " $B_1$ ", " $B_2$ ", " $C_1$ ", " $C_2$ ",
" $A_3$ ", " $B_3$ ", " $C_3$ ", " $O_A$ ", " $O_B$ ", " $O_C$ "
};
pair[] points = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc};
pair[] ll = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc};
for (int i=0; i<names.length; ++i)
dot(names[i], points[i], dir(ll[i]));
[/asy]
The claim here is that $B_3, O_A,$ and $C_3$ are collinear so that the three lines will concur at the centroid of $\bigtriangleup A_3B_3C_3$ .
One way to prove this is by noting that $BB_3C_1C_3$ is a parallelogram, since both $C_1C_3$ and $BB_3$ are perpendicular to $AC$ (proving $BB_3 \perp AC$ is already discussed in the first solution above), and $BB_3 = C_1C_3 = k \cdot AC$ , where $k = \frac{AC}{CC_2}$ . Therefore, $O_A$ is the midpoint of $B_3C_3$ as well, as desired. $\blacksquare$</details>
<details><summary>Solution 3 (Homothety and a generalization)</summary>[asy]
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(130), B = dir(210), C = dir(-30);
pair A2 = rotate(-90, A) * ((B + A)/2);
pair B1 = rotate(90, B) * ((A + B)/2);
pair B2 = rotate(-90, B) * ((C + 2 * B)/3);
pair C1 = rotate(90, C) * ((B + 2 * C)/3);
pair C2 = rotate(-90, C) * ((A + 7 * C)/8);
pair A1 = rotate(90, A) * ((C + 7 * A)/8);
pair A3 = A1 + A2 - A, B3 = B1 + B2 - B, C3 = C1 + C2 - C;
pair Oa = (B + C1)/2, Ob = (C + A1)/2, Oc = (A + B1)/2;
draw(A--A2--B1--B--cycle^^A2--A3^^B1--B3, blue+linewidth(1));
draw(B--B2--C1--C--cycle^^B2--B3^^C1--C3, deepcyan+linewidth(1));
draw(C--C2--A1--A--cycle^^C2--C3^^A1--A3, heavymagenta+linewidth(1));
draw(A--B1^^A--C2^^Ob--Oc^^C2--B1^^C3--B3, black+dashed);
string[] names = {" $A$ ", " $B$ ", " $C$ ", " $A_1$ ", " $A_2$ ", " $B_1$ ", " $B_2$ ", " $C_1$ ", " $C_2$ ",
" $A_3$ ", " $B_3$ ", " $C_3$ ", " $O_A$ ", " $O_B$ ", " $O_C$ "
};
pair[] points = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc};
pair[] ll = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc};
for (int i=0; i<names.length; ++i)
dot(names[i], points[i], dir(ll[i]));
[/asy]
Using this solution, we can generalize the problem by removing condition (b). Here we will show $\bigtriangleup A_3B_3C_3$ and $\bigtriangleup O_AO_BO_C$ are homothetic, so the three lines concur at the centre of the homothety.
FIrst, $O_CO_B || B_1C_2$ since $O_CO_B$ is a midline of $\bigtriangleup AB_1C_2$ .
Second, it's easy to see that $B_1C_2C_3B_3$ is a parallelogram. So, $B_1C_2 || B_3C_3$ .
Therefore, $O_CO_B || B_3C_3$ . Similarly, $O_AO_B || A_3B_3$ and $O_AO_C || A_3C_3$ . Therefore the two triangles $\bigtriangleup A_3B_3C_3$ and $\bigtriangleup O_AO_BO_C$ are homothetic as desired. $\blacksquare$</details> | [
"Let $M_A,M_B,M_C$ be the midpoints of $BC,CA,AB$ respectively, let $G$ be the centroid of $ABC$ , let $K$ be the symmedian point of $ABC$ , and $K_A,K_B,K_C$ its projections on $BC,CA,AB$ .\nSince the barycentric coordinates of $K$ are $(a^2:b^2:c^2)$ it follows that $|\\vec{KK_A}|:BC=|\\vec{KK_B... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 292,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2743549.json"
} |
Let $n$ be a natural number, with the prime factorisation
\[ n = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r} \] where $p_1, \ldots, p_r$ are distinct primes, and $e_i$ is a natural number. Define
\[ rad(n) = p_1p_2 \cdots p_r \] to be the product of all distinct prime factors of $n$ . Determine all polynomials $P(x)$ with rational coefficients such that there exists infinitely many naturals $n$ satisfying $P(n) = rad(n)$ . | Since $0<rad (n) \leq n$ for all $n\in \mathbb{Z}$ , $0<P(n)\leq n$ for infinitely many $n$ . This leads to $\lim_{n\to \infty}\frac{P(n)}{n} \leq 1$ and hence $\deg P(x)\leq 1$ .**Case 1:** $P(x)\equiv c \in \mathbb{Q}$ , it's easy to see that $c$ is square-free is all we need.**Case 2:** $\deg P(x)=1$ , let $P(x)=\frac{ax+b}{c}$ for some $a,b,c \in \mathbb{Z}$ and $ a,c>0$ . For $b\ne 0$ Since $P(n)=rad(n)$ for infinitely many $n$ , $rad(n) | b$ for infinitely many $n$ . Thus, for infinitely many $n$ , $an=c.rad(n)-b\leq c.|b|+|b|$ , a contradiction! $P(x)$ must be of the form $\frac{ax}{c}$ . For infinitely many $n$ , $an=c.rad(n)$ or $c=a\frac{n}{rad(n)}$ so $a|c$ and we can consider $a$ as $1$ , $P(x)=\frac{x}{c}$ . For every positive integer $c=p_1^{\alpha_1}\dots p_m^{\alpha_m}$ , and consider: $$ S:=\{c.p: p \;\text{runs in the set of all prime number which is not}\; p_1,\dots,p_m\} $$ $S$ is a infinite set and $P(n)=rad(n)$ for all $n\in S$ which means $P(x)=\frac{x}{c}, c\in \mathbb{N}$ is all polynomials we need to find. | [
"When you have zero effort in making test problems. \n<blockquote>\nLet $n$ be a positive integer, with prime factorization $$ n = p_1^{e_1}p_2^{e_2} \\cdots p_r^{e_r} $$ for distinct primes $p_1, \\ldots, p_r$ and $e_i$ positive integers. Define $$ rad(n) = p_1p_2\\cdots p_r, $$ the product of all distin... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 70,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2743550.json"
} |
Let $a$ and $b$ be two positive reals such that the following inequality
\[ ax^3 + by^2 \geq xy - 1 \] is satisfied for any positive reals $x, y \geq 1$ . Determine the smallest possible value of $a^2 + b$ .
*Proposed by Fajar Yuliawan* | <blockquote> Let $a$ and $b$ be two positive reals such that the inequality
\[ ax^3 + by^2 \geq xy - 1 \] holds for any positive reals $x, y \geq 1$ . Determine the smallest possible value of $a^2 + b$ .
</blockquote>
---------------------
The answer is $\boxed{\frac{2}{3\sqrt{3}}}$ , which is achieved by $(a,b) = \left( \frac{1}{\sqrt[4]{108}}, \frac{1}{2\sqrt{3}} \right)$ . To show that this holds, note that we have $a^2 b^3 = \frac{1}{432}$ . Therefore, for any $x, y \ge 1$ , we have
\begin{align*}
ax^3 + by^2 &= 2 \cdot \frac{ax^3}{2} + 3 \cdot \frac{by^2}{3} + 1 - 1
&\ge 6 \sqrt[6]{\frac{a^2b^3 x^6 y^6}{2^2 3^3}} - 1
&\ge 6xy \sqrt[6]{\frac{a^2 b^3}{2^2 3^3}} - 1
&= xy - 1
\end{align*}
It suffices to show that $\frac{2}{3\sqrt{3}}$ is indeed the minimum.
Before that, call $(a,b)$ *cool* if $(a,b)$ satisfies the statement given above.
We will present two solutions, where the first solution is more natural, by investigating the behavior of the polynomial, without knowing what the answer might be and the second solution is heavily dependent with the equality case in mind. **<span style="color:#00f">Solution 1. (Analyze polynomial)</span>**
Let us write $P(x,y) = ax^3 + by^2 - xy + 1$ . We know that $P(x,y) \ge 0$ for all $x,y \ge 1$ .
The main idea is to investigate the behavior of the multivariable polynomial $P(x,y)$ as a quadratic polynomial in $y$ .**<span style="color:#f00">Claim 01.</span>** Fix $x = \ell \ge 1$ . We have $Q_{\ell}(y) = P(\ell,y) \ge 0$ for all $y \in \mathbb{R} \setminus (0,1)$ .
*Proof.* Indeed, note that if $y \le 0$ , we have
\[ P(\ell,y) = a\ell^3 + by^2 - \ell y + 1 \ge a\ell ^3 + by^2 + 1 > 0 \]
and if $y \ge 1$ , we have $P(\ell,y) \ge 0$ from the condition of the problem.
-----------
Now, we consider two cases:
- Suppose that there exists $\ell$ such that $Q_{\ell}(y)$ has two distinct real roots, which should be on interval $(0,1]$ by Claim 01. This implies
\[ \frac{\ell}{2b} < \frac{\ell + \sqrt{\ell^2 - 4b(a \ell^3 + 1)}}{2b} \le 1 \]
This means that $1 \le \ell < 2b \implies b > \frac{1}{2}$ .
Therefore, we must have $a^2 + b > \frac{1}{2} > \frac{2}{3\sqrt{3}}$ .
- Else, $Q_{\ell}(y) \ge 0$ for all $y \in \mathbb{R}$ and $\ell \ge 1$ .This implies that $Q_{\ell}(y)$ has at most one real root, i.e. $Q_{\ell}(y)$ has non-positive discriminant. We will then get
\[ 4b(a x^3 + 1) \ge x^2, \ \forall x \ge 1 \]
Now, call $(a,b)$ nice if $4b(ax^3 + 1) \ge x^2$ for all $x \ge 1$ . We will prove that nice implies cool. Indeed, if $(a,b)$ is nice, then
\[ ax^3 + by^2 = (ax^3 + 1) + by^2 - 1 \ge \frac{x^2}{4b} + by^2 - 1 \ge 2 \sqrt{\frac{x^2}{4b} \cdot by^2} - 1 = xy - 1 \]
which implies that $(a,b)$ is cool. It suffices to minimize $a^2 + b$ if $(a,b)$ is nice.
Let us denote $f(x) = 4abx^3 - x^2 + 4b$ . We could see that $(a,b)$ nice is essentially saying that $\min_{x \ge 1} f(x) \ge 0$ .
Now, note that $f(x) = 4ab x^3 - x^2 + 4b$ has two critical points ( $f'(x) = 0$ ), i.e. when $x = 0$ and $x = \frac{1}{6ab}$ .
[list]
- Suppose that $\frac{1}{6ab} < 1$ , then we have
\[ a^2 + b > a^2 + \frac{1}{6a} = a^2 + 2 \cdot \frac{1}{12a} \ge 3 \sqrt[3]{a^2 \cdot \frac{1}{144a^2}} = \sqrt[3]{\frac{3}{16}} > \frac{2}{3\sqrt{3}} \]
- Else, $\frac{1}{6ab} \ge 1$ , which means $\min_{x \ge 1} f(x) = f \left( \frac{1}{6ab} \right)$ . Therefore, we have
\[ f \left( \frac{1}{6ab} \right) \ge 0 \implies a^2 b^3 \ge \frac{1}{432} \]
This gives us
\[ a^2 + b = a^2 + 3 \cdot \frac{b}{3} \ge 4 \sqrt[4]{\frac{a^2 b^3}{27}} \ge \frac{2}{3\sqrt{3}} \]
Equality holds if and only if $a^2 = \frac{b}{3}$ and $a^2 b^3 = \frac{1}{432}$ . Solving which gives us $b = \frac{1}{2\sqrt{3}}$ and $a = \frac{1}{\sqrt[4]{108}}$ , which we have shown to work at the beginning of the solution.
[/list]
[/list]
----------------------------**<span style="color:#00f">Solution 2. (Tricky Manipulation)</span>**
We consider two possible cases:
[list]
- Suppose $\max(a,b) \ge 1$ , then we have $a^2 + b > 1 > \frac{2}{3\sqrt{3}}$ .
- Else, we must have $a, b < 1$ . Take $x^3 = \frac{2}{a}$ and $y^2 = \frac{3}{b}$ . Clearly, $x,y > 1$ . Note that
\[ 6^6 = (ax^3 + by^2 + 1)^6 \ge (xy)^6 = \frac{2^2 \cdot 3^3}{a^2 \cdot b^3} \]
This implies $a^2 b^3 \ge \frac{1}{432}$ . Therefore, we have
\[ a^2 + b = a^2 + 3 \cdot \frac{b}{3} \ge 4 \sqrt[4]{\frac{a^2 b^3}{27}} \ge \frac{2}{3\sqrt{3}} \]
<details><summary>Short Note on Motivation for the second solution</summary>The hardest part of this solution is to find the equality case. We want $a,b$ such that $ax^3 + by^2 + 1 \ge xy$ . One pair of $(a,b)$ could be easily found by assuming equality holds for applying AM-GM the following way, i.e. by carefully choosing degree, we want something like AM-GM on
\[ \frac{ax^3}{2} + \frac{ax^3}{2} + \frac{by^2}{3} + \frac{by^2}{3} + \frac{by^2}{3} + 1 \]
because the product of these $6$ terms is of the form $cx^6 y^6$ for some constant $c$ , which is "good" for us. Therefore, finding $a,b$ such that $a^2 b^3 \cdot 2^2 3^3 = 6^6$ should gives us what we want. Now, this motivates the rest part of the solution since minimizes $a^2 + b$ if $a^2 b^3$ is known is an easy task.</details> | [
"Here's an elementary solution using calculus.\n\nDefine $f(x,y)=ax^3+by^2-xy+1$ . Taking partial derivatives, this is stationary at $3ax^2-y=0$ and $2by-x=0$ , that is at $f(0,0)=1$ and $f\\left(\\tfrac{1}{6ab},\\tfrac{1}{12ab^2}\\right)=1-\\tfrac{1}{432a^2b^3}$ . The latter is clearly the minimum, and si... | [
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} |
A $3 \times 3 \times 4$ cuboid is constructed out of 36 white-coloured unit cubes. Then, all six of the cuboid's sides are coloured red. After that, the cuboid is dismantled into its constituent unit cubes. Then, randomly, all said unit cubes are constructed into the cuboid of its original size (and position).
a) How many ways are there to position eight of its corner cubes so that the apparent sides of eight corner cubes are still red? (Cube rotations are still considered distinct configurations, and the position of the cuboid remains unchanged.)
b) Determine the probability that after the reconstruction, all of its apparent sides are still red-coloured. (The cuboid is still upright, with the same dimensions as the original cuboid, without rotation.)
<details><summary>Notes</summary>The problem might have multiple interpretations. We agreed that this problem's wording was a bit ambiguous. Here's the original Indonesian version:
Suatu balok berukuran $3 \times 3 \times 4$ tersusun dari 36 kubus satuan berwarna putih. Kemudian keenam permukaan balok diwarnai merah. Setelah itu, balok yang tersusun dari kubus-kubus satuan tersebut dibongkar. Kemudian, secara acak, semua kubus satuan disusun lagi menjadi balok seperti balok semula.
a) Ada berapa cara menempatkan kedelapan kubus satuan yang berasal dari pojok sehingga kedelapan kubus di pojok yang tampak tetap berwarna merah? (Rotasi kubus dianggap konfigurasi yang berbeda, namun posisi balok tidak diubah.)
b) Tentukan probabilitas balok yang tersusun lagi semua permukaannya berwarna merah. (Balok tegak tetap tegak dan balok tetap dalam suatu posisi.)</details> | a) I think the answer is $3^88!$ although I'm not so sure
b)The corner cubes must permute, and so also the edge cubes and the face cubes. So we have a probability of $$ \frac{3^8\cdot 8!2^{16}16!4^{10}10!24^{2}2!}{24^{36}\cdot 36!} $$ by considering the possible permutations of the four classes (corner, edge, face, internal) and their possible rotations which preserve the red sides. | [] | [
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Let $AB$ be the diameter of circle $\Gamma$ centred at $O$ . Point $C$ lies on ray $\overrightarrow{AB}$ . The line through $C$ cuts circle $\Gamma$ at $D$ and $E$ , with point $D$ being closer to $C$ than $E$ is. $OF$ is the diameter of the circumcircle of triangle $BOD$ . Next, construct $CF$ , cutting the circumcircle of triangle $BOD$ at $G$ . Prove that $O,A,E,G$ are concyclic.
(Possibly proposed by Pak Wono) | [
"Chinese Western Mathematical Olympiad 2006, Problem 6\nhttps://artofproblemsolving.com/community/c6h118089p670265"
] | [
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} | |
Given positive odd integers $m$ and $n$ where the set of all prime factors of $m$ is the same as the set of all prime factors $n$ , and $n \vert m$ . Let $a$ be an arbitrary integer which is relatively prime to $m$ and $n$ . Prove that:
\[ o_m(a) = o_n(a) \times \frac{m}{\gcd(m, a^{o_n(a)}-1)} \] where $o_k(a)$ denotes the smallest positive integer such that $a^{o_k(a)} \equiv 1$ (mod $k$ ) holds for some natural number $k > 1$ . | <blockquote>Given positive odd integers $m$ and $n$ where $\text{rad}(m) = \text{rad}(n)$ , and $n \vert m$ . Let $a$ be an arbitrary integer which is relatively prime to $m$ and $n$ . Prove that:
\[ o_m(a) = o_n(a) \times \frac{m}{\gcd(m, a^{o_n(a)}-1)} \] where $o_k(a)$ denotes the smallest positive integer such that $a^{o_k(a)} \equiv 1$ (mod $k$ ) holds for some natural number $k > 1$ .</blockquote>**<span style="color:#f00">Claim 01.</span>** $o_n(a) \mid o_m(a)$ *Proof.* Note that $n \mid m \mid a^{o_m(a)} - 1 \implies o_n(a) \mid o_m(a)$ .
--------**<span style="color:#f00">Claim 02.</span>** $\frac{m}{\gcd(m,a^{o_n(a)} - 1)} \mid \frac{o_m(a)}{o_n(a)}$ .
*Proof.* It suffices to prove this $\nu_p(m) - \min \{ \nu_p(m) , \nu_p(a^{o_n(a)} - 1) \} \le \nu_p(o_m(a)) - \nu_p(o_n(a))$ for every prime $p \mid n$ .
There are two cases:
- If $\nu_p(m) \le \nu_p(a^{o_n(a)} - 1)$ , then it suffices to prove that $\nu_p(o_m(a)) - \nu_p(o_n(a)) \ge 0$ , which must be true as $o_n(a) \mid o_m(a)$ .
- If $\nu_p(m) > \nu_p(a^{o_n(a)} - 1)$ , then as $p$ is odd, by LTE,
\begin{align*}
\nu_p \left( \frac{o_m(a)}{o_n(a)} \right) &= \nu_p(a^{o_n(a) \cdot \frac{o_m(a)}{o_n(a)}} - 1) - \nu_p(a^{o_n(a)} - 1)
&= \nu_p(a^{o_m(a)} - 1) - \nu_p(a^{o_n(a)} - 1)
&\stackrel{m \mid a^{o_m(a)} - 1}{\ge} \nu_p(m) - \nu_p(a^{o_n(a)} - 1)
\end{align*}
which is what we wanted.
**<span style="color:#f00">Claim 03.</span>** $o_m(a) \mid o_n(a) \cdot \frac{m}{\gcd(m,a^{o_n(a)} - 1)}$ .
*Proof.* We aim to prove that $m \mid a^{o_n(a) \cdot \frac{m}{\gcd(m,a^{o_n(a)} - 1)}} - 1$ . Therefore, for all prime $p \mid m$ , as $p$ must be odd, by LTE, we have
\begin{align*}
\nu_p( a^{o_n(a) \cdot \frac{m}{\gcd(m,a^{o_n(a)} - 1)}} - 1)&= \nu_p(a^{o_n(a)} - 1) + \nu_p \left( \frac{m}{\gcd(m,a^{o_n(a)} - 1)} \right)
&= \nu_p(m) + \nu_p \left( \frac{a^{o_n(a) } - 1}{\gcd(m,a^{o_n(a)} - 1)} \right) \ge \nu_p(m)
\end{align*}
This gives us
\[ m \mid a^{o_n(a) \cdot \frac{m}{\gcd(m,a^{o_n(a)} -1}}-1 \implies o_m(a) \mid o_n(a) \cdot \frac{m}{\gcd(m,a^{o_n(a)} - 1)} \]
which is what we wanted.
------------
Combining both lemma above, we conclude that
\[ o_m(a) = o_n(a) \cdot \frac{m}{\gcd(m,a^{o_n(a)} - 1)} \]**<span style="color:#f00">Remark.</span>** A standard idea in proving $A = B$ in number theory is to prove that $A \mid B$ and $B \mid A$ . Now, use the property of order and LTE to sort everything out. | [
"Let $p$ be a prime dividing $n$ . Note that $o_n(a)|o_m(a)$ and $v_p(a^{ko_n(a)}-1)=v_p(a^{o_n(a)}-1)+v_p(k)$ .\nTherefore if $v_p(a^{o_n(a)}-1)\\geq v_p(m)$ we don't need any factors of $p$ in $k$ but otherwise we must have $p^{v_p(m)-v_p(a^{o_n(m)}-1)}=p^{v_p(\\frac{m}{\\gcd(m,a^{o_n(a)}-1)})}$ The... | [
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For each natural number $n$ , let $f(n)$ denote the number of ordered integer pairs $(x,y)$ satisfying the following equation:
\[ x^2 - xy + y^2 = n. \]
a) Determine $f(2022)$ .
b) Determine the largest natural number $m$ such that $m$ divides $f(n)$ for every natural number $n$ . | [
"Gosh, this is just Turkey 2000 TST just change into 2022:\nhttps://artofproblemsolving.com/community/q2h396239p2203313"
] | [
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In a nonisosceles triangle $ABC$ , point $I$ is its incentre and $\Gamma$ is its circumcircle. Points $E$ and $D$ lie on $\Gamma$ and the circumcircle of triangle $BIC$ respectively such that $AE$ and $ID$ are both perpendicular to $BC$ . Let $M$ be the midpoint of $BC$ , $N$ be the midpoint of arc $BC$ on $\Gamma$ containing $A$ , $F$ is the point of tangency of the $A-$ excircle on $BC$ , and $G$ is the intersection of line $DE$ with $\Gamma$ . Prove that lines $GM$ and $NF$ intersect at a point located on $\Gamma$ .
(Possibly proposed by Farras Faddila) | <details><summary>Solution</summary>[asy]
size(10cm);
defaultpen(fontsize(9pt));
pair A = dir(130), B = dir(215), C = dir(-35), O = circumcenter(A, B, C);
pair A1 = -A;
pair H = orthocenter(A, B, C);
pair E = foot(A, B, C) * 2 - H;
pair I = incenter(A, B, C), D1 = foot(I, B, C), N = dir(90), P = dir(270), E1 = foot(I, C, A), F1 = foot(I, A, B), X = extension(E1, F1, B, C);
pair G = extension(N, X, P, D1), D = extension(I, D1, G, E), M = (B + C)/2;
pair F = 2 * M - D1;
pair XXX = extension(G, M, N, F);
pair Ia = 2 * P - I;
pair EE = extension(P,E, B,C);
draw(A--B--C--cycle, red+linewidth(1));
draw(circumcircle(A, B, C), heavymagenta);
draw(circumcircle(B, C, I), blue);
draw(incircle(A, B, C), gray);
draw(A--E^^N--P^^I--D, heavycyan);
draw(G--XXX^^N--XXX^^G--P^^E--A1^^D--Ia, gray+dashed);
draw(D--G^^A--A1, orange);
draw(P--A1^^A--Ia, magenta);
markscalefactor=0.008;
draw(rightanglemark(A1, P, Ia));
draw(P--EE^^B--EE, grey);
draw(circumcircle(D1, P, D), green);
string[] names = {" $A$ ", " $B$ ", " $C$ ", " $D$ ", " $E$ ", " $F$ ", " $I$ ", " $G$ ", " $M$ ", " $N$ ", " $P$ ", " $Q$ ", " $A_1$ ", " $O$ ", " $I_A$ ", " $E'$ ", " $X$ "};
pair[] points = {A, B, C, D, E, F, I, G, M, N, P, D1, A1, O, Ia, EE, XXX};
pair[] ll = {A, B, C, D, E, F, dir(315), G, dir(225), N, P, D1, A1, dir(180), Ia, EE, XXX};
int pt = names.length;
for (int i=0; i<pt; ++i)
dot(names*, points[i], dir(ll[i]));
[/asy]
Let $Q$ be the touchpoint of the incircle with $BC$ , $P$ be the midpoint of arc $BC$ opposite of $N$ , $A_1$ be the antipode of $A$ on $\odot(ABC)$ , and $I_A$ be the $A$ -excenter of $\bigtriangleup ABC$ . We prove the following claim:
<span style="color:#f00">**Claim**</span> $G, Q, P$ are collinear.
[i]Proof.* First we will prove $\measuredangle DPE = 90^{\circ}$ . This can be proven by simple angle-chasing, but one simple way here is by using $\phi$ , the reflection map across $PN$ . Note that $\phi(E) = A_1$ and $\phi(D) = I_A$ . Since $A, P, I_A$ are collinear, therefore $\measuredangle DPE = \measuredangle I_APA_1 = \measuredangle APA_1 = 90^{\circ}$ . Then use inversion with centre $P$ and radius $PB$ . $\Gamma$ will be mapped to line $BC$ , $E$ will be mapped to a point $E' = PE \cap BC \in BC$ , and $G$ will be mapped to $G'$ , which is the intersection of $BC$ and $\odot(DPE')$ . Since $\measuredangle DPE' = \measuredangle DPE = 90^{\circ}$ , therefore $\measuredangle DG'E' = \measuredangle DPE' = 90^{\circ}$ as well. So, $G'$ is the projection of $D$ to $BC$ , which means $G' = Q$ . Therefore, $P, Q, G$ are collinear. $\square$ Let $GM \cap \Gamma = X$ , and $NX \cap BC = F_1$ . We have, by *Butterfly Theorem*, $MQ = MF_1$ , meaning that $F_1$ is the touchpoint of $A$ -excircle on $BC$ . Therefore, $F_1 = F$ , and $GM \cap NF = X \in \Gamma$ , as desired. $\blacksquare$ .</details>
<details><summary>Remark</summary>$G$ is a point with a lot of properties (intersection of $\odot(ABC)$ and circle with diameter $AI$ ) and has already appeared in many other problems.</details> | [
"I like this problem :D**<span style=\"color:#f00\">Claim 1:</span>** $G$ is the A-sharkydevil point.**<span style=\"color:#f00\">Proof:</span>** Let $F'$ the touch point of the incircle of $\\triangle ABC$ with $BC$ and let $N'$ the midpoint of the minor arc $BC$ of $\\Gamma$ using I-E lemma we do an... | [
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Let $A$ be a subset of $\{1,2,\ldots,2020\}$ such that the difference of any two distinct elements in $A$ is not prime. Determine the maximum number of elements in set $A$ . | I think the answer is $\boxed{505}$ . We can form $A$ as: $A=\{1,5,9,...,2017\}$ , $A=\{2,6,10,...,2018\}$ $A=\{3,7,11,...,2019\}$ and $A=\{4,8,12,...,2020\}$ Assume that $|A|>505$ . We will seperate $\{1,2,...,2020\}$ into 505 subsets which form as $\{4k+1,4k+2,4k+3,4k+4\}, k=0,1,2,...,504$ . Because $|A|>505$ , so by Dirichlet, there are at least 2 number would be in one of 505 subsets. WLOG, call it $4m+a$ and $4m+b$ . Because $1\leq |a-b| \leq 3$ so the difference between 2 numbers would be prime, contradicts. | [
"any solution?\n",
"How about $|a - b| = 1$ ? @above",
"If I'm not mistake at most $5$ of $20$ consecutive numbers can be element of $A$ , which $\\frac{1}{4}$ of $20$ . So the result follows immediately. (Answer is $505$ )",
"The answer is $505$ , first order the set and let $n$ be the size of ... | [
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} |
Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
\[ f(a^2) - f(b^2) \leq (f(a)+b)(a-f(b)) \] for all $a,b \in \mathbb{R}$ . | Let $P(a,b)$ be the given assertion. $P(0,0)\Rightarrow f(0)^2\le0\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(x^2)\le xf(x)$ $P(0,x)\Rightarrow-f(x^2)\le-xf(x)\Rightarrow f(x^2)\ge xf(x)$ So $f(x^2)=xf(x)$ , hence $f$ is odd. Now $P(x,y)$ gives the new assertion $Q(x,y):f(x)f(y)\le xy$ . $Q(x,-y)\Rightarrow-f(x)f(y)\le-xy\Rightarrow f(x)f(y)\ge xy$ So equality holds in $Q(x,y)$ , so $f(x)=\frac1{f(1)}x$ after setting $y=1$ . Then $x=1$ yields the solutions $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ , which both work. | [
"[MEMO 2021 T-1](https://artofproblemsolving.com/community/c6h2665236p23091906)",
"Let $P(a,b)$ denote the assertion for this functional inequality. $P(0,0)\\implies (f(0))^2\\leq 0$ and hence, $f(0)=0$ .\nNow, $P(a,a)\\implies (f(a))^2\\leq a^2$ . $P(0,a)$ and $P(a,0)$ gives $a\\cdot f(a)=f(a^2)$ which... | [
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"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2744245.json"
} |
Non-zero polynomials $P(x)$ , $Q(x)$ , and $R(x)$ with real coefficients satisfy the identities $$ P(x) + Q(x) + R(x) = P(Q(x)) + Q(R(x)) + R(P(x)) = 0. $$ Prove that the degrees of the three polynomials are all even. | <blockquote>Non-zero polynomials $P(x)$ , $Q(x)$ , and $R(x)$ with real coefficients satisfy the identities $$ P(x) + Q(x) + R(x) = P(Q(x)) + Q(R(x)) + R(P(x)) = 0. $$ Prove that the degrees of the three polynomials are all even.</blockquote> $$ P(x)= a_n \cdot x^n + ... + a_2 \cdot x^2 + a_1 \cdot x + a_0 $$ $$ Q(x)= b_m \cdot x^m + ... + b_2 \cdot x^2 + b_1 \cdot x + b_0 $$ $$ a_n,b_m \neq 0 $$ because $P(x)$ and $Q(x)$ are polynomials different from $0$ .
Suppose $m> n$ . $$ R(x) = -P(x)-Q(x) \implies deg R =m $$ $$ P(Q(x)) + Q(R(x)) + R(P(x)) = 0. $$ $$ deg(P(Q(x))=mn,deg(Q(R(x))= m^2,deg(R(P(x)) = mn $$ Note that $m^2> mn$ .Then the coefficient in front of $m^2$ should be $0$ .
we calculate the coefficient before $x^{m^2}$ . $P(Q(x)) + R(P(x))$ does not include $x^{m^2}$ . $$ Q(R(x))= b_m(R(x))^m + b_{m-1}(R(x))^{m-1} + ... + b_0 $$ So the coefficient before $x^{m^2}$ is $b_m \cdot (-b_m)^m$ $\implies b_m \cdot (-b_m)^m = 0$ .Contradiction.
a similar $n>m$ case is also a contradiction.
So $degP=degQ$ . $\implies$ similarity $degP = deg Q$ .
So we have the following. $$ \boxed{degP=deg Q =deg R} $$ Let's say $n$ is an odd number. $$ P(x)= a_n \cdot x^n + ... + a_2 \cdot x^2 + a_1 \cdot x + a_0 $$ $$ Q(x)= b_n \cdot x^n + ... + b_2 \cdot x^2 + b_1 \cdot x + b_0 $$ $$ R(x)= c_n \cdot x^n + ... + c_2 \cdot x^2 + c_1 \cdot x + c_0 $$ $$ a_n=a,b_n=b,c_n=c,a+b+c=0 $$ The coefficient in front of $x^{n^2}$ is $0$ .So we have the following expression. $$ \boxed{ab^n + bc^n + ca^n = 0} $$ at least 2 of $a,b,c$ have the same sign.Suppose $a$ and $b$ . $1)a,b > 0 \implies ab^n = b(a+b)^n+(a+b)a^n > a^{n+1}+b^{n+1} \implies $ Contradiction. $2)a,b<0 $ This is the same as above.So $n$ is a evan number. $\blacksquare$ | [
"This one was a really good free problem. \nWLOG $deg(P)>deg(Q),deg(R)$ , then from $$ P+Q+R=0 $$ we'll get that $P=-(Q+R)$ which is a contradiction since they're non-zero. Then WLOG $deg(Q)=deg(P)=n$ , $deg(R)=m$ . But it's easy to see that from $$ P(Q)+Q(R)+R(P)=0 $$ we get: $$ P(Q)=-(Q(R)+R(P)) \\... | [
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"end_of_proof": true,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783537.json"
} |
A ten-level $2$ -tree is drawn in the plane: a vertex $A_1$ is marked, it is connected by segments with two vertices $B_1$ and $B_2$ , each of $B_1$ and $B_2$ is connected by segments with two of the four vertices $C_1, C_2, C_3, C_4$ (each $C_i$ is connected with one $B_j$ exactly); and so on, up to $512$ vertices $J_1, \ldots, J_{512}$ . Each of the vertices $J_1, \ldots, J_{512}$ is coloured blue or golden. Consider all permutations $f$ of the vertices of this tree, such that (i) if $X$ and $Y$ are connected with a segment, then so are $f(X)$ and $f(Y)$ , and (ii) if $X$ is coloured, then $f(X)$ has the same colour. Find the maximum $M$ such that there are at least $M$ permutations with these properties, regardless of the colouring. | Make the question general such that we are working on $n$ -level $2$ -tree.Notice that we have $3$ kinds of points in $2-tree$ .
$(1)$ Vertices connected with $2$ segments (Starting vertex)
$(2)$ Vertices connected with $1$ segments (Last level vertices of tree)
$(3)$ vertices connected with $3$ segments (middle vertices)
It is easy to see that last vertices ( $J_i$ ) has property $f(J_i)=J_j$ . Since $J_i,J_{i+1}$ ( $i$ is odd here )are connected to the same vertex, we have $f(J_i)=J_{j+1}$ . In more informal way, pairs must stay pairs ( possibly swapping each other).Notice that after we permutate all $J$ s, other points are permutated too ( there are no more extra permutations after that). So, it is enough to work on our last level only. And "pairs of pairs" $J_{i}J_{i+1}, J_{i+2}J_{i+3}$ ( where $ i \equiv 1 \mod 4)$ must stay same.
There are total $2^{n-3}$ pairs of pairs,
Any sets pairs of pairs have additional symmetry, giving lower bound as $M \ge 2^{2^{n-3}}$ , now we show equality case.
To make a construction for equality case, we must have such pairs of pairs such that they have one symmetry only ( $0001, 0111, 0101 $ and their symmetry where $0,1$ shows colors), call these $A,B,C$ , to not have symmetry, we must have $AB,BC,CA$ as combinations, call them $X,Y,Z$ . to not have symmetry, we must have $XY,YZ,ZX$ , call them $N,L,R$ . We can repeat this prosses, each prosses reduces number of points $2$ times. Since we have $3$ points, reduce the number of points to $4$ , , call the last combinations $U,V,W$ , select $U,V,W,U$ , which is obviously there are no more symmetries.
Since our question asked for special case $n=10$ , our answer is $M \ge 2^{128}$ , hence we are done!. $\blacksquare$ | [
"if I'm not mistaken, the answer is $M=g(n)=2^{2^{n-3}}$ for $n$ -level trees ( $n\\geq 3$ ).\nfirst of all it is clear that all such \"permutations\" are in fact uniquely achievable from the main tree by the way of \"swapping\", where we define a \"swap\" as taking the two subtrees descending from a vertex of a... | [
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"boxed": false,
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"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783538.json"
} |
In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$ , and a point $M$ on the segment $CN$ so that $AB = BM = CM$ . Point $K$ is the reflection of $N$ in line $MD$ . The line $MK$ meets the segment $AD$ at point $L$ . Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$ . Prove that $\angle CPM = \angle DPL$ . | <blockquote>In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$ , and a point $M$ on the segment $CN$ so that $AB = BM = CM$ . Point $K$ is the reflection of $N$ in line $MD$ . The line $MK$ meets the segment $AD$ at point $L$ . Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$ . Prove that $\angle CPM = \angle DPL$ .</blockquote>
We define and calculate the angle. $$ \angle MBC = \angle MCB = \alpha, \angle MDC = \beta \implies MCD = 180 -2\beta,\angle MBA = 2\beta - 2\alpha \implies \angle MAB = 90 + \alpha - \beta,\angle MAD = 90 -\beta \implies \angle MPD = 90 -\beta $$ $$ \angle MND =\alpha,\angle MDN = \beta - \alpha \implies \angle NMD = 180 - \beta \implies \angle MNK = \angle CNK = \angle CPK = 90 -\beta $$ So we have the following expression. $$ \boxed{\angle MPD = \angle CPK = 90 -\beta} $$ So we need to prove that $PM$ is the bisector of $\angle LPK$ .For this it is enough to prove that $\boxed{\frac{PL}{LM} = \frac{PK}{MK} = \frac{PK}{MN}}$ . $$ \frac{MN}{LM}= \frac{PK}{PL} \iff (\frac{MN}{LM})^2= (\frac{PK}{PL})^2 $$ $\angle MLN =2\beta - \alpha$ .For the $MLN$ triangle. (*) $$ \frac{MN}{LM} = \frac{sin(2\beta - \alpha)}{sin \alpha} \implies (\frac{MN}{LM})^2 =(\frac{sin(2\beta - \alpha)}{sin \alpha})^2 $$ $\angle DAP = \gamma,\angle AMD = 90 + \alpha \implies \angle APD = 90 + \alpha \implies \angle PDA = \angle PDL =90 - \alpha - \gamma $ Let $R$ be the radius of $(AMD)$ . $$ PL^2 = DP^2 + DL^2 - 2DP\cdot DL \cdot cos(90 - \alpha - \gamma) $$ $DP = 2R sin\gamma,DM =2R cos \beta,DN = \frac{Rsin 2\beta}{sin \alpha} $ $$ \frac{DL}{sin \beta } = \frac{DM}{sin (2\beta - \alpha)} \implies DL = \frac{Rsin 2\beta}{sin (2\beta - \alpha)} $$ $$ PL^2 = R^2(4(sin\gamma)^2 + (\frac{sin2\beta}{sin (2\beta - \alpha)})^2 - 2 \cdot 2sin \gamma \cdot \frac{sin2\beta}{sin(2\beta - \alpha)} \cdot sin(\alpha + \gamma)) $$ $$ \angle PDK = \angle PDM + \angle KDM = \gamma + \beta - 90 + \beta - \alpha = \gamma + 2\beta - \alpha - 90 $$ $$ PK^2 = DK^2 + DP^2 - 2 \cdot DP \cdot DK cos(\gamma + 2\beta - \alpha - 90) = DN^2 + DP^2 - 2 \cdot DP \cdot DN cos(\gamma + 2\beta - \alpha - 90)= R^2((\frac{sin2\beta}{sin\alpha})^2 + 4(sin\gamma)^2 - 2 \cdot 2sin \gamma \cdot \frac{sin 2\beta }{sin \alpha} \cdot cos(\gamma + 2\beta - \alpha - 90) $$ $$ (\frac{PK}{PL})^2= \frac{(sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin\alpha)^2 - 4 \cdot sin \gamma \cdot sin\alpha \cdot sin 2\beta \cdot cos(\gamma + 2\beta - \alpha - 90)}{(sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin2\beta- \alpha)^2 - 4 \cdot sin \gamma \cdot sin2\beta \cdot sin (2\beta - \alpha) \cdot sin(\alpha+ \gamma)} \cdot (\frac{sin(2\beta - \alpha)}{sin \alpha})^2 =\frac{(sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin\alpha)^2 - 4 \cdot sin \gamma \cdot sin\alpha \cdot sin 2\beta \cdot cos(\gamma + 2\beta - \alpha - 90)}{(sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin2\beta- \alpha)^2 - 4 \cdot sin \gamma \cdot sin2\beta \cdot sin (2\beta - \alpha) \cdot sin(\alpha+ \gamma)} \cdot (\frac{MN}{LM})^2 $$ hence it suffices to prove the following. $$ (sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin\alpha)^2 - 4 \cdot sin \gamma \cdot sin\alpha \cdot sin 2\beta \cdot cos(\gamma + 2\beta - \alpha - 90)= (sin2\beta)^2+ 4 \cdot (sin\gamma)^2 \cdot (sin2\beta- \alpha)^2 - 4 \cdot sin \gamma \cdot sin2\beta \cdot sin (2\beta - \alpha) \cdot sin(\alpha+ \gamma) $$
which is also equally strong as follows $$ sin\gamma \cdot (sin2\beta- \alpha)^2 - sin2\beta \cdot sin (2\beta - \alpha) \cdot sin(\alpha+ \gamma) = sin\gamma \cdot (sin\alpha)^2 - sin\alpha \cdot sin 2\beta \cdot sin(\gamma + 2\beta - \alpha ) $$ . $$ sin(\alpha+ \gamma ) = sin \alpha \cdot cos \gamma + sin \gamma \cdot cos \alpha $$ . $$ sin(\gamma+ 2\beta - \alpha ) = sin \gamma \cdot cos(2\beta -\alpha) + cos\gamma \cdot sin(2\beta - \alpha) $$ using these two formulas, let us look at the numbers before $sin \gamma$ above. $$ (sin2\beta- \alpha)^2 - sin2\beta \cdot sin (2\beta - \alpha) \cdot cos \alpha = (sin\alpha)^2 - sin \alpha \sin 2\beta cos(2\beta - \alpha) $$ . $$ (sin2\beta- \alpha)^2 - (sin\alpha)^2 = (sin(2\beta- \alpha) - sin \alpha)( sin(2\beta- \alpha) + sin \alpha) = 2sin (\beta - \alpha)cos \beta \cdot 2sin \beta cos(\beta - \alpha) = sin2\beta \cdot sin(2\beta - 2\alpha) = sin 2\beta \cdot sin(2\beta - \alpha - \alpha) = sin2 \beta \cdot(sin(2\beta - \alpha) \cdot cos\alpha - sin 2\beta sin \alpha cos(2\beta - \alpha) $$ . So the numbers before $sin \gamma$ are equal. Now let's look at the numbers before $cos \gamma$ $$ -sin2\beta \cdot sin(2\beta - \alpha) \sin \alpha = - sin \alpha \cdot (sin 2\beta) sin(2\beta - \alpha) $$ .The proof is over. $\blacksquare$ | [
"Let $\\angle MBC=\\angle MCB=\\alpha$ . Then, $\\angle MKD=\\angle MND=\\angle MCB=\\alpha$ .\nLet $\\angle NDM=\\angle KDM=\\beta$ . Since $DM\\perp NK$ , we find that $\\angle DKN=90^\\circ-\\beta$ . Hence, $\\angle MNK=\\angle MKN=90^\\circ-\\alpha-\\beta$ . Since $|CM|=|AB|=|CD|$ , we find that $\\angl... | [
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"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783539.json"
} |
A polynomial $f(x)$ with real coefficients of degree greater than $1$ is given. Prove that there are infinitely many positive integers which cannot be represented in the form \[f(n+1)+f(n+2)+\cdots+f(n+k)\]
where $n$ and $k$ are positive integers. | Assume the leading coefficient of $f$ is positive. Let $f_k(n)=f(n)+\cdots+f(n+k-1)$ , so we want to show that there are infinitely many positive integers not representable as $f_k(n)$ for some $(k,n)$ .
Consider some large interval $[1,N]$ and suppose $f_k(n) \in [1,N]$ . This clearly implies that $n$ is $O(N^{0.5})$ . Furthermore, we can check (by estimating with an integral, for instance) that if $k~N^{0.4}$ , then $f(1)~N^{1.2}$ , hence $k$ must be $O(N^{0.4})$ , so there are $O(N^{0.9})$ pairs $(k,n)$ that could possibly work. $\blacksquare$ | [
"Reposting my solution from yesterday:\nSuppose that $f$ has degree $d \\ge 2$ . Clearly, we may assume that the leading coefficient of $f$ is positive. In that case $f(n)+C \\gg n^d$ for all $n$ where $C$ is a suitable constant depending only on $f$ and hence\n\\[f(n+1)+\\dots+f(n+k) +Ck \\gg (n+1)^d+... | [
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"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783540.json"
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In triangle $ABC$ , a point $M$ is the midpoint of $AB$ , and a point $I$ is the incentre. Point $A_1$ is the reflection of $A$ in $BI$ , and $B_1$ is the reflection of $B$ in $AI$ . Let $N$ be the midpoint of $A_1B_1$ . Prove that $IN > IM$ . | $\angle ABI = \angle CBI = \angle A_1BI \implies $ $B,C,A_1$ collinear.Similarity $A,C,B_1$ collinear. $$ AI= A_1I = x , BI = B_1I = y $$ $$ IN = \frac{1}{2} \cdot \sqrt{2(x^2+y^2)- A_1B_1^2} $$ $$ IM = \frac{1}{2} \cdot \sqrt{2(x^2+y^2)- AB^2} $$ it suffices to prove $AB > A_1B_1$ $\angle ACB = \alpha,AB =c,BC = a,CA =b$ $$ A_1B_1^2 = (c-b)^2+(a-c)^2+2(c-b)(a-c) \cdot cos\alpha,AB^2 = a^2+ b^2 - 2ab cos\alpha $$ $$ c^2-2cb + b^2 + a^2 -2ac + c^2 + 2(ca - c^2- ab +bc) cos \alpha < a^2+ b^2 - 2ab cos\alpha $$ $$ 2c^2(1-cos \alpha) + 2bc(cos \alpha -1)+ 2ac(cos\alpha-1) <0 $$ $$ (1-cos\alpha) > 0 \implies 2bc - 2c^2+ 2ac >0 \implies a+b >c $$ $\blacksquare$ | [
"First note that $A_1,B_1$ lie on $BC,AC$ and $AI=A_1I, BI=B_1I$ . calculating the medians in two triangles $AIB, A_1IB_1$ , we see that $$ 4IN^2=2AI^2+2BI^2-A_1B_1^2, 4IM^2=2AI^2+2BI^2-AB^2 $$ so its enough to prove $AB>A_1B_1$ , Set the middles of arcs $ACB, AC$ on the circumcircle as $T,R$ . by a spi... | [
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"boxed": false,
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"n_reply": 14,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783541.json"
} |
Do there exist two bounded sequences $a_1, a_2,\ldots$ and $b_1, b_2,\ldots$ such that for each positive integers $n$ and $m>n$ at least one of the two inequalities $|a_m-a_n|>1/\sqrt{n},$ and $|b_m-b_n|>1/\sqrt{n}$ holds? | For each $i$ , draw a square of side length $1/\sqrt{i}$ with sides parallel to the coordinate axes centered at $(a_i,b_i)$ . Then no squares overlap, since for $m>n$ we have $\max\{|a_m-a_n|,|b_m-b_n|\}>1/\sqrt{n}>1/(2\sqrt{n})+1/(2\sqrt{m})$ . On the other hand, the sum of the areas of the squares diverges, hence their centers cannot be contained inside a bounded box, which clearly implies the conclusion. $\blacksquare$ **Remark:** Hm this did not feel very difficult. I guess the main motivation was the following: this condition that "at least one of the two following inequalities holds" is really hard to do anything with, but the condition does imply that $(a_m-a_n)^2+(b_m-b_n)^2>1/n$ for all $m>n$ . This lends to an obvious geometric interpretation: if we draw a circle of radius $1/\sqrt{n}$ around each point $(a_n,b_n)$ , then any $(a_m,b_m)$ for $m>n$ can't lie inside this circle. But once we get this idea to think about points in the 2d plane it's obvious that we can replace these circles with squares. And then we probably want to have the squares not overlap at all, instead of this weird "points that come after can't lie inside the square", but this turns out to be very easy to do (the idea to take a smaller shape shows up fairly often). | [
"This is a really nice one. \n\nPlot all $(a_i, b_i)$ on the coordinate plane. Center a square with side length $1/\\sqrt{i}$ and sides parallel to the coordinate axes at each $(a_i,b_i).$ For any $m > n$ we have $|a_m-a_n|>1/\\sqrt{n} > 1/(2\\sqrt{n}) + 1/(2\\sqrt{m})$ and/or $|b_m-b_n|>1/\\sqrt{n} > ... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783542.json"
} |
Three parallel lines $L_1, L_2, L_2$ are drawn in the plane such that the perpendicular distance between $L_1$ and $L_2$ is $3$ and the perpendicular distance between lines $L_2$ and $L_3$ is also $3$ . A square $ABCD$ is constructed such that $A$ lies on $L_1$ , $B$ lies on $L_3$ and $C$ lies on $L_2$ . Find the area of the square. | Uh oh... That $90^\circ$ angle between $A,B,C$ screams for complex come on! For sure unnecessary, but if first post can be trig, then this is fine as well.
*Solution:* Toss this onto the complex number with $B$ as origin. By the problem condition, assume $c = m + 3i$ for some real $m$ . Then $A$ is precisely $90^\circ$ anti-clockwise rotation of $C$ at $B$ . Thus, we can get $a = i \cdot c = -3 + mi$ . But now since $a \in L_1$ , we get $m = 6$ . Apply Pythagoras once to get that side length of square is $\sqrt{6^2 + 3^2}$ . Thus the area shall be $\boxed{45}$ which is the desired answer. $\blacksquare$ | [
"Say the side length is $x$ , now say $\\angle (L_3, AB)=\\theta$ , this means $\\angle (BC, L_3)=90^{\\circ}- \\theta$ , now you have $x\\cos{\\theta}=3$ and $x\\sin{\\theta}=6$ , squaring and adding them you get $x^2=45$ .",
"18 gang :( ",
" $\\frac{x^2}{2}=3\\times\\frac{15}{2}$ but i somehow calcul... | [
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"path": "Contest Collections/2022 Contests/2022 IOQM India/2796887.json"
} |
Ria writes down the numbers $1,2,\cdots, 101$ in red and blue pens. The largest blue number is equal to the number of numbers written in blue and the smallest red number is equal to half the number of numbers in red. How many numbers did Ria write with red pen? | $If$ the largest blue number is $x$ and the number of numbers written in blue is also $x$ then $x+1$ is the smallest number written in red(since every number before that was written in blue) $101-x$ is the number of numbers written in red
then by the question,
$$ x+1=\frac{101-x}{2} $$ solving which gives,
$$ x=33 $$ $$ x+1=34 $$ and the answer is $68$ | [
"I lost 2 marks in this just because I got 101-33=78 :wallbash_red: ",
"its $68$ , the idea is to keep the min of one thing to be as close to the max of the other thing.",
"Let $x$ be the number of blue pens. Then the number of red pens will be $101-x$ . **Since the largest blue pen marking will be the num... | [
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} |
Consider the set $\mathcal{T}$ of all triangles whose sides are distinct prime numbers which are also in arithmetic progression. Let $\triangle \in \mathcal{T}$ be the triangle with least perimeter. If $a^{\circ}$ is the largest angle of $\triangle$ and $L$ is its perimeter, determine the value of $\frac{a}{L}$ . | A lengthier approach.
Ofcourse the triangle is a $3,5,7$ triangle. Thus using heron's formula, the area of the traingle is $$ \frac{\sqrt{(15)(9)(1)(5)}}{4} = \frac{1}{2} \times 5 \times 3 \times \sin (\angle \text{opposite to 7}). $$ Ofcourse the angle opposite to $7$ is the largest. $$ \Longrightarrow \sin \theta = \frac{\sqrt{3}}{2}. $$ Thus the angle is either $60^{\circ}$ or $120^{\circ}$ . Ofcourse it can't be $60^{\circ}$ as then the total interior angle of the $\bigtriangleup$ will not be $180^{\circ}$ . Thus the angle is $120^{\circ}$ . And the perimeter is $15$ . So the required answer is $\frac{120}{15} = \boxed{8}$ . | [
"Notice that triangle with sides $3$ , $5$ and $7$ is required triangle $\\triangle$ belonging to the set $\\mathcal{T}$ with least perimeter $L$ that is 15. So, we basically need to find angle opposite to side with length $7$ . By LoC, we get $\\cos(a)=-1/2$ <span style=\"font-size:50%\">and you mess... | [
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"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796889.json"
} |
Consider the set of all 6-digit numbers consisting of only three digits, $a,b,c$ where $a,b,c$ are distinct. Suppose the sum of all these numbers is $593999406$ . What is the largest remainder when the three digit number $abc$ is divided by $100$ ? | $a+b+c = 22$ so yesh | [
"...\\[98\\]",
"Let the set denoted by S\nThen we have 3^6=729 numbers belong to S.\nNow if we look at the central tendency we get the mean (593999406÷729)=814814\nClearly a,b,c are 8,4,1 \nThen the largest reminder is 84",
"We get the expression:\n(a+b+c)(1+10+100+1000+10000)(3.3.3.3.3)= $593999406$ \nSimpli... | [
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} |
In parallelogram $ABCD$ , the longer side is twice the shorter side. Let $XYZW$ be the quadrilateral formed by the internal bisectors of the angles of $ABCD$ . If the area of $XYZW$ is $10$ , find the area of $ABCD$ | I will just post a solution because I am mastermind who blundered it in exam even after seeing the bash on exam.
Let $\overline{AB}$ be the longer side of the parallelogram. Let $\angle DAB = 2\theta$ . One can angle chase to find that $XYZW$ is a rectangle. Let $B$ bisector meet $C$ bisector at $X$ and $Y$ be the intersection of $A$ bisector and $B$ bisector.
From $\triangle XBC$ , $\overline{XB} = x\sin(\theta)$ . From $\triangle AYB$ , $\overline{XY} = 2x\sin(\theta)$ . With $\overline{XB}$ and $\overline{XY}$ . one can deduce $X$ is the midpoint of $\overline{YB}$ . Now one can compute side length of the rectangle as $x\sin(\theta)$ . Similarly, one can find the other one to be $x\cos(\theta)$ . By the problem condition
\[x\sin(\theta) \cdot x\cos(\theta) = 10 \iff \sin(2\theta) = \frac{20}{x^2}\]
Now $2 [\triangle BCD]$ is the area of parallelogram. One can just use the $\frac{1}{2}ab\sin(C)$ formula now to get
\[\text{Desired Area} = 2 \times \frac{1}{2}x\cdot 2x \cdot \sin(2\theta) = \fbox{40}\]
and thus we are done. $\blacksquare$ | [
"The key is that the angle bisectors meet on the midpoint of the opposite sides.\nand also form a rectangle.\nfrom here it is easy",
"i just assumed that the parallelogram is a rectangle and from there it is easy",
"@numbertheorydog @mathlearner2357\neven if you dont, its still a simple question, you just have ... | [
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"path": "Contest Collections/2022 Contests/2022 IOQM India/2796891.json"
} |
Let $x,y,z$ be positive real numbers such that $x^2 + y^2 = 49, y^2 + yz + z^2 = 36$ and $x^2 + \sqrt{3}xz + z^2 = 25$ . If the value of $2xy + \sqrt{3}yz + zx$ can be written as $p \sqrt{q}$ where $p,q \in \mathbb{Z}$ and $q$ is squarefree, find $p+q$ . | create a triangle $ABC$ , let $O$ be a point inside $ABC$ such that $\angle{AOB}=\frac{\pi}{2}$ , $\angle{AOC}=\frac{2\pi}{3}$ and $\angle{BOC}=\frac{5\pi}{6}$ , now denote $AO=y, BO=x, OC=z$ , and $AB=7,AC=6,BC=5$ . now $[AOC]+[BOC]+[AOC]=[ABC]$ , so $\frac{xy}{2}+\frac{yz\sin{\frac{2\pi}{3}}}{2}+\frac{xz\sin{\frac{5\pi}{6}}}{2}=6\sqrt{6}$ , now we want $2xy+\sqrt{3}yz+zx$ , so multiply by $4$ both sides , so we get $2xy+\sqrt{3}yz+zx=24\sqrt{6}$ , and hence $p+q=30$ . this was the first problem i did in the test , thanks to CMI :P | [
"Cosine identities go brrrrrrrr.... and I obviously didn't notice them in the test. $30$ is the answer.",
"More like [CMI 2019/5](https://artofproblemsolving.com/community/c6h1944124p13395044)"
] | [
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 IOQM India/2796892.json"
} |
Find the number of maps $f: \{1,2,3\} \rightarrow \{1,2,3,4,5\}$ such that $f(i) \le f(j)$ whenever $i < j$ . | When $f(1)=1$ , you have 15 possible cases and similarly, number of such maps for a fixed $f(1)$ is a triangular number, and summing all of them up you get $15+10+6+3+1=35$ . | [
"\\[\\sum_{i=1}^{5} x_{i}=3\\] which gives \\[{5+3-1}\\choose{3}\\]",
"@bove That's C++ right? And [c] documenting?",
"Sol:- $1 \\leq f(1)<f(2)+1<f(3)+2\\leq 7$ . So we just need to select 3 numbers from $1,2..,7$ . So ans is $C(7,3)=35$ ",
"By stars and bars there are 35 ways to pick a set of 3 values fr... | [
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"boxed": false,
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"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796893.json"
} |
For any real number $t$ , let $\lfloor t \rfloor$ denote the largest integer $\le t$ . Suppose that $N$ is the greatest integer such that $$ \left \lfloor \sqrt{\left \lfloor \sqrt{\left \lfloor \sqrt{N} \right \rfloor}\right \rfloor}\right \rfloor = 4 $$ Find the sum of digits of $N$ . | The first nest of the floor and square root implies $$ 16 \leq \left \lfloor \sqrt{\left \lfloor \sqrt{N} \right \rfloor} \right \rfloor < 25. $$ The next nest of floor and square root implies $$ 256 \leq \left \lfloor \sqrt{N} \right \rfloor < 625 $$ Thus, $$ 256^2 \leq N < 625^2. $$ The maximum value is $625^2 - 1 = 390624$ for a total of $\boxed{24}.$ | [
"hahaha just take $5^8-1$ and obviously its sum of digits is $25$ . ",
"<blockquote>hahaha just take $5^8-1$ and obviously its sum of digits is $25$ .</blockquote>\n\nLol, same mistake done here"
] | [
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"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796894.json"
} |
A $12 \times 12$ board is divided into $144$ unit squares by drawing lines parallel to the sides. Two rooks placed on two unit squares are said to be non-attacking if they are not in the same column or same row. Find the least number $N$ such that if $N$ rooks are placed on the unit squares, one rook per square, we can always find $7$ rooks such that no two are attacking each other. | fill $6$ rows/columns with $6*12$ rooks and place another in an empty row/col. | [
"easiest problem on the test\n",
"how to interpret this? i did not get the wording",
"You have to find the least number of rooks which when placed on the board in any manner will always result in 7 non attacking rooks. \nSo we fill any six columns/rows completely with 12*6=72 rooks(this is basicaly the worst wa... | [
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"path": "Contest Collections/2022 Contests/2022 IOQM India/2796895.json"
} |
Let $P_0 = (3,1)$ and define $P_{n+1} = (x_n, y_n)$ for $n \ge 0$ by $$ x_{n+1} = - \frac{3x_n - y_n}{2}, y_{n+1} = - \frac{x_n + y_n}{2} $$ Find the area of the quadrilateral formed by the points $P_{96}, P_{97}, P_{98}, P_{99}$ . | We claim that the area of the convex quadrilateral formed by $P_{i}, P_{i +1}, P_{i + 2}, P_{i+3}$ is the same as that formed by $P_{i + 1}, P_{i + 2}, P_{i +3}, P_{i +4}$ .
Let $\vec{v}_i = \begin{pmatrix}x_i y_i\end{pmatrix}$ for all $i\ge 0$ and $M =\begin{pmatrix}-3/2 & 1/2 -1/2 & -1/2\end{pmatrix}$ and let $Q_i$ be the convex quadrilateral formed by $P_i, P_{i+1}, P_{i +2}, P_{i + 3}$ . The given recurrences imply $\vec{v}_{n + 1} = M\vec{v}_n$ . In particular, the linear map $M$ sends $Q_i$ to $Q_{i + 1}$ . Thus
\[ [Q_{i+1}] = \text{det}(M)\cdot [Q_i].\]
However, $\text{det}(M) = (-1/2)(-3/2) - (-1/2)(1/2) = 1$ , so $[Q_{i + 1}] = [Q_i]$ , as desired.
Note that $P_0 = (3,1)$ , $P_1 = (-4, -2)$ , $P_2 = (5,3)$ , and $P_3 = (-6, -4)$ . By the shoelace formula, the area of the convex quadrilateral formed by these points is $8$ (alternatively note that the region is a parallelogram). Thus $[Q_0] = 8$ , and finally $[Q_{96}] = \boxed{8}$ | [
"Scam, apparently the coordinates of those points are just those of the first four and you get $8$ as the answer.",
"Nope they are not those of the first four. However the area remains the same",
"Why this problem has been declared bonus by HBCSE?",
"Can anyone please clarify why this problem has been decla... | [
"origin:aops",
"2022 Contests",
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"answer_score": 1040,
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"path": "Contest Collections/2022 Contests/2022 IOQM India/2796896.json"
} |
Suppose that $P$ is the polynomial of least degree with integer coefficients such that $$ P(\sqrt{7} + \sqrt{5}) = 2(\sqrt{7} - \sqrt{5}) $$ Find $P(2)$ . | Say $t_1=\sqrt{7}+\sqrt{5}$ Since , $P(\sqrt{7}+\sqrt{5})=2(\sqrt{7}-\sqrt{5})$ . We try to write the RHS in terms of $t_1$ . It is easy to see through rationalisation that $t_1=\frac{2}{\sqrt{7}-\sqrt{5}}\implies \sqrt{7}-\sqrt{5}=\frac{2}{t_1}$ . Now , we define : $Q(t)=tP(t)-4$ such that $Q(t_1)=0$ from the question statement , $P$ is a polynomial of integer coefficients $\implies Q(t)$ is also a polynomial of integer coefficients . We want to generate a polynomial of least degree of integer coefficients whose one solution is $t_1$ . We do this by : $(t_1-\sqrt{7})^2=\sqrt{5}^2$ $\implies t_1^2-2\sqrt{7}t_1+2=0\implies (t_1^2+2)^2=28t_1^2\implies t_1^4-24t_1^2+4=0$ We conclude that $t_1$ is a zero of $Q(t)=\lambda(t^4-24t^2+4)$ for some integer $\lambda$ . $\implies P(t)=\frac{\lambda(t^4-24t^2)+4(\lambda+1)}{t}$ Since , $P$ is a polynomial , we must have the constant term vanished ,i.e, $4(\lambda+1)=0\implies \lambda =-1$ Hence , our required polynomial is $\boxed{P(t)=24t-t^3}$ | [
"yes, you try $deg(P)=2$ , it doesn't work, you try $deg(P)=3$ and you get $24x-x^3$ so $P(2)=40$ .",
" $P(\\sqrt7 + \\sqrt5) = 2(\\sqrt7 - \\sqrt5)$ \nwe have : $(\\sqrt7 + \\sqrt5)(\\sqrt7 - \\sqrt5) = 2 $ now we have : $P(\\sqrt7 + \\sqrt5) = (\\sqrt7 + \\sqrt5)(\\sqrt7 - \\sqrt5)^2 $ \nWe also have... | [
"origin:aops",
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"answer_score": 1036,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796897.json"
} |
In how many ways can four married couples sit in a merry-go-round with identical seats such that men and women occupy alternate seats and no husband seats next to his wife? | just seat 4 ppl of the same gender in $(4-1)!$ ways and u get $2$ choices to seat another (as seating one of em fixes the whole arrangement).
So, $3!*2$ . | [
"I had already solve this problem from isi tomato book and in exam I did, mistake , very disappointing "
] | [
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"2022 Contests",
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"answer_score": 6,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 IOQM India/2796898.json"
} |
For positive integers $a$ and $b$ , if the expression $\frac{a^2+b^2}{(a-b)^2}$ is an integer, prove that the expression $\frac{a^3+b^3}{(a-b)^3}$ is an integer as well. | Let $\gcd(a,b)=d$ . Then, $\frac{(da)^2+(db)^2}{(da-db)^2} = \frac{a^2+b^2}{(a-b)^2}$ and $\frac{(da)^3+(db)^3}{(da-db)^3} = \frac{a^3+b^3}{(a-b)^3}$ so it suffices to solve the problem for $a,b$ such that $\gcd(a,b)=1$ . Assume WLOG, $a>b$ . Now, note that if $a-b\geq 2$ there exists a prime $p\mid a-b$ . Then,
\begin{align*}
p & \mid a-b
p&\mid a^2-b^2
p& \mid a^2+b^2 \ \ \ \ \ \ \ \ \text{(by the assumed divisibility)}
p &\mid 2a^2 , 2b^2
\end{align*}
If $p>2$ then, $p\mid a^2,b^2$ which implies that $p\mid a,b$ which is a contradiction to the fact that $\gcd(a,b)=1$ . Thus, $p=2$ . But then, $a,b$ must both be odd (if they are both even they have a common factor of 2). But we know,
\[a^2 + b^2 \equiv 1 +1 \equiv 2\pmod{4} \text{ for all odd integers } a,b\]
But clearly, $4\mid (a-b)^2$ which is a contradiction to the assumption that $\frac{a^2+b^2}{(a-b)^2}$ is an integer. Thus, our assumption must have been false and there exists no positive integers $a,b$ such that $\gcd(a,b)=1$ , $a-b >1$ and $\frac{a^2+b^2}{(a-b)^2}$ is an integer.
Thus, $\frac{a^2+b^2}{(a-b)^2}$ is an integer if and only if $a-b=1$ from which it is clear that $\frac{a^3+b^3}{(a-b)^3}$ is also clearly an integer. We are done. | [
"WLOG $(a,b)=1$ . We have $$ (a-b)^2|a^2+b^2-(a-b)^2=2ab $$ Also, $gcd((a-b)^2, a)= gcd((a-b)^2, b)=1$ . Hence, $(a-b)^2|2$ . Therefore, $a-b=\\pm 1$ . The rest follows.",
" $$ (a-b)^2|a^2+b^2-(a-b)^2=2ab \\Rightarrow \\frac{2ab}{(a-b)^2}\\in \\mathbb{Z}\\Rightarrow \\frac{2a^2b-2ab^2}{(a-b)^3}\\in \\mathb... | [
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"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802471.json"
} |
For a real number $a$ , $[a]$ denotes the largest integer not exceeding $a$ .
Find all positive real numbers $x$ satisfying the equation $$ x\cdot [x]+2022=[x^2] $$ | Let $[x]=a$ and $x=a+b$ where $0\leq b<1$ . We have $$ (a+b)a+2022=[(a+b)^2]\Rightarrow a^2+ab+2022=a^2+[2ab+b^2]\Rightarrow ab+2022=[2ab+b^2] $$ Clearly, RHS is an integer. Hence, $ab$ should be an integer too. Then, $[2ab+b^2]=2ab+[b^2]=2ab$ . So the original equation becomes $$ ab+2022=2ab\Rightarrow 2022=ab\Rightarrow b=\frac{2022}a $$ Since $1>b$ , we find that $a\ge 2023$ .
Clearly, all numbers in the form $k+\frac{2022}k$ works where $k\ge 2023, k\in\mathbb{Z}$ . So, we are done. | [
"we see that if 2022 is whole and [x^2] is whole too then x*[x] is whole too. meanwhile , if x is natural we get x^2+2022=x^2 which is imossible. Then let for satisfying x let's take x=b/a and x is not included to Z or b is not divisible by a ; a,b are natural and [x]=a. then we get that a+1>b/a>a<==>a^2<b<a^2+a so... | [
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"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802473.json"
} |
Each of the $29$ people attending a party wears one of three different types of hats. Call a person *lucky* if at least two of his friends wear different types of hats. Show that it is always possible to replace the hat of a person at this party with a hat of one of the other two types, in a way that the total number of lucky people is not reduced. | <details><summary>Solution</summary>Consider the obvious graph interpretation, where vertices represent people, edges represent friendships and we write $0,1$ or $2$ on each of the vertices. Call the graph $G$ and for a vertex $v \in V(G)$ , we denote $d(v)$ and $N(v)$ as the degree of $v$ and the set of neighbours of $v$ , respectively. Also let $f : V(G) \to \{0,1,2\}$ be the function that indicates the colours of vertices.
Assume to the contrary. If some vertex has degree $0$ or $1$ , then the conclusion is obvious, so assume that each degree is at least $2$ .
Now we construct a directed graph $H$ whose vertices are the vertices of $G$ .
For some $u,v \in V(G)$ , if $uv \in E(G)$ and $f(w) = f(t) \neq f(u) \ \forall w,t \in N(v), w,t \neq u$ , draw the edge $u \to v$ in $H$ .
Denote by $d_i(v)$ the indegree of $v$ and $d_j(v)$ the outdegree of $v$ in graph $H$ .
By the assumption, we have $d_i(v) \le 2$ and $d_j(v) \ge 2$ for all $v \in V(H)$ . Moreover, $d_i(v) = 2$ only if $d(v) = 2$ .
Thus we get that $d(v) = 2 \ \forall v \in V(G)$ , so $G$ consists of disjoint cycles. In particular, there exists a cycle whose length is not divisible by $3$ since $3 \nmid 29$ , and it is not hard to prove the result in this case, so we get our desired contradiction and we are done. $\square$</details> | [
"@hakN How do you obtain that $d_j(v) \\geq 2$ for every $v$ ? We do indeed have $d_i(v) \\leq 2$ and $d_i(v) + d_j(v) \\geq 2$ but it is quite unclear how to show $d_j(v) \\geq 2$ . Also, the claim \" $d_i(v) = 2$ only if $d_i(v) + d_j(v) = 2$ \" is correct but does not really fit with $d_j(v) \\geq 2$... | [
"origin:aops",
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} |
Given a convex quadrilateral $ABCD$ such that $m(\widehat{ABC})=m(\widehat{BCD})$ . The lines $AD$ and $BC$ intersect at a point $P$ and the line passing through $P$ which is parallel to $AB$ , intersects $BD$ at $T$ . Prove that $$ m(\widehat{ACB})=m(\widehat{PCT}) $$ | $AB\cap CD=Q$ We will use method of moving points.
Take $QBCD$ fixed. Animate $A$ over $QB$ . Denote $R(XY,ZT)$ as reflection of $XY$ to $ZT$ .
\[f:A\rightarrow AC\rightarrow R(AC,BC)\rightarrow R(AC,BC)\cap BQ\]
\[g: A\rightarrow AD\rightarrow AD\cap BC=T\rightarrow T(QB)_{\infty}\cap BD=P\rightarrow PC\cap QB\] $f,g$ has degree $2$ . $i)A=B$ $f: B\rightarrow BC\rightarrow BC\rightarrow B$ $g: B\rightarrow BD\rightarrow B\rightarrow B\rightarrow B$ So they are same at $A=B$ . $ii)A=Q$ Let the parallel line from $C$ to $QB$ intersect $BD$ at $S$ . $f:Q\rightarrow QC\rightarrow Q'C\rightarrow QB_{\infty}$ $g: Q\rightarrow QD\rightarrow C\rightarrow C(QB)_{\infty}\cap BD=S\rightarrow SC\cap QB=QB_{\infty}$ So they are same at $A=Q$ .
Let the parallel from $D$ to $BC$ intersect $QB$ at $E$ and the parallel from $C$ to $BD$ intersect $QB$ at $F$ . $iii)A=E$ $f: E\rightarrow EC\rightarrow FC\rightarrow F$ $g: E\rightarrow ED\rightarrow BC_{\infty}\rightarrow (BC)_{\infty}(QB)_{\infty}\cap BD=BD_{\infty}\rightarrow F$ Thus $f,g$ are same at $3$ points as desired. $\blacksquare$ | [
"Let the line passing through $D$ and parallel to $AB$ intersects $BP$ and $AT$ at $K$ and $L$ , respectively. We have $$ \\frac{|DK|}{|AB|}=\\frac{|PD|}{|PA|}=\\frac{|PD|}{|PA|}\\cdot\\frac{|PT|}{|DL|}\\cdot\\frac{|DL|}{|PT|}=\\frac{|PD|}{|PA|}\\cdot\\frac{|AP|}{|AD|}\\cdot\\frac{|DL|}{|PT|}= $$ $$ ... | [
"origin:aops",
"2022 Contests",
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"answer_score": 170,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802490.json"
} |
Each of the $n$ students writes one of the numbers $1,2$ or $3$ on each of the $29$ boards. If any two students wrote different numbers on at least one of the boards and any three students wrote the same number on at least one of the boards, what is the maximum possible value of $n$ ? | <details><summary>Solution</summary>The answer is $3^{28}$ .
Instead of writing a number in each board, assume that each student chooses a $29$ digit number where each digit is either $1,2$ or $3$ .
Constructing an example is not hard. Assume that the students wrote every $29$ digit number starting with $1$ . This gives a total of $3^{28}$ students and clearly satisfies the condition.
Let's consider all the numbers that could have been chosen. Group some numbers if we can derive one of them by adding the same number to each digit of another number (we consider the digits in modulo $3$ so $2+2=1$ ). For example, one of the group is $$ 11\cdots 1, 22\cdots 2, 33\cdots 3 $$ It is easy to check that each group is consist of $3$ numbers. Hence, there are $3^{28}$ groups in total. Now, just see that two students cannot choose their numbers from the same group. Therefore, there are at most $3^{28}$ students, done.</details> | [] | [
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"answer_score": 28,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802498.json"
} |
Let $c$ be a real number. If the inequality $$ f(c)\cdot f(-c)\ge f(a) $$ holds for all $f(x)=x^2-2ax+b$ where $a$ and $b$ are arbitrary real numbers, find all possible values of $c$ . | I claim the answer is $c=\pm 1/2$ . This is easily seen to work. Next, note that $f(c)f(-c)\ge f(a)$ iff $g(b)\ge 0$ where
\[
g(b) = b^2 + (2c^2-1)b +c^4 -4a^2c^2 + a^2.
\]
We want the quadratic $g(b)\ge 0$ for all choices of $a,b$ . This holds iff its discriminant is non-positive: $(2c^2-1)^2 - 4(c^4+a^2-4a^2c^2) = (4a^2-1)(4c^2-1)\le 0$ . It is now clear that unless $4c^2=1$ , one can choose $a$ such that ${\rm sgn}(4a^2-1)={\rm sgn}(4c^2-1)$ and make the discriminant positive, which would yield a contradiction. Thus, $c=\pm 1/2$ as claimed. | [] | [
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} |
In a triangle $\triangle ABC$ with $\angle ABC < \angle BCA$ , we define $K$ as the excenter with respect to $A$ . The lines $AK$ and $BC$ intersect in a point $D$ . Let $E$ be the circumcenter of $\triangle BKC$ . Prove that
\[\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}.\] | We need to prove that $\frac{|KA|}{|KD|}+\frac{|KA|}{|KE|}=1$ .
Simple Angle-Chasing gives us $\angle ADC=\frac{C-B}2$ . Hence, $$ \frac{|KA|}{|KD|}=\frac{|AC|}{|CD|}=\frac{\sin{ADC}}{\sin{CAD}}=\frac{\sin{(\frac{C-B}2)}}{\cos{(\frac A2)}} $$ Also, $$ \frac{|KA|}{|KE|}=\frac{2|KA|}{2R_{BKC}}=\frac{2|KA|\cdot\sin{(\frac B2)}}{|KC|}=\frac{2\cos{(\frac C2)\cdot \sin{(\frac B2)}}}{\cos{(\frac A2)}} $$ Therefore, it suffices to show that $$ 1=\frac{\sin{(\frac{C-B}2)}}{\cos{(\frac A2)}}+\frac{2\cos{(\frac C2)\cdot \sin{(\frac B2)}}}{\cos{(\frac A2)}}=\frac{\sin{(\frac{C+B}2)}}{\cos{(\frac A2)}}\Leftrightarrow \sin{\left(\frac{C+B}2\right)}=\cos{\left(\frac A2\right)} $$ , which is correct. Done. | [
" \n we know that if $ (A,B;C,D)=-1 $ then $ \\frac{2}{AB}=\\frac{1}{CB}+\\frac{1}{DB}$ ( with signed distance)\n just consider $T$ the symmetric of $K$ in $E$ which is on $(E)$ ...\n\nMy regards**RH HAS**"
] | [
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"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802503.json"
} |
Find all prime numbers $p$ such that the number $$ 3^p+4^p+5^p+9^p-98 $$ has at most $6$ positive divisors. | <blockquote>See that $p=2,3,5$ works. Let $p\ge 7$ .
See that $3^p+4^p\equiv 5^p+9^p\equiv -98\equiv 0\pmod{7}$ as $p$ is odd.
Also, $11|3^p+4^p+5^p+9^p-98$ as $p$ is odd and not divisible by $5$ .
Since this number has at most $6$ positive divisors, we find that $3^p+4^p+5^p+9^p-98\leq 11^2\cdot 7=847$ . But, this contradicts the fact that $p\ge 7$ , done.</blockquote>
how did you find $11|3^p+4^p+5^p+9^p-98??$ | [
"See that $p=2,3,5$ works. Let $p\\ge 7$ .\nSee that $3^p+4^p\\equiv 5^p+9^p\\equiv -98\\equiv 0\\pmod{7}$ as $p$ is odd.\nAlso, $11|3^p+4^p+5^p+9^p-98$ as $p$ is odd and not divisible by $5$ .\nSince this number has at most $6$ positive divisors, we find that $3^p+4^p+5^p+9^p-98\\leq 11^2\\cdot 7=84... | [
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"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802505.json"
} |
Represent $\frac{1}{2021}$ as a difference of two irreducible fractions with smaller denominators.
*(Proposed by Bogdan Rublov)* | <blockquote>Represent $\frac{1}{2021}$ as a difference of two irreducible fractions with smaller denominators.</blockquote>
From $\frac ab-\frac cd=\frac 1{2021}$ , we get $ad-bc=\frac{bd}{2021}$ And so $(b,d)\in\{(2021u,v),(u,2021v),(43u,47v),(47u,43v)\}$ I dont know what is the order relation you use when you speak of "smallest" pair of integers.
If we suppose that smallest is the one for which $\max(|b|,|d|)$ is smallest, then we have $(b,d)=(43,47)$ or $(b,d)=(47,43)$ The first gives for example $\boxed{\frac{11}{43}-\frac{12}{47}=\frac 1{2021}}$ and the second $\boxed{\frac{35}{47}-\frac{32}{43}=\frac 1{2021}}$ (and a lot of others).
If your order relation is different, just give it to us and we'll adapt the result.
| [] | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 1"
] | {
"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 1/2764246.json"
} |
There are $n$ sticks which have distinct integer length. Suppose that it's possible to form a non-degenerate triangle from any $3$ distinct sticks among them. It's also known that there are sticks of lengths $5$ and $12$ among them. What's the largest possible value of $n$ under such conditions?
*(Proposed by Bogdan Rublov)* | <details><summary>Solution</summary>The answer is $6$ . Clearly, all further lengths must be at least $8$ and at most $16$ to form a non-degenerate triangle with $5$ and $12$ .
In particular, $5$ is the smallest length. Now, if $n$ is the next smallest length, then the largest length can be at most $n+4$ to form a triangle with $n$ and $5$ . Hence, there can be at most $5$ lengths that are distinct from $5$ , and hence at most $6$ in total.
Finally, clearly $6$ is possible e.g. by choosing $5,8,9,10,11,12$ or $5,12,13,14,15,16$ (or indeed $5,n,n+1,n+2,n+3,n+4$ for any $8 \le n \le 12$ ).</details> | [
"<details><summary>Answer</summary>Answer --> 6</details>"
] | [
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} |
You are given $n$ not necessarily distinct real numbers $a_1, a_2, \ldots, a_n$ . Let's consider all $2^n-1$ ways to select some nonempty subset of these numbers, and for each such subset calculate the sum of the selected numbers. What largest possible number of them could have been equal to $1$ ?
For example, if $a = [-1, 2, 2]$ , then we got $3$ once, $4$ once, $2$ twice, $-1$ once, $1$ twice, so the total number of ones here is $2$ .
*(Proposed by Anton Trygub)* | <blockquote>You are given $n$ not necessarily distinct real numbers $a_1, a_2, \ldots, a_n$ . Let's consider all $2^n-1$ ways to select some nonempty subset of these numbers, and for each such subset calculate the sum of the selected numbers. What largest possible number of them could have been equal to $1$ ?
For example, if $a = [-1, 2, 2]$ , then we got $3$ once, $4$ once, $2$ twice, $-1$ once, $1$ twice, so the total number of ones here is $2$ .</blockquote>
Let us consider the $2^n$ subsets (including the empty subset, whose sum of elements is zero).
If all elements are zero, the number of subsets whose sum is $1$ is zero.
If at least one element $x_0$ is nonzero :
If a subset with sum $1$ contains $x_0$ , then the same subset without $x_0$ has a sum $\ne 1$ If a subset with sum $1$ does not contain $x_0$ , then the same subset with $x_0$ has a sum $\ne 1$ So the number of subsets with sum $1$ is $\le$ the number of subsets with sum $\ne 1$ And so the requested number is $\le\frac 122^n$ And $2^{n-1}$ indeed can be reached, for example with $\{1,0,0,...,0\}$ Hence the answer $\boxed{2^{n-1}}$ | [] | [
"origin:aops",
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"answer_score": 1058,
"boxed": true,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 1/2764254.json"
} |
In some magic country, there are banknotes only of values $3$ , $25$ , $80$ hryvnyas. Businessman Victor ate in one restaurant of this country for $2024$ days in a row, and each day (except the first) he spent exactly $1$ hryvnya more than the day before (without any change). Could he have spent exactly $1000000$ banknotes?
*(Proposed by Oleksii Masalitin)* | [] | [
"origin:aops",
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"2022 Kyiv City MO Round 1"
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"boxed": false,
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} | |
Consider $5$ distinct positive integers. Can their mean be
a)Exactly $3$ times larger than their largest common divisor?
b)Exactly $2$ times larger than their largest common divisor?
| <blockquote>Consider $5$ distinct positive integers. Can their mean be
a)Exactly $3$ times larger than their largest common divisor?</blockquote>
Yes, choose for example $1,2,3,10,14$ <blockquote>b)Exactly $2$ times larger than their largest common divisor?</blockquote>
Yes, choose for example $1,2,3,4,10$ | [
"<blockquote><blockquote>Consider $5$ distinct positive integers. Can their mean be \n\na)Exactly $3$ times larger than their largest common divisor?</blockquote>\nYes, choose for example $1,2,3,10,14$ </blockquote>\nI'm not sure, I understand. It seems to me that the largest common divisor is $1$ while the ... | [
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In triangle $ABC$ $\angle B > 90^\circ$ . Tangents to this circle in points $A$ and $B$ meet at point $P$ , and the line passing through $B$ perpendicular to $BC$ meets the line $AC$ at point $K$ . Prove that $PA = PK$ .
*(Proposed by Danylo Khilko)* | [] | [
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What's the largest number of integers from $1$ to $2022$ that you can choose so that no sum of any two different chosen integers is divisible by any difference of two different chosen integers?
*(Proposed by Oleksii Masalitin)* | [] | [
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$2022$ teams participated in an underwater polo tournament, each two teams played exactly once against each other. Team receives $2, 1, 0$ points for win, draw, and loss correspondingly. It turned out that all teams got distinct numbers of points. In the final standings the teams were ordered by the total number of points.
A few days later, organizers realized that the results in the final standings were wrong due to technical issues: in fact, each match that ended with a draw according to them in fact had a winner, and each match with a winner in fact ended with a draw. It turned out that all teams still had distinct number of points! They corrected the standings, and ordered them by the total number of points.
Could the correct order turn out to be the reversed initial order?
*(Proposed by Fedir Yudin)* | [] | [
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What's the smallest possible value of $$ \frac{(x+y+|x-y|)^2}{xy} $$ over positive real numbers $x, y$ ? | Let $x, y$ be positive real numbers. Prove that $$ \frac{(x+y+|x-y|)^2}{xy}\geq 4 $$ $$ \frac{(3x+y+|x-2y|)^2}{xy} \geq 24 $$ <details><summary>*</summary><blockquote>What's the smallest possible value of $$ \frac{(x+y+|x-y|)^2}{xy} $$ over positive real numbers $x, y$ ?</blockquote></details> | [
"wlog cheese guys??? \nby symmetry $ x \\geq y$ Then inequality just $\\frac{4x^2}{xy}$ Then $\\frac{4x^2}{xy} \\geq \\frac{4x^2}{x^2} = 4$ "
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For any reals $x, y$ , show the following inequality: $$ \sqrt{(x+4)^2 + (y+2)^2} + \sqrt{(x-5)^2 + (y+4)^2} \le \sqrt{(x-2)^2 + (y-6)^2} + \sqrt{(x-5)^2 + (y-6)^2} + 20 $$ *(Proposed by Bogdan Rublov)* | In the cartesian system of coordinates $xOy$ consider the variable point $M(x,y)$ , ( $x,y\in\mathbb{R}$ ) and the fixed points: $A(-4,-2);B(5,-4);C(2,6);D(5,6)$ . $\sqrt{(x+4)^2 + (y+2)^2}=MA;\;\sqrt{(x-5)^2 + (y+4)^2}=MB$ ; $\sqrt{(x-2)^2 + (y-6)^2}=MC;\;\sqrt{(x-5)^2 + (y-6)^2}=MD$ .
We observe: $AC=BD=10$ .
The requested inequality becomes: $MA+MB\le MC+MD+AC+BD\quad(1)$ .
Applying the triangle inequality in $\triangle{MAC}$ and $\triangle{MBD}$ results: $MC+AC\ge MA$ ; $MD+BD\ge MB$ .
Adding, results the inequality $(1)$ . $AC\cap BD=\{E\}$ , with the coordinates $E(5,10)$ .
The equality in $(1)$ occurs for $M\equiv E\Longleftrightarrow \begin{cases}x=5;y=10.\end{cases}$ | [
"<blockquote>For any reals $x, y$ , show the following inequality: $$ \\sqrt{(x+4)^2 + (y+2)^2} + \\sqrt{(x-5)^2 + (y+4)^2} \\le \\sqrt{(x-2)^2 + (y-6)^2} + \\sqrt{(x-5)^2 + (y-6)^2} + 20 $$ *(Proposed by Bogdan Rublov)*</blockquote>\nHardest algebra ever?\nis it?\n",
"@above there's something called sarcasm a... | [
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Let $AL$ be the inner bisector of triangle $ABC$ . The circle centered at $B$ with radius $BL$ meets the ray $AL$ at points $L$ and $E$ , and the circle centered at $C$ with radius $CL$ meets the ray $AL$ at points $L$ and $D$ . Show that $AL^2 = AE\times AD$ .
*(Proposed by Mykola Moroz)* | [
"Dear Mathlinkers,\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=530917\n\ninspire a new approach...\n\nSincerely\nJean-Louis"
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Let's call integer square-free if it's not divisible by $p^2$ for any prime $p$ . You are given a square-free integer $n>1$ , which has exactly $d$ positive divisors. Find the largest number of its divisors that you can choose, such that $a^2 + ab - n$ isn't a square of an integer for any $a, b$ among chosen divisors.
*(Proposed by Oleksii Masalitin)* | Let $n = p_1 \cdot p_2 \dots p_k$ where $p_i$ are primes and $k \geq 1$ . Then $d = 2^k$ and we claim that the answer is $2^{k-1}$ .
It is clear that we can't choose two divisors $a,b$ of $n$ so that $ab = n$ , so this means we can choose at most $\frac{d}{2} = 2^{k-1}$ divisors.
Now if we choose the divisors that have $p_1$ in them, then we have chosen exactly $2^{k-1}$ divisors and now we show the property holds:
Let $n = a \cdot t$ , we then have $a^2 + ab - n = a(a + b - t)$ . If we assume this to be a square, then each prime in $a$ has to divide $a+b-t$ since $a$ is squarefree. So we get $a \mid b - t$ . But, $p_1$ divides both $a$ and $b$ , but it doesn't divide $t$ since $n$ is squarefree, thus we reached a contradiction. | [] | [
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$n\ge 2$ teams participated in an underwater polo tournament, each two teams played exactly once against each other. A team receives $2, 1, 0$ points for a win, draw, and loss correspondingly. It turned out that all teams got distinct numbers of points. In the final standings, the teams were ordered by the total number of points.
A few days later, organizers realized that the results in the final standings were wrong due to technical issues: in fact, each match that ended with a draw according to them in fact had a winner, and each match with a winner in fact ended with a draw. It turned out that all teams still had distinct number of points! They corrected the standings and ordered them by the total number of points.
For which $n$ could the correct order turn out to be the reversed initial order?
*(Proposed by Fedir Yudin)* | [] | [
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Does there exist a quadratic trinomial $ax^2 + bx + c$ such that $a, b, c$ are odd integers, and $\frac{1}{2022}$ is one of its roots? | <blockquote>Does there exist a quadratic trinomial $ax^2 + bx + c$ such that $a, b, c$ are odd integers, and $\frac{1}{2022}$ is one of its roots?</blockquote>
No, since $2022^2c+2022b+a=0$ implies $a$ is even
| [
" $2022^2c+2022b+a=0 \\implies a$ is evan $\\implies$ contradiction",
" this problem is very easy"
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You are given $2n$ distinct integers. What's the largest integer $C$ such that you can always form at least $C$ pairs from them, so that no integer is in more than one pair, and the sum of integers in each pair is a composite number?
*(Proposed by Anton Trygub)* | Bump! I think answer is $C=n-1$ (look at parity) | [
"<blockquote>Bump! I think answer is $C=n-1$ (look at parity)</blockquote>\n\nBump!"
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Diagonals of a cyclic quadrilateral $ABCD$ intersect at point $P$ . The circumscribed circles of triangles $APD$ and $BPC$ intersect the line $AB$ at points $E, F$ correspondingly. $Q$ and $R$ are the projections of $P$ onto the lines $FC, DE$ correspondingly. Show that $AB \parallel QR$ .
*(Proposed by Mykhailo Shtandenko)* | Cute problem.
By angle chase, $\measuredangle PCD = \measuredangle FCP$ and $\measuredangle PDC = \measuredangle EDP$ . If $W$ is the projection of $P$ onto $CD$ , it is easy to see that $PR=PW=PQ$ . Another angle chase shows that $\measuredangle DEF = \measuredangle EFC$ . Now, if $CF || DE$ , the desired statement is obvious. Otherwise, denote by $X$ the intersection point of $CF$ and $DE$ and observe that both lines are perpendicular to the bisector of $\angle CXD$ . | [
"I assume you mean the projections of $E,F$ .\nIf so, since $EFQR$ is cyclic of diameter $EF$ , it suffices to probe that the two angles at the base of the quadrilateral are equal. Indeed $\\measuredangle QFE=\\measuredangle CFB=\\measuredangle CPB=\\measuredangle DPA=\\measuredangle DEA=\\measuredangle REF$ ,... | [
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For any nonnegative reals $x, y$ show the inequality $$ x^2y^2 + x^2y + xy^2 \le x^4y + x + y^4 $$ . | For any nonnegative reals $x, y$ show the inequality $$ x^2y^2 + x^2y + xy^2 \le x^3y^2 + x^2 + y^3 $$ | [
" $(6x^4y+2x+5y^4)+(5x^4y+6x+2y^4)+(2x^4y+5x+6y^4)\\ge 13(x^2y^2+x^2y+xy^2)$ ",
"Adding a dummy third variable makes the inequality look less weird: $x^4y+y^4z+z^4x \\ge x^2y^2z + x^2yz^2 + xy^2z^2$ . The rest of the proof is a simple application of Cauchy with weights $(\\frac{6}{13}, \\frac{5}{13}, \\frac{2}{... | [
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There is a black token in the lower-left corner of a board $m \times n$ ( $m, n \ge 3$ ), and there are white tokens in the lower-right and upper-left corners of this board. Petryk and Vasyl are playing a game, with Petryk playing with a black token and Vasyl with white tokens. Petryk moves first.
In his move, a player can perform the following operation at most two times: choose any his token and move it to any adjacent by side cell, with one restriction: you can't move a token to a cell where at some point was one of the opponents' tokens.
Vasyl wins if at some point of the game white tokens are in the same cell. For which values of $m, n$ can Petryk prevent him from winning?
*(Proposed by Arsenii Nikolaiev)* | [] | [
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The teacher wrote $5$ distinct real numbers on the board. After this, Petryk calculated the sums of each pair of these numbers and wrote them on the left part of the board, and Vasyl calculated the sums of each triple of these numbers and wrote them on the left part of the board (each of them wrote $10$ numbers). Could the multisets of numbers written by Petryk and Vasyl be identical? | [] | [
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Let $H$ and $O$ be the orthocenter and the circumcenter of the triangle $ABC$ . Line $OH$ intersects the sides $AB, AC$ at points $X, Y$ correspondingly, so that $H$ belongs to the segment $OX$ . It turned out that $XH = HO = OY$ . Find $\angle BAC$ .
*(Proposed by Oleksii Masalitin)* | By the way, the property $XH=OY$ is enough for the result. The argumentation could be the following.
Triangles $AXH$ and $AOY$ are equal because $XH=OY$ , $\angle XAH=\angle YAO$ and these triangles have the same length of the altitude from $A$ . Thus, $R=AH$ . At the same time $R=\frac{a}{2\sin\alpha}$ and $AH=\frac{a}{\tan \alpha}$ and as a consequence $\cos \alpha=\frac{1}{2}$ . | [] | [
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You are given $n\ge 4$ positive real numbers. It turned out that all $\frac{n(n-1)}{2}$ of their pairwise products form an arithmetic progression in some order. Show that all given numbers are equal.
*(Proposed by Anton Trygub)* | [] | [
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Find the smallest integer $n$ for which it's possible to cut a square into $2n$ squares of two sizes: $n$ squares of one size, and $n$ squares of another size.
*(Proposed by Bogdan Rublov)* | Bump! Is answer $n=20$ ? | [
"Could you tell me where you get the problems from?",
"<blockquote>Could you tell me where you get the problems from?</blockquote>\n\nThey are published in Ukrainian on http://matholymp.com.ua.",
"<blockquote>Bump! Is answer $n=20$ ?</blockquote>\n\nBump! Has anyone got the official solutions? I can't find the... | [
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a) Do there exist positive integers $a$ and $d$ such that $[a, a+d] = [a, a+2d]$ ?
b) Do there exist positive integers $a$ and $d$ such that $[a, a+d] = [a, a+4d]$ ?
Here $[a, b]$ denotes the least common multiple of integers $a, b$ . | a) No. Note that otherwise one would have
\[
\frac{a(a+d)}{(a,d)} = \frac{a(a+2d)}{(a,2d)},
\]
where we used $(a,a+d)=(a,d)$ and $(a,a+2d)=(a,2d)$ via Euclidean algorithm. Now, note that $(a,2d)=k(a,d)$ with $k\in\{1,2\}$ . The case $k=1$ is clearly contradictory, whereas for $k=2$ , we obtain $2a+2d=a+2d$ , not possible.
b) Yes, sufffices to take $(a,d)=(2d,d)$ where $d\in\mathbb{N}$ is arbitrary. (In fact, it appears that an analogous analysis shows this is the unique such family.) | [
"deleted ....\n@below pointed out already",
"@above\nb) is wrong for $a=2d$ we have $[a, a+d] =[2d, 3d]=6d, [a,a+4d]=[2d,6d]=6d$ Solution in a) has error because $\\gcd(a,2b) = \\gcd(a,b) $ only for odd $a$ "
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There is a central train station in point $O$ , which is connected to other train stations $A_1, A_2, \ldots, A_8$ with tracks. There is also a track between stations $A_i$ and $A_{i+1}$ for each $i$ from $1$ to $8$ (here $A_9 = A_1$ ). The length of each track $A_iA_{i+1}$ is equal to $1$ , and the length of each track $OA_i$ is equal to $2$ , for each $i$ from $1$ to $8$ .
There are also $8$ trains $B_1, B_2, \ldots, B_8$ , with speeds $1, 2, \ldots, 8$ correspondently. Trains can move only by the tracks above, in both directions. No time is wasted on changing directions. If two or more trains meet at some point, they will move together from now on, with the speed equal to that of the fastest of them.
Is it possible to arrange trains into stations $A_1, A_2, \ldots, A_8$ (each station has to contain one train initially), and to organize their movement in such a way, that all trains arrive at $O$ in time $t < \frac{1}{2}$ ?
*(Proposed by Bogdan Rublov)* | [] | [
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In triangle $ABC$ the median $BM$ is equal to half of the side $BC$ . Show that $\angle ABM = \angle BCA + \angle BAC$ .
*(Proposed by Anton Trygub)* | Let $N$ be the midpoint of $BC$ and let $D$ be the reflection of $N$ wrt $B$ . Then $BD=BM=BN$ , so $DM\perp MN\parallel AB$ , so $\angle ABM=\angle ABD=\angle BCA+\angle BAC$ , as desired. | [
"Double the median, i.e let $N \\in BM$ be such that $MB = MN$ . Then $ABCN$ is a parallelogram and $BN = BC$ , from which the result follows easily."
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Fedir and Mykhailo have three piles of stones: the first contains $100$ stones, the second $101$ , the third $102$ . They are playing a game, going in turns, Fedir makes the first move. In one move player can select any two piles of stones, let's say they have $a$ and $b$ stones left correspondently, and remove $gcd(a, b)$ stones from each of them. The player after whose move some pile becomes empty for the first time wins. Who has a winning strategy?
As a reminder, $gcd(a, b)$ denotes the greatest common divisor of $a, b$ .
*(Proposed by Oleksii Masalitin)* | [] | [
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Find all triples $(a, b, c)$ of positive integers for which $a + [a, b] = b + [b, c] = c + [c, a]$ .
Here $[a, b]$ denotes the least common multiple of integers $a, b$ .
*(Proposed by Mykhailo Shtandenko)* | <details><summary>Solution</summary>W.l.o.g. $\text{gcd}(a,b,c)=1$ since otherwise we can just divide it out.
If $p \mid a,b$ , then $p \mid a+[a,b]-[c,a]=c$ , contradiction!
Similarly, we can deduce that the three numbers are pairwise coprime. But then $[a,b]=ab$ etc. so the equations just become $a+ab=b+bc=c+ca$ .
Hence $a \mid c+1, b \mid a+1, c \mid b+1$ .
If w.l.o.g. $a=\max(a,b,c)$ , then this shows that $c=a$ or $c=a-1$ .
If $c=a$ , then $a=b=c=1$ which is indeed a solution.
If $c=a-1$ , we have $b \ge a-2$ , so $b=a-2$ or $b=a-1$ or $b=a$ . From here it is easy to get that the only solution of the divisibilities is $(3,1,2)$ but it does not satisfy the equations.
Hence the only solution in general is $a=b=c$ for arbitrary $a$ .</details> | [
"Note that $a\\mid a+[a,b]=c+[c,a]\\implies a\\mid c$ . Therefore $a \\mid c \\mid b \\mid a$ , or $a=b=c$ , which works."
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Monica and Bogdan are playing a game, depending on given integers $n, k$ . First, Monica writes some $k$ positive numbers. Bogdan wins, if he is able to find $n$ points on the plane with the following property: for any number $m$ written by Monica, there are some two points chosen by Bogdan with distance exactly $m$ between them. Otherwise, Monica wins.
Determine who has a winning strategy depending on $n, k$ .
*(Proposed by Fedir Yudin)* | Claim: Monica wins if and only if $n\leq k$ .
If $n\geq k+1$ , then Bogdan picks the origin and the points $(x_i,0)$ ( $1\leq i\leq k$ ) where the $x_i$ are the specified distances.
If $n\leq k$ then it suffices for Monica to pick the distances $1,2,2^2,...,2^{k-1}$ . Assume Bogdan could win.
Consider the graph with the vertices picked by Bogdan and the edges are between vertices with distances belonging to those specified by Monica (if a certain distance appears more than once, pick only one example at random). There can't be any cycle, because $2^t>2^{t-1}+\cdots +1$ for all $t$ , and all edge lengths are distinct.
Therefore the graph is a union of trees. Each of them satisfies $E_i+1=V_i$ , which means that in the total of the graph we have $n=V\geq E+1\geq k+1$ . However this is a contradiction with $n\leq k$ , so we are done. | [] | [
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Nonzero real numbers $x_1, x_2, \ldots, x_n$ satisfy the following condition: $$ x_1 - \frac{1}{x_2} = x_2 - \frac{1}{x_3} = \ldots = x_{n-1} - \frac{1}{x_n} = x_n - \frac{1}{x_1} $$ Determine all $n$ for which $x_1, x_2, \ldots, x_n$ have to be equal.
*(Proposed by Oleksii Masalitin, Anton Trygub)* | <blockquote>Nonzero real numbers $x_1, x_2, \ldots, x_n$ satisfy the following condition: $$ x_1 - \frac{1}{x_2} = x_2 - \frac{1}{x_3} = \ldots = x_{n-1} - \frac{1}{x_n} = x_n - \frac{1}{x_1} $$ Determine all $n$ for which $x_1, x_2, \ldots, x_n$ have to be equal.</blockquote>
Let us look for cases where all $x_i$ are not equal. WLOG (since cyclic) consider $x_2\ne x_1$ .
Let $u>0$ such that the common value in the equalities is $u-\frac 1u$ Let $S_n=u^{-n}-(-u)^n$ It is easy to establish (or simply check) that $x_k=\frac{S_{k-2}x_1+S_{k-1}}{S_{k-1}x_1+S_k}$ $x_{n+1}=x_1$ is $S_{n-1}x_1+S_{n}=S_{n}x_1^2+(S_{n+1}-S_{n-1})x_1+S_n=0$ Which is $S_n(x_1^2+(\frac 1u-u)x_1-1)=0$ Which is $S_n(x_1-x_2)(x_1-u+\frac 1u)=0$ But $x_1-u+\frac 1u\ne 0$ (in order $x_2$ be defined) and $x_1-x_2\ne 0$ So $S_n=0$ , which is $u^{2n}=(-1)^n$ and so $n$ even and $u=1$ and $x_{k+1}=\frac 1{x_k}$ , which indeed fits, whatever is $x_1\notin\{-1,0,1\}$ Hence the answer $\boxed{\text{Any odd }n}$ | [] | [
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Points $D, E, F$ are selected on sides $BC, CA, AB$ correspondingly of triangle $ABC$ with $\angle C = 90^\circ$ such that $\angle DAB = \angle CBE$ and $\angle BEC = \angle AEF$ . Show that $DB = DF$ .
*(Proposed by Mykhailo Shtandenko)* | [] | [
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Find all triples $(a, b, c)$ of positive integers for which $a + (a, b) = b + (b, c) = c + (c, a)$ .
Here $(a, b)$ denotes the greatest common divisor of integers $a, b$ .
*(Proposed by Mykhailo Shtandenko)* | Solved it ORALLY..
WLOG $gcd(a, b, c) =1$ , now if a prime $p \mid gcd(a, b) $ , then $p \mid c$ , which is not possible, so $a, b, c$ are pairwise coprime, hence condition is $a+1 = b+1 =c+1$ , implying $a=b=c$ , so only solutions that work are $(k, k, k) $ for all natural $k$ ... | [] | [
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$2022$ points are arranged in a circle, one of which is colored in black, and others in white. In one operation, The Hedgehog can do one of the following actions:
1) Choose two adjacent points of the same color and flip the color of both of them (white becomes black, black becomes white)
2) Choose two points of opposite colors with exactly one point in between them, and flip the color of both of them
Is it possible to achieve a configuration where each point has a color opposite to its initial color with these operations?
*(Proposed by Oleksii Masalitin)* | No. Label the positions of the points consecutively $1,-1,1,-1,\dots$ and let $\ell_i,i=1,\dots,2022$ be the values od these labels. Set $c_i:=1$ if the color of the $i$ -th point is white, otherwise $c_i=-1$ . Consider $S:=\sum_{i=1}^{2022}\ell_i c_i.$ This value is invariant under the allowed recolorings. But $S$ has different values for the initial and desired configurations. | [] | [
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Let $\omega$ denote the circumscribed circle of triangle $ABC$ , $I$ be its incenter, and $K$ be any point on arc $AC$ of $\omega$ not containing $B$ . Point $P$ is symmetric to $I$ with respect to point $K$ . Point $T$ on arc $AC$ of $\omega$ containing point $B$ is such that $\angle KCT = \angle PCI$ . Show that the bisectors of angles $AKC$ and $ATC$ meet on line $CI$ .
*(Proposed by Anton Trygub)* | <blockquote>Let $\omega$ denote the circumscribed circle of triangle $ABC$ , $I$ be its incenter, and $K$ be any point on arc $AC$ of $\omega$ not containing $B$ . Point $P$ is symmetric to $I$ with respect to point $K$ . Point $T$ on arc $AC$ of $\omega$ containing point $B$ is such that $\angle KCT = \angle PCI$ . Show that the bisectors of angles $AKC$ and $ATC$ meet on line $CI$ .
</blockquote>
Let $D$ be intersection of $\overline{CI}$ and the angle bisector of $\angle AKC$ . Let $I_B$ be the $B$ -excenter of $\triangle ABC$ and let $M$ be the midpoint of arc $AC$ , not containing $B$ . Let $Q$ be reflection of $I$ over $D$ .**Claim.** $I_BPQC$ is cyclic.
*Proof.* By homothety, it is sufficient to show that $\measuredangle MKD=90^\circ$ . Firstly, observe that $AIKD$ is cyclic since $\measuredangle DIA=90^\circ+\tfrac{1}{2}\measuredangle CBA=\tfrac{1}{2}\measuredangle CKA=\measuredangle DKA$ . Now indeed, \begin{align*}
\measuredangle MKD=\measuredangle MKA+\measuredangle AKD=\measuredangle IBA+\measuredangle AID=90^\circ,
\end{align*}as desired. $\square$ Now, \begin{align*}
\measuredangle DMK=\measuredangle QI_BP=\measuredangle QCP=\measuredangle ICP=\measuredangle TCK=\measuredangle TMK,
\end{align*}which means that $\overline{TM}$ passes through $D$ , we are done. $\blacksquare$ | [
"<blockquote>Am I tired or something but this problem also does not make any sense for me, could you check your problem.</blockquote>\n\nSorry, I checked and everything seems right, what exactly is your problem?",
"<blockquote><blockquote>Let $\\omega$ denote the circumscribed circle of triangle $ABC$ , $I$ ... | [
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Positive reals $x, y, z$ satisfy $$ \frac{xy+1}{x+1} = \frac{yz+1}{y+1} = \frac{zx+1}{z+1} $$ Do they all have to be equal?
*(Proposed by Oleksii Masalitin)* | $$ xyz + yz + x + 1 = xy^2 + xy + y + 1,
xyz + xy + y + 1 = y z^2 + yz + z + 1,xyz + xy + z + 1 =z x^2 + zx + x + 1 $$ Summing all equations and after canceling we get $$ xy^2 + yz^2+zx^2=3xyz $$ By AM - GM , we have $$ xy^2 + yz^2+zx^2\geq 3xyz $$
So $ x = y = z.$ <details><summary>Very nice.</summary><blockquote>After multiplying we get :
xyz + xy + z + 1 = x^2*z + xz + x + 1
xyz + yz + x + 1 = x*y^2 + xy + y + 1
xyz + xy + y + 1 = y * z^2 + yz + z + 1
Summing all equations and after canceling we get:
3*xyz = z* x^2 + x*y^2 + y*z^2
Since x, y and z are positive real numbers we have:
z*x^2 + x*y^2 + y*z^2 >= 3* xyz ( By AM - GM inequality ), so x = y = z.</blockquote></details>
| [
" $\\frac{xy+1}{yz+1}=\\frac{x+1}{y+1} \\to \\frac{y(x-z)}{yz+1}=\\frac{x-y}{y+1}$ same way $\\frac{z(y-x)}{zx+1}=\\frac{y-z}{z+1},\\frac{x(z-y)}{xy+1}=\\frac{z-x}{x+1}$ If we multiply it, then we get $ \\frac{xyz(x-z)(y-x)(z-y)}{(xy+1)(yz+1)(zx+1)}= \\frac{(x-y)(y-z)(z-x)}{(x+1)(y+1)(z+1)}$ or $(x-y)(y-z)(z-x)... | [
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Let $AH_A, BH_B, CH_C$ be the altitudes of triangle $ABC$ . Prove that if $\frac{H_BC}{AC} = \frac{H_CA}{AB}$ , then the line symmetric to $BC$ with respect to line $H_BH_C$ is tangent to the circumscribed circle of triangle $H_BH_CA$ .
*(Proposed by Mykhailo Bondarenko)* | [asy]
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Nice geo, :D
Assume that the circle $(AH_BH_C)$ touches $B'C'$ the reflection of $BC$ across $H_BH_C$ and let it touch $B'C'$ in point $D'$ .
When we reflect the circle across $H_BH_C$ we know that the new circle must touch $BC$ , let it touch in point $D$ .
Now notice the following: $$ \angle H_CDH_B = \angle H_CD'H_B = \angle H_CAH_B = \alpha $$ $$ \angle H_CHH_B = 180 - \alpha = \angle H_CH'H_B $$ where $H'$ is the reflection of $H$ across $H_BH_C$ . The two conditions layed out give us that $H_CH'H_BD$ is cyclic, impliy that $H$ is the orthocenter of $\triangle H_CH_BD$ and that $H'$ is the orthocenter of $\triangle H_CD'H_B$ . This implies that $D',H',H,D$ all belong to a line.
From here we get that $\angle H_CH_BD = \gamma$ and that $\angle H_BH_CD = \beta$ , these two imply that $H_BD \parallel AB$ and that $H_CD \parallel AC$ , this implies that $AH_B =DH_C$ and that $DH_B=AH_C$ .
From the angle conditions we have that $H_CDB \sim H_BCD \sim ACB$ , which gives us the following relation: $$ \frac{AC}{AB}=\frac{H_BC}{H_BD} = \frac{H_BC}{AH_C} $$ which implies the following condtion $\frac{H_BC}{AC} = \frac{AH_C}{AB}$ .
Thus if the last condition holds, we have the tangency condition. | [
"Let $D$ lie on $BC$ s.t. $AC//DH_C$ .\nBy the condition, we have $DH_B//AB$ .\nThis implies $(AH_BH_C)$ is symmetric to $(DH_BH_C)$ wrt $H_BH_C$ .\nSince $AH\\bot BC$ , we have the tangent at $D$ wrt $(DH_BH_C)$ is parallel to $BC$ , which implies $BC$ is tangent to $(DH_BH_C)$ .\nSo we're done... | [
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Prime $p>2$ and a polynomial $Q$ with integer coefficients are such that there are no integers $1 \le i < j \le p-1$ for which $(Q(j)-Q(i))(jQ(j)-iQ(i))$ is divisible by $p$ . What is the smallest possible degree of $Q$ ?
*(Proposed by Anton Trygub)* | <blockquote>Prime $p>2$ and a polynomial $Q$ with integer coefficients are such that there are no integers $1 \le i < j \le p-1$ for which $(Q(j)-Q(i))(jQ(j)-iQ(i))$ is divisible by $p$ . What is the smallest possible degree of $Q$ ?
*(Proposed by Anton Trygub)*</blockquote>
<details><summary>Theorem</summary>see [here](https://laikhanhhoangchuyenndu.blogspot.com/2021/12/a-useful-lemma.html)</details>
<details><summary>About the deg of Q</summary>Assume $Q$ is the polynomial with the smallest degree satisfied the statement**<span style="color:#f00">Claim 1:</span>** $deg(Q) = p-2$ <span style="color:#00f">Proof:</span> $\bullet$ $deq(Q)>p-1$ . Then use the identity $x^p \equiv x \textrm{(mod p)}$ , we get a contradiction with $deg(Q)$ is smallest $\bullet$ $deq(Q)=p-1$ . Then we can use the identity $x^{p-1} \equiv 1 \textrm{(mod p)}$ because the problem is just care about $\overline{1,p-1}$ $\bullet$ $deg(Q)<p-2$ . We have two case to kill:**<span style="color:#0f0">Case 1:</span>** $(Q(i),p)=1, \forall i=\overline{1,p-1}$ We have $\left\{Q(1),Q(2),..,Q(p-1) \right\}$ and $\left\{ 1\times Q(1),2\times Q(2),...,(p-1)\times Q(p-1)\right\}$ are two full residues mod p
So by Wilson theorem, we get: $\left\{\begin{matrix}Q(1)\times Q(2)\times ...\times Q(p-1)\equiv -1 (\textrm{mod p})
&
& (1.Q(1))\times (2.Q(2))... \times ((p-1).Q(p-1))\equiv - 1 (\textrm{mod p})
\end{matrix}\right.$ But $$ (1.Q(1))\times (2.Q(2))...\times ((p-1).Q(p-1)) \equiv (1.2...(p-1))\times \left [ Q(1).Q(2)...Q(p-1) \right ]\equiv (-1).(-1) \equiv 1 (\textrm{mod p}) $$
a contradiction**<span style="color:#0f0">Case 2:</span>** Exist $i: Q(i) \vdots p$ Because $deg(Q)<p-2$ , $deg(xQ(x)) \leq p-2$ .
So we have the identity: $\sum_{i=0}^{p-1}iQ(i)\equiv 0 (\textrm{mod p}) \leftrightarrow \sum_{i=1}^{p-1}iQ(i)\equiv 0 (\textrm{mod p}), (1)$
But exist $i:Q(i) \vdots p$ and $\left\{ 1\times Q(1),2\times Q(2),...,(i-1)Q(i-1),(i+1)Q(i+1),...(p-1)\times Q(p-1)\right\} \sim (1,2,...x-1,x+1,...,p-1)$
From $(1)$ , we get: $1+2+...+(p-1)-x \equiv 0 (\textrm{mod p}) \leftrightarrow x\equiv 0 (\textrm{mod p}) $ , a contradiction
Q.E.D</details>
<details><summary>About the form of Q</summary>I haven't got any ideas, can someone help me ?
one years back, is it just $Q(x)=x^{p-2}+1$ ? :rotfl:</details>
P/s: My 409-th post. This is a beautiful number for me :-D | [] | [
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Initially memory of computer contained a single polynomial $x^2-1$ . Every minute computer chooses any polynomial $f(x)$ from its memory and writes $f(x^2-1)$ and $f(x)^2-1$ to it, or chooses any two distinct polynomials $g(x), h(x)$ from its memory and writes polynomial $\frac{g(x) + h(x)}{2}$ to it (no polynomial is ever erased from its memory). Can it happen that after some time, memory of computer contains $P(x) = \frac{1}{1024}(x^2-1)^{2048} - 1$ ?
*(Proposed by Arsenii Nikolaiev)* | <blockquote>Initially memory of computer contained a single polynomial $x^2-1$ . Every minute computer chooses any polynomial $f(x)$ from its memory and writes $f(x^2-1)$ and $f(x)^2-1$ to it, or chooses any two distinct polynomials $g(x), h(x)$ from its memory and writes polynomial $\frac{g(x) + h(x)}{2}$ to it (no polynomial is ever erased from its memory). Can it happen that after some time, memory of computer contains $P(x) = \frac{1}{1024}(x^2-1)^{2048} - 1$ ?
*(Proposed by Arsenii Nikolaiev)*</blockquote>
No.
Easy to prove that for $u=\frac{1+\sqrt 5}2$ , root of $u^2-1=u$ , any function in memory is such that $f(u)=u$ And easy to prove that $P(u)=u$ is wrong : it would be $u^{2048}=1024(u+1)$ which is wrong since $u^{2048}>2^{1024}>2^{12}=1024(3+1)>1024(u+1)$ | [] | [
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Find the largest $k$ for which there exists a permutation $(a_1, a_2, \ldots, a_{2022})$ of integers from $1$ to $2022$ such that for at least $k$ distinct $i$ with $1 \le i \le 2022$ the number $\frac{a_1 + a_2 + \ldots + a_i}{1 + 2 + \ldots + i}$ is an integer larger than $1$ .
*(Proposed by Oleksii Masalitin)* | <details><summary>Answer</summary>$k = 1011$</details>
<details><summary>Solution</summary>Let $b_i = \frac{a_1 + a_2 + \ldots + a_i}{1 + 2 + \ldots + i}$ .
Now, note that $k = 1011$ is achievable by letting $a_i = 2i~~\forall i \leq 1011$ and completing the permutation in any way. This will have $b_i = 2$ for $i \leq 1011$ .
It remains to show the upper bound. I claim that at most 1 $i \geq 1011$ can have $b_i$ be an integer larger than 1.
First note that $b_i \leq \frac{n + (n-1) + \ldots + (n-i+1)}{1 + 2 + \ldots + i} = \frac{n + n - i + 1}{1 + i} \leq \frac{3034}{1012} < 3$ for $i \geq 1011$ , so the only possible value of $b_i$ is 2.
Now note that if $b_i \leq 2$ then $b_{i+1} < 2$ . We can see this because $\frac{a_{i+1}}{i+1} \leq \frac{2022}{i+1} \leq \frac{2022}{1012} < 2$ and as $b_{i+1}$ is the mediant of $b_i$ and $\frac{a_{i+1}}{i+1}$ this implies $b_{i+1} < 2$ . This means that if $b_i = 2$ for any $i \geq 1011$ then all further $b_j, j > i$ will be less than $2$ and cannot be integers.
Thus the maximum number of $b_i$ that are integers larger than $1$ , is the first $1010$ and at most $1$ from the remaining terms. Thus $k \leq 1011$ as desired.</details> | [] | [
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Let $ABCD$ be the cyclic quadrilateral. Suppose that there exists some line $l$ parallel to $BD$ which is tangent to the inscribed circles of triangles $ABC, CDA$ . Show that $l$ passes through the incenter of $BCD$ or through the incenter of $DAB$ .
*(Proposed by Fedir Yudin)* | Do I miss something or the problem is not true. Take any triangle $ABC$ with $BA\neq BC$ , let $D$ be reflection of $B$ over $AC$ . For this quadrilateral, the statement does not hold even though $ABCD$ has incircle and there is line $l$ parallel to $BD$ tangent to incircles of triangles $ABC,CDA$ by symmetry. | [
"<blockquote>Do I miss something or the problem is not true. Take any triangle $ABC$ with $BA\\neq BC$ , let $D$ be reflection of $B$ over $AC$ . For this quadrilateral, the statement does not hold even though $ABCD$ has incircle and there is line $l$ parallel to $BD$ tangent to incircles of triangles... | [
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"answer_score": 18,
"boxed": false,
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"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 2/2769235.json"
} |
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that \[ f(a-b)f(c-d) + f(a-d)f(b-c) \leq (a-c)f(b-d) \] for all real numbers $a, b, c,$ and $d$ . | We claim the only solution is given by $f(x)=x$ or $f(x)=0$ . Clearly these work.
Assume $f$ is not identically $0$ . If we plug in $a=c$ , we get
\[f(a-d)\left[f(a-b)+f(b-a)\right]\leq 0\]
Letting $y=a-d$ and $x=b-a$ (and noting these are independent), we get
\[f(y)(f(x)+f(-x))\leq 0\]
Assuming that $f(x)\neq -f(-x)$ , we get that $f(y)$ must have a constant sign as $y$ ranges over $\mathbb R$ . If we take $y$ such that $f(y)\neq 0$ , However, this means that $f(x)+f(-x)$ shares that sign as well, a contradiction. Thus, $f(x)=-f(-x)$ .
Now, we plug in $a=b$ . Then, we get
\[f(a-d)f(a-c)\leq(a-c)f(a-d)\]
Thus, letting $x=a-d$ and $y=a-c$ , we get
\[f(x)\left[f(y)-y\right]\leq 0\]
However, we also know that
\[0\leq -f(x)=\left[f(y)-y\right]=f(-x)\left[f(y)-y\right]\leq 0\]
which implies everything is an equality. As $f$ is not identically $0$ , we get $f(y)=y$ for all $y\in\mathbb R$ , as desired. | [
" $P(a,a,a,a)$ gives $f(0)=0$ .\n\nIf there's a $x$ such that $f(x)<0$ . $P(a,a,c,d)$ gives $f(a-d)f(a-c)<=(a-c)f(a-d)$ \nNow if we take $d$ such that $a-d=x$ we have $f(a-c)<=a-c$ Of we take $d$ such that $f(a-d)>0$ we have $f(a-c)>=a-c$ So $f(x)=x$ If there is no $x$ such that $f(x)<0$ the... | [
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"answer_score": 44,
"boxed": false,
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"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2804292.json"
} |
The PMO Magician has a special party game. There are $n$ chairs, labelled $1$ to $n$ . There are $n$ sheets of paper, labelled $1$ to $n$ .
- On each chair, she attaches exactly one sheet whose number does not match the number on the chair.
- She then asks $n$ party guests to sit on the chairs so that each chair has exactly one occupant.
- Whenever she claps her hands, each guest looks at the number on the sheet attached to their current chair, and moves to the chair labelled with that number.
Show that if $1 < m \leq n$ , where $m$ is not a prime power, it is always possible for the PMO Magician to choose which sheet to attach to each chair so that everyone returns to their original seats after exactly $m$ claps. | Really, the problem is asking for us to create a bunch of cycles (of length $\ge 2$ ) on $n$ vertices such that every vertex is part of exactly one cycle and such that the lcm of cycle lengths is $m$ . If $m$ is not a prime power, say its $m = st$ where $s,t$ are coprime and $> 1$ , since $m > st - s - t$ , there exist nonnegative integers $x,y$ so that $n = sx + ty$ by the chicken mc nugget theorem. Take $x$ cycles of length $s$ and $y$ cycles of length $t$ so the lcm of cycle lengths is indeed $m$ and we use all $n$ vertices.
In fact, if $m$ was a prime power of a prime say $p^k$ with $p$ not dividing $n$ , then clearly this is not possible since each cycle length must be divisible by $p$ but $n$ is not. But if we had $p | n$ , then take a cycle of length $p^k$ , then $\frac{n-p^k}{p}$ cycles of length $p$ , which works.
thanks @below - fixed now | [
"@above I think you made a typo when you wrote $m=sx+ty$ . It should be $n=sx+ty$ . Nice solution btw.",
"oh wait pmo is done already?"
] | [
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 Philippine MO/2804299.json"
} |
Call a lattice point *visible* if the line segment connecting the point and the origin does not pass through another lattice point. Given a positive integer $k$ , denote by $S_k$ the set of all visible lattice points $(x, y)$ such that $x^2 + y^2 = k^2$ . Let $D$ denote the set of all positive divisors of $2021 \cdot 2025$ . Compute the sum \[ \sum_{d \in D} |S_d| \]
Here, a lattice point is a point $(x, y)$ on the plane where both $x$ and $y$ are integers, and $|A|$ denotes the number of elements of the set $A$ . | If $(x, y)$ is a visible lattice point, then $x$ and $y$ are integers such that $gcd(x, y) = 1$ .
If $gcd(x, y) = 1$ , then $gcd(x, k^2) = gcd(y, k^2) = 1$ . So there exists an integer $m$ such that $y \equiv mx \pmod{k^2}$ . Then $m^2 \equiv -1 \pmod{k^2} \implies m^2 \equiv -1 \pmod{k}$ .
If a prime number $p \equiv 3 \pmod{4}$ divides $k$ , then $m^2 \equiv -1 \pmod{p}$ . But $-1$ is not a quadratic residue modulo $p$ , contradiction.
Thus, if $k$ is odd, then $k$ must only contain prime divisors that are $1 \pmod {4}$ .
Note that $2021 \cdot 2025 = 3^4 \cdot 5^2 \cdot 43 \cdot 47$ . The only prime factor that is $1 \pmod{4}$ is $5$ .
Thus, $\sum_{d \in D} |S_d| = |S_1| + |S_5| + |S_{25}|$ .
It's easy to check that the only solutions of $x^2 + y^2 = 1$ are $(1, 0), (-1, 0), (0, -1), (0, 1)$ . Since $gcd(0, 1) = 1$ , we have $|S_1| = 4$ .
For $k > 1$ , $gcd(0, k) = k > 1$ , so $x, y \ne 0, k$ .
It's easy to check that the only solutions of $x^2 + y^2 = 25$ are $(4, 3), (-4, 3), (4, -3), (-4, -3), (3, 4), (-3, 4), (3, -4), (-3, -4)$ . Since $gcd(3, 4) = 1$ , we have $|S_5| = 8$ .
It's easy to check that the only solutions of $x^2 + y^2 = 625$ are $(24, 7), (-24, 7), (24, -7), (-24, -7), (7, 24), (-7, 24), (7, -24), (-7, -24)$ . Since $gcd(7, 24) = 1$ , we have $|S_{25}| = 8$ .
Therefore, $\sum_{d \in D} |S_d| = |S_1| + |S_5| + |S_{25}| = 4 + 8 + 8 = \boxed{20}$ . | [
"We can note that the visible condition rewrites to $(x,y)$ is *visible* if and only if $\\gcd(x,y)=1$ .\n\nThus, we're asked to find the number of primitive solutions to $x^2+y^2=k^2$ as $k$ ranges over the divisors of $2021\\cdot 2025$ .\n\nWe first assume $x=0$ (where the case $y=0$ is identical, and... | [
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"boxed": true,
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"path": "Contest Collections/2022 Contests/2022 Philippine MO/2804304.json"
} |
Let $\triangle ABC$ have incenter $I$ and centroid $G$ . Suppose that $P_A$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$ , and $Q_A$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$ . Define $P_B$ , $P_C$ , $Q_B$ , and $Q_C$ similarly. Show that $P_A, P_B, P_C, Q_A, Q_B,$ and $Q_C$ lie on a circle whose center is on line $IG$ . | Let $M_A, M_B, M_C$ be the midpoints of $BC, CA, AB$ respectively and let $I_A, I_B, I_C$ be the $A, B, C$ -excenters of $\triangle ABC$ respectively.
Note that $\angle BP_AC = 90^{\circ}$ , so $M_AB = M_AC = M_AP_A$ . So $\angle BM_AP_A = 180^{\circ} - 2\angle CBI_A = 180^{\circ} - 2\left(90^{\circ} - \frac{\angle ABC}{2}\right) = \angle ABC$ . Hence, $M_AP_A \parallel AB$ , and since $M_AM_B \parallel AB$ , we get $P_A \in M_AM_B$ .
It can be proved analogously that $Q_B \in M_AM_B, Q_A, Q_C \in M_CM_A$ , and $P_B, P_C \in M_BM_C$ .
Then $M_AQ_B = M_AM_B + M_BQ_B = M_CA + M_BA = M_CQ_C + M_AM_C = M_AQ_C$ . Similarly, we can show that $M_BP_A = M_BP_C$ and $M_CQ_A = M_CP_B$ .
Thus, the angle bisector of $\angle M_BM_AM_C$ is the perpendicular bisector of both $P_AQ_A$ and $Q_BQ_C$ . Similarly, we can show that the angle bisector of $\angle M_CM_BM_A$ is the perpendicular bisector of both $P_CP_A$ and $P_BQ_B$ , and the angle bisector of $\angle M_AM_CM_B$ is the perpendicular bisector of both $Q_AP_B$ and $P_CQ_C$ .
Let $I_M$ be the incenter of $\triangle M_AM_BM_C$ . Then $I_MP_A = I_MQ_A = I_MP_B = I_MQ_B = I_MQ_C = I_MP_C$ , making $I_M$ the desired center of the six points. A homothety centered at $G$ takes $\triangle ABC$ to $\triangle M_AM_BM_C$ , and this same homothety must take $I$ to $I_M$ . Thus, $I_M$ lies on line $IG$ , and we are done. | [
"Let $I_A,I_B,I_C$ be the excenter of $ABC$ and $X,Y,Z$ be the midpoint of $BC,AC,AB$ .\n\n :arrow: Taylor circle of $I_AI_BI_C$ gives $P_CQ_CP_AQ_AP_BQ_B$ are on a circle.\n\n :arrow: $P_CQ_C//I_AI_B$ :arrow: $XQ_B//AB\\Rightarrow X,Y,Q_B \\text{ collinear}$ :arrow: $XP_C=XQ_B$ :arrow: If $J$ ... | [
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"answer_score": 72,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2804308.json"
} |
Find all positive integers $n$ for which there exists a set of exactly $n$ distinct positive integers, none of which exceed $n^2$ , whose reciprocals add up to $1$ . | Let $S_n = \{a_1, a_2, \dots, a_n\}$ be a set of $n$ distinct positive integers where $a_1 < a_2 <\dots< a_n \le n^2$ and $\sum_{i=1}^n \frac{1}{a_i} = 1$ .
For $n = 1$ , let $S_1 = \{1\}$ .
For $n = 2$ , $\frac{1}{a_1} + \frac{1}{a_2} = 1 \iff (a_1 - 1)(a_2 - 1) = 1 \implies (a_1, a_2) = (2, 2)$ . But $a_1 \ne a_2 $ , contradiction. So $S_2$ does not exist.
For $n \ge 3$ , we will show $S_n$ exists.
If $n$ cannot be expressed as $m^2+m$ for some integer $m \ge 2$ , then let $S_n = \{k(k+1) | 1 \le k \le n - 1\} \cup \{n\}$ . Then $\sum_{k=1}^{n-1} \frac{1}{k(k+1)} + \frac{1}{n} = 1 - \frac{1}{n} + \frac{1}{n} = 1$ .
If $n = m^2+m$ for some integer $m \ge 2$ (so $n \ge 6$ ), then let $S_n = \{k(k+1) | 2 \le k \le n - 3\} \cup \{3, 8, 24, n-2\}$ . Then $\sum_{k=2}^{n-3} \frac{1}{k(k+1)} + \frac{1}{n-2} + \frac{1}{3} + \frac{1}{8} + \frac{1}{24} = \frac{1}{2} - \frac{1}{n-2} + \frac{1}{n-2} + \frac{1}{2} = 1$ .
Since $3, 8, 24, 5, 10, 26$ cannot be expressed as $m^2+m$ for some integer $m \ge 2$ , then $S_n$ contains $n$ distinct positive integers.
Thus, the answer is $\boxed{\text{all positive integers} \: n \ne 2}$ . | [
"<details><summary>'Solution'</summary>We claim that all $n \\neq 2$ work. \nIt's not hard to see that $\\frac{1}{a_1} + \\frac{1}{a_2} = 1$ has no solution in positive integers by multiplying and factoring the resulting expression.\nTo prove that all other $n$ work, we will construct the numbers $a_1$ to ... | [
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} |
In $\triangle ABC$ , let $D$ be the point on side $BC$ such that $AB+BD=DC+CA.$ The line $AD$ intersects the circumcircle of $\triangle ABC$ again at point $X \neq A$ . Prove that one of the common tangents of the circumcircles of $\triangle BDX$ and $\triangle CDX$ is parallel to $BC$ . | Let $E$ and $F$ be the intersections of the angle bisectors of $\angle DXB$ and $\angle DXC$ with $(DXB)$ and $(DXC)$ respectively ( $E, F \ne D$ ). Let $Y$ and $Z$ be the projections from $E$ and $F$ to $BC$ respectively. Let $I$ be the incenter of $\triangle ABC$ and $P$ be the foot from $I$ to $BC$ .
Note that $D$ is the $A$ -extouch point of $\triangle ABC$ . It is well known that $BD = CP$ and $CD = BP$ .
We see that $E$ and $F$ are the arc midpoints of arcs $\overarc{DB}$ and $\overarc{DC}$ , so $EB = ED$ and $FC = FD$ . Thus, $Y$ and $Z$ are the midpoints of line segments $BD$ and $CD$ respectively.
Then $\angle EDY = \angle EDB = \angle EXB = \frac{1}{2}\angle AXB = \frac{1}{2}\angle ACB = \angle ICP$ . Together with $\angle DYE = \angle CPI = 90^{\circ}$ , we get $\triangle EDY \sim \triangle ICP$ . Similarly, we can show that $\triangle FDZ \sim \triangle IBP$ .
So $EY = IP \cdot \frac{YD}{PC} = IP \cdot \frac{YD}{BD} = \frac{1}{2}IP$ and $FZ = IP \cdot \frac{DZ}{BP} = IP \cdot \frac{DZ}{CD} = \frac{1}{2}IP = EY$ . Hence, $EFZY$ is a rectangle and $EF \parallel BC$ .
Then, $\angle FED = \angle EDB = \angle EBD$ and $\angle EFD = \angle FDC = \angle FCD$ , so $EF$ is tangent to $(DXB)$ and $(DXC)$ , and we are done.
| [
"Notice that $D$ is the $\\text{A - excenter touch point}$ in $\\triangle ABC$ , and perform a $\\mathcal{I} \\left (D, -\\sqrt{DB\\cdot DC} \\right) $ inversion to finish.\n"
] | [
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 Philippine MO/2805001.json"
} |
Let $a, b,$ and $c$ be positive real numbers such that $ab + bc + ca = 3$ . Show that \[ \dfrac{bc}{1 + a^4} + \dfrac{ca}{1 + b^4} + \dfrac{ab}{1 + c^4} \geq \dfrac{3}{2}. \] | $$ bc=x,ab=z,ac=y $$ $\frac{bc}{1+a^4} = \frac{x^3}{x^2+y^2z^2}= x - \frac{xy^2z^2}{x^2+y^2z^2}$ $$ \sum \frac{xy^2z^2}{x^2+y^2z^2} \leq \sum \frac{xy^2z^2}{2xyz} \leq \frac{3}{2} $$ $\blacksquare$ | [
"I can't believe I had to do this again.**Claim:** $\\tfrac{1}{1 + c^2} \\geq 1 - \\tfrac c2$ for $c \\geq 0$ . **Proof:** Just expand, and get $2 \\geq (2-c)(1 + c^2) \\iff c(c-1)^2 \\geq 0$ . Equality when $c = 0, 1$ . \n\nHence, $\\tfrac{1}{1 + x^4} \\geq \\tfrac{2 - x^2}{2}$ with equality at $x = 1$ fo... | [
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"boxed": false,
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"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2805004.json"
} |
The set $S = \{1, 2, \dots, 2022\}$ is to be partitioned into $n$ disjoint subsets $S_1, S_2, \dots, S_n$ such that for each $i \in \{1, 2, \dots, n\}$ , exactly one of the following statements is true:
(a) For all $x, y \in S_i$ , with $x \neq y, \gcd(x, y) > 1.$ (b) For all $x, y \in S_i$ , with $x \neq y, \gcd(x, y) = 1.$ Find the smallest value of $n$ for which this is possible. | Since there is no solution still, I will post sketch of my solution. I hope I didn't make any mistake.
The answer is $14$ . To construct $S_i$ s, let's first construct $S'_i$ s. Let $p_i$ be $i$ th prime number. $S'_i$ be set of numbers that are less than $2023$ and divisible by $p_i$ , for $i=1,2,...,12$ . $S'_{13}=\{41\cdot 43,43\cdot 47,41\cdot 47\}$ . $S'_{14}$ is set of all prime powers less than $2023$ $($ including $1$ st power, I mean $p,p^2,...,p^r)$ and $1$ . Then take any number lies in $[1,2022]$ and remove it from all sets that it's in, except $1$ of them, so each number will be at most $1$ set. After this prosses we will get $S_i$ from $S'_i$ . Also since $41\cdot 53>2022$ we get all numbers are placed in exactly $1$ of the sets. It's obvious that all sets holds $1$ of statements, so we are done with construction. Main part of the problem is to prove $13$ is not valid. Using induction we can show that if we divide set $T_{n}=\{1,p_1\cdot p_2,p_1\cdot p_3,...,p_1\cdot p_{n},p_2\cdot p_3,...,p_2\cdot p_{n},...,p_{n-1}\cdot p_{n}\}$ into disjoint sets such that all sets holds exaclty $1$ of given statements, then we need at least $n-1$ sets. Observe that $T_{15}\subseteq S \implies n\geq 14$ . So we are done! | [
" $S_{14}'$ seems to have $1, 43$ and $43^2$ , which are not in any of the lower $S_i'$ , so $S_{14}$ fails the condition: ie $gcd(1,43)=1$ and $gcd(43,43^2)=43$ .\n\nThe official solutions doesn't seem to have a complete proof either, I can sketch one at some point if there is enough interest for me to b... | [
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"boxed": false,
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"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2805005.json"
} |
Given quadrilateral $ABCD$ inscribed into a circle with diagonal $AC$ as diameter. Let $E$ be a point on segment $BC$ s.t. $\sphericalangle DAC=\sphericalangle EAB$ . Point $M$ is midpoint of $CE$ . Prove that $BM=DM$ . | <details><summary>Solution</summary>Let $P=AE\cap (ABCD)$ $\textbf{Claim:}$ $M$ on $DP$ $\emph{Proof.}$ Let $M'=DP\cap EC$ . Thus $\angle CEP=\angle AEB=\frac \pi 2 -\angle EAB=\frac \pi 2 -\angle DAC=\angle ACD=\angle APD$ . So $\triangle M'PE$ is isosceles. Because $\angle APC=\angle ADC$ , $\triangle EPC$ is right-angled. Therefore $M'$ is the midpoint of $EC$ . So $M=M'$ , $P$ on $DM$ . $\Box $ So $\angle BDP=\angle BAP=\angle BAE=\angle DAC=\angle DBC$ . Thus $BM=DM$ as desired. $\Box $</details> | [
"Let $AE\\cap (ABCD)=D'$ . Then, because $M$ is the midpoint of $CE$ , $\\angle MD'C = \\angle DAC$ , so $D,M,D'$ collinear. $\\triangle DCM \\cong BD'M$ which finishes the proof.",
" $AE$ and $(ABCD)$ intersection $X$ . $XD$ and $BC$ intersection $Y$ .Very easy $Y$ is midpoint of $CE$ ",
"... | [
"origin:aops",
"2022 Contests",
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"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Poland - Second Round/2779163.json"
} |
Let $n$ be an positive integer. We call $n$ $\textit{good}$ when there exists positive integer $k$ s.t. $n=k(k+1)$ . Does there exist 2022 pairwise distinct $\textit{good}$ numbers s.t. their sum is also $\textit{good}$ number?
| <blockquote>Let $n$ be an positive integer. We call $n$ $\textit{good}$ when there exists positive integer $k$ s.t. $n=k(k+1)$ . Does there exist 2022 pairwise distinct $\textit{good}$ numbers s.t. their sum is also $\textit{good}$ number?</blockquote>
Yes,
Choose $2021$ any parwise distinct good numbers. Let $S$ be their sum.
Note that each good number is even and so is $S$ Juste write $S+(\frac S2-1)\frac S2=\frac S2(\frac S2+1)$ LHS is a sum of $2022$ good numbers (obviously paitrwise distinct since $(\frac S2-1)\frac S2$ is greater than $S$ and so different of each existing number)
RHS is a good number.
Q.E.D.
| [
"an example can be constructed for this problem or for any natural number in general",
"This problem was proposed by **[Burii](https://artofproblemsolving.com/community/user/100466)**.",
"Let $G$ be the set of all good numbers.\n\nObviously we must have that if $n\\in G$ then $n$ must be even.\n<span styl... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Poland - Second Round/2779164.json"
} |
$n$ players took part in badminton tournament, where $n$ is positive and odd integer. Each two players played two matches with each other. There were no draws. Each player has won as many matches as he has lost. Prove that you can cancel half of the matches s.t. each player still has won as many matches as he has lost. | <details><summary>Click to expand</summary>Define a graph, where there is a directed edge $A \rightarrow B$ , when $A$ won 2 times with $B$ , and undirected edge $A - B$ , when $A$ won one time and $B$ also won one time. Note that for every pair of vertices there exist exactly one (un)directed edge. Now let's deleted directed cycles, i.e. for a directed cycle $A_1 \rightarrow A_2 \rightarrow \ldots \rightarrow A_m \rightarrow A_1$ , we cancel one of the matches $A_i \rightarrow A_{i+1}$ , where $A_{n+1} = A_1$ . Then we can ignore the cycle, and just delete it. Repeat this process. We can see that it terminates and also at the end there cannot be any directed edges. Suppose not, let $A_1\rightarrow A_2 \rightarrow \ldots \rightarrow A_k$ be the longest directed path, by the assumption we know that there exists $A_r$ such that $A_k \rightarrow A_r$ , because $A_k$ lost to $A_{k-1}$ . We know that $r\neq 1,2,\ldots,k-2$ , because it would form a cycle, so $A_1\rightarrow A_2\rightarrow\ldots\rightarrow A_k \rightarrow A_r$ is a longer that the longest, contradiction. So we left with an undirected graph with even degree of every vertex. So for every component there exist a Euler cycle, so for every Euler cycle $B_1 - B_2 - \ldots - B_s - B_1$ , cancel matches $B_i - B_{i+1}$ , where $B_{s+1} = B_1$ . In that way we canceled exactly have matches.</details> | [
"I think its solvable easly with Ore's theorem [https://en.wikipedia.org/wiki/Ore%27s_theorem](https://en.wikipedia.org/wiki/Ore%27s_theorem). Credits to Franek W."
] | [
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"path": "Contest Collections/2022 Contests/2022 Poland - Second Round/2779165.json"
} |
Find all real quadruples $(a,b,c,d)$ satisfying the system of equations $$ \left\{ \begin{array}{ll}
ab+cd = 6
ac + bd = 3
ad + bc = 2
a + b + c + d = 6.
\end{array} \right. $$ | Adding the first two equations yields $(a+d)(b+c)=9.$ Since $a+d+b+c=6,$ \begin{align*}(a+d)(6-(a+d))=9&\implies (a+d)^2-6(a+d)+9=0&\implies (a+d-3)^2=0\end{align*} and $a+d=b+c=3.$ Similarly, $a+c=\{2,4\}$ and $a+b=\{1,5\}.$ Let $a+c=k_1,b+d=6-k_1,a+b=k_2,c+d=6-k_2$ by Vieta. We see $k_1=a+c=a+3-b$ and $k_2=a+b$ so $a=\frac{k_1+k_2-3}{2}$ and $b=\frac{-k_1+k_2+3}{2}.$ Letting $k_1=2,4$ and $k_2=1,5,$ we have $$ (a,b,c,d)=(0,1,2,3),(2,3,0,1),(1,0,3,2),(3,1,2,0). $$ We can test the solutions to find that they all work. $\square$ | [
"<blockquote>Find all real quadruples $(a,b,c,d)$ satisfying the system of equations $$ \\left\\{ \\begin{array}{ll}\nab+cd = 6 \nac + bd = 3 \nad + bc = 2 \na + b + c + d = 6.\n\\end{array} \\right.\t $$ </blockquote>\nCute problem. \nWe have $(a + d)(b + c) = (ab + cd) + (ac + bd) = 9$ and $(a + d) + (b + ... | [
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Given a cyclic quadriteral $ABCD$ . The circumcenter lies in the quadriteral $ABCD$ . Diagonals $AC$ and $BD$ intersects at $S$ . Points $P$ and $Q$ are the midpoints of $AD$ and $BC$ . Let $p$ be a line perpendicular to $AC$ through $P$ , $q$ perpendicular line to $BD$ through $Q$ and $s$ perpendicular to $CD$ through $S$ . Prove that $p,q,s$ intersects at one point. | Let $p,q$ cut $BD,AC$ again at $E,F$ . It is enough to prove $EF \parallel CD$ by orthocenter $\Leftrightarrow \frac{SC}{SD} = \frac{CF}{DE}$ but $\frac{CF}{DE} = \frac{CQ}{DP} = \frac{BC}{AD} = \frac{SC}{SD} \ \blacksquare.$ | [
"This problem was proposed by **[Burii](https://artofproblemsolving.com/community/user/100466)**.",
"Let $p$ meet $AC$ at $K$ and $q$ meet $BD$ at $T$ and $p,q$ meet at $X$ and $XS$ meet $DC$ at $H$ . we want to prove $\\angle XHD = \\angle 90$ . we will prove $XTHD$ is cyclic. $\\angle T... | [
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Positive integers $a,b,c$ satisfying the equation $$ a^3+4b+c = abc, $$ where $a \geq c$ and the number $p = a^2+2a+2$ is a prime. Prove that $p$ divides $a+2b+2$ . | Solved with **[proxima1681](https://artofproblemsolving.com/community/user/849747)**.
We first of all start by finding the parity of $a,b,c$ . Clearly $a$ must be odd because $p$ is prime. Now one can check that $c$ being odd is forced. The parity of $c$ and $a$ combined gives $b$ even.
Rearranging the equation we get
\[c(ab-1)=a^3+4b \implies ab-1 \mid a^3+4b\]
We now do some algebra do eliminate $b$ from Right-Hand-Side.
\begin{align*}
ab-1 &\mid a(a^3+4b)-4(ab-1)
ab-1 &\mid a^4+4
ab-1 &\mid (a^2+2a+2)(a^2-2a+2)
ab-1 &\mid p(a^2-2a+2)
\end{align*}
We would show that $\gcd(ab-1,p) \ne 1$ . Assume for sake of contradiction that $\gcd(ab-1,p) = 1$ . Then we get
\[ab-1 \mid a^2-2a+2\]
We do some more algebra to get
\begin{align*}
ab-1 &\mid a^2-2a+2
ab-1 &\mid a(a^2-2a+2)-(a^3+4b)
ab-1 &\mid -2a^2+2a-4b
\end{align*}
From here, we divide by 2 on RHS because $ab-1$ is odd.
\begin{align*}
ab-1 &\mid -a^2+a-2b
ab-1 &\mid -a^2+a-2b+(a^2-2a+2)
ab-1 &\mid 2b+a-2
\end{align*}
Now, one can clearly bound this by $ab-1 \le 2b+a-2 \iff (b-1)(a-2) \le 1$ . This restricts $(a,b)$ . Only possibilities from this inequality are $(a,b)= (1,b), (3,2)$ . Putting $(a,b)=(3,2)$ in original equation yields $c=7$ . For $(a,b)=(1,b)$ , putting $a=1$ and solving by factoring yields either $(b,c)=(6,5)$ or $(b,c)=(2,9)$ . We now write down all solutions.
\[(a,b,c)=(1,2,9) \quad (1,6,5) \quad (3,2,7)\]
Notice that none of the solutions follow $a \ge c$ . This is the desired contradiction.
Therefore we now get $\gcd(p,ab-1)=p \implies p \mid ab-1$ .The finish is just a little more algebra.
\begin{align*}
a^2+2a+2 &\mid ab-1
a^2+2a+2 &\mid (a^2+2a+2)+2(ab-1)
a^2+2a+2 &\mid a^2+2a+2ab
a^2+2a+2 &\mid a(a+2+2b)
\end{align*}
As $\gcd(a^2+2a+2,a)=1$ , we are done. $\blacksquare$ | [
"Interesting. \n\nWe have $$ a\\geq c=\\frac{a^3+4b}{ab-1}. $$ The inequality actually gives us that $a^2b\\geq a^3+4b+a>a^3\\implies b>a$ , which is useful later on. The divisibility gives us that $$ ab-1 \\mid (a^3+4b)a-4(ab-1)=a^4+4. $$ Observe that $p=a^2+2a+2\\mid a^4+4$ since $$ a^2+2a+2\\mid (a^2+... | [
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Let be given a parallelogram $ ABCD$ and two points $ A_1$ , $ C_1$ on its sides $ AB$ , $ BC$ , respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$ . Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$ . Prove that $ \angle PDA \equal{} \angle QBA$ . | Nice angle chasing exercise! :)
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[/asy]**Lemma 1.** Points $A_1, B, C, Q$ lie on a circle.
*Proof.* We have $$ \angle A_1QC=\angle A_1QP+\angle PQC=\angle A_1AP+\angle PC_1B=180^{\circ}-\angle A_1BC_1 \Longrightarrow A_1, B, C, Q \, \text{are concyclic}. \, \square $$ **Lemma 2.** Points $A, Q, A_2$ are collinear where $A_2$ is the second intersection of $(CC_1P)$ with $CD$ .
*Proof.* Note that from $A_2C \parallel A_1B$ we have $$ \angle AQP+\angle A_2QP=\angle PA_1B+\left(180^{\circ}-\angle PCA_2\right)=180^{\circ} \Longrightarrow A, Q, A_2 \, \text{are collinear}. \, \square $$ **Lemma 3.** Points $D, A_2, P, A$ are concyclic.
*Proof.* We have $$ \angle APA_2=\angle A_2CC_1=180^{\circ}-\angle ADA_1 \Longrightarrow A, D, A_2, P \, \text{are concyclic}. \, \square $$ As $$ \angle PDA=\angle PA_2Q=\angle QCA_1=\angle QBA, $$ we are done. | [
"<span style=\"color:darkblue\">From [**Virgil Nicula**'s extension](http://www.mathlinks.ro/viewtopic.php?p=1486021#1486021) we have $ \\frac {\\sin \\angle ABQ}{\\sin \\angle CBQ} \\equal{} \\frac {AA_1}{CC_1}$ Denote by $ E$ the intersection of the lines $ AD$ and $ CP$ . Then $ \\dfrac{\\sin \\angle ADP}{... | [
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A finite set $M$ of real numbers has the following properties: $M$ has at least $4$ elements, and there exists a bijective function $f:M\to M$ , different from the identity, such that $ab\leq f(a)f(b)$ for all $a\neq b\in M.$ Prove that the sum of the elements of $M$ is $0.$ | <blockquote>A finite set $M$ of real numbers has the following properties: $M$ has at least $4$ elements, and there exists a bijective function $f:M\to M$ , different from the identity, such that $ab\leq f(a)f(b)$ for all $a\neq b\in M.$ Prove that the sum of the elements of $M$ is $0.$ </blockquote>
(Not very sure if this is right, but....)
Borrowing notation from the above solutions, let $a_i$ be the elements of $M$ . Now $\sum_{i=1}^{n} {a_i}^2 = \sum_{i=1}^{n} {b_i}^2$ , and $\sum_{i=1}^{n} {a_i} = \sum_{i=1}^{n} {b_i}$ . Squaring the second equation and subtracting it from the first, we get that $\sum_ {1 \leq i < j \leq n} a_ia_j= \sum_ {1 \leq i < j \leq n} b_ib_j$ . Now it is given that $a_ia_j \leq b_ib_j$ , for all $ i \neq j$ . Summing up all such inequalities, and using the fact the sum taken twice at a time is equal, equality must hold in all inequalities. So $a_ia_j =b_ib_j$ for all $i \neq j$ . Now consider these $3$ equations:
\begin{align*}
a_ia_j =b_ib_j
a_ja_k =b_jb_k
a_ia_k =b_ib_k
\end{align*}
So multiplying first 2 and using third gives that ${a_j}^2 = {b_j}^2$ . So $a_j = \pm b_j$ for all $j$ .
<details><summary>Claim</summary>If for any element, $a_i= b_i$ , then $a_i=0$ .</details>
<details><summary>Proof</summary>Assume there is such an $a_i$ Firstly, all elements cannot map to each other, since the function is different from the identity, So there is another nonzero element such that $a_j= -b_j$ . Multiplying these two we get $a_i=0$ .</details>
So we get that $a_k= -b_k$ , for all $k$ . Summing it up, and using that $\sum_{i=1}^{n} {a_i} = \sum_{i=1}^{n} {b_i}$ , we get that $\sum_{i=1}^{n} {a_i}=0$ , and we're done
| [
"Let $M=\\{a_1,a_2,...,a_n\\}$ and $f(a_i)=b_i$ for all $1\\le i\\le n$ . Then $\\sum_{1\\le i<j\\le n}a_ia_j\\le \\sum_{1\\le i<j\\le n}b_ib_j= \\sum_{1\\le i<j\\le n }a_ia_j$ .So $a_ia_j=b_ib_j$ , for all $i\\ne j$ . And it means that $c=\\frac{b_i}{a_i}$ for all $i$ , where $c$ is constant. So $\\s... | [
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At first, on a board, the number $1$ is written $100$ times. Every minute, we pick a number $a$ from the board, erase it, and write $a/3$ thrice instead. We say that a positive integer $n$ is *persistent* if after any amount of time, regardless of the numbers we pick, we can find at least $n$ equal numbers on the board. Find the greatest persistent number. | We claim that the greatest $n$ is $67$ .
First, we show that we can always find at least $67$ equal numbers. Suppose that at some point that there are at most $66$ of each number, and that there are $m$ numbers written on the board. Then, note that each number written on the board is of the form $\frac1{3^i}$ for some integer $i$ . Let $a_i$ be the number of times $\frac1{3^i}$ is written on the board for any $1\le i\le m$ . Now note that $\sum_{i=1}^m\frac{a_i}{3^i}=100$ , since the sum of the numbers on the board is always the same. Then: $$ 100=\sum_{i=1}^m\frac{a_i}{3^i}\le\sum_{i=1}^m\frac{66}{3^i}\le\sum_{i=1}^\infty\frac{66}{3^i}=99 $$ which is a contradiction.
Second, we show that some sequence of choices lead to no more than $67$ equal numbers. Now first, we choose $33$ of the $1$ 's on the board, erase them, and instead write $99$ $\frac13$ 's. Next, choose $32$ of the $\frac13$ 's, erase them, and instead write $96$ $\frac19$ 's. <details><summary>etc.</summary>Next, choose $29$ of the $\frac19$ 's, erase them, and instead write $87$ $\frac1{81}$ 's. Next, choose $20$ of the $\frac1{81}$ 's, erase them, and instead write $60$ $\frac1{243}$ 's.</details> At this point, we have $67$ each of $1,\frac13,\ldots,\frac1{81}$ and $60$ of $\frac1{243}$ . | [
"gg nice problem\n<details><summary>Solution</summary>Clearly any number that ever appears on the board is of the form $1/3^{i}$ for some $i$ . Further the sum of the numbers on the board, is fixed, it is always $100$ . Letting $a_i$ denote the number of copies of $1/3^{i}$ at a certain fixed moment we have... | [
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Let $ABCD$ be a convex quadrilateral and let $O$ be the intersection of its diagonals. Let $P,Q,R,$ and $S$ be the projections of $O$ on $AB,BC,CD,$ and $DA$ respectively. Prove that \[2(OP+OQ+OR+OS)\leq AB+BC+CD+DA.\] | By symmetry, we only need to prove $2(OP+OR) \le AD+BC$ . It follows from these facts:
- Let the angle bisectors of $\angle AOB$ and $\angle COD$ be $OK$ and $OL$ ( $K,L$ lie on $AB,CD$ , respectively). Then $OP \le OK$ , $OQ \le OL$ .
- Let $M,N$ be the midpoint of $AB,CD$ , then $MN=|\overrightarrow{MN}|=|\tfrac{1}{2}(\overrightarrow{AD}+\overrightarrow{BC})| \le \tfrac{1}{2}(AD+BC)$ .
- In a triangle $XYZ$ , the projection of the midpoint of $YZ$ on the angle bisector of $\angle YXZ$ lies outside the triangle (or on the side $YZ$ ). (Obvious.) In this configuration, $|\text{the projection of } MN \text{ on the line } KOL| \ge KL$ .
Therefore, $2(OP+OR) \le 2(OK+OL)=2KL \le 2 \cdot |\text{the projection of } MN \text{ on the line } KOL| \le 2MN \le AD+BC$ . | [] | [
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For every positive integer $N\geq 2$ with prime factorisation $N=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ we define \[f(N):=1+p_1a_1+p_2a_2+\cdots+p_ka_k.\] Let $x_0\geq 2$ be a positive integer. We define the sequence $x_{n+1}=f(x_n)$ for all $n\geq 0.$ Prove that this sequence is eventually periodic and determine its fundamental period. | <blockquote><blockquote>Let $N$ be a composite number such that $N=p_1\cdots p_m$ , with $p_i$ being prime. Then the function is basically $1+\sum_{i=1}^m p_i$ .
Since $m$ is $\geq 2$ and $p_i\geq 2$ for all $i$ , this is bounded by $f(N)\leq 1+2(m-1)+\frac{N}{2^{m-1}}$ (note that $\frac{N}{2^m}\geq 2$ since $N\geq 2^m$ ); this is basically equivalent to maximizing $x_1+x_2+\cdots +x_m$ given $x_1\cdots x_m=N$ with $x_i\in [2,+\infty)$ (and $N\geq 2^m$ ).
Furthermore, we have $1+2(m-1)+\frac{N}{2^{m-1}}<N-1$ for all $m>2$ , because $N(1-\frac {1}{2^{m-1}})\geq 2^m(1-\frac{1}{2^{m-1}})=2^m-2$ By an easy induction, this is greater than $2m$ for all $m> 3$ , implying the result. For $m=2$ the wanted inequality is equivalent to $3+\frac{N}{2}<N\iff N>6$ . For $m=3$ the inequality is equivalent to $5+\frac N4<N-1\iff N>8$ .
Furthermore, if $N$ is prime, $f(N)=N+1$ is composite, and so $f(f(N))=f(N+1)<(N+1)-1=N$ if $N+1>8\iff N>7$ .
Therefore, for all $N>8$ one of $f(N)$ and $f(f(N))$ is less than $N$ . Therefore, every orbit eventually goes to a number which is at most $8$ .
Now, since $7\rightarrow 8\rightarrow 6,2\rightarrow 3\rightarrow 4\rightarrow 5\rightarrow 6\rightarrow 6$ , it follows that the unique cycle (apart from $f(1)=1$ , which anyway has period $1$ ) is $f(6)=1+2+3=6$ , which has a fundamental period of $1$ .
Also,</blockquote>
how you found that $f(8)=6$ ?? $f(8)= 1+(2)(3)=7$ </blockquote>
I'm sorry I had a mistake. By correcting it we only have the two cycles $6\rightarrow 6$ and $7\rightarrow 8$ , so the fundamental period is $1$ or $2$ depending on the starting numbers, so I guess we can say overall that the period is $2$ . | [
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Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that all real numbers $x$ and $y$ satisfy \[f(f(x)+y)=f(x^2-y)+4f(x)y.\] | Let $P(x,y)$ be the assertion.
<span style="color:#f00">Claim 1</span>: $f(0)=0$ Proof: $P(0,0)$ gives $f(f(0))=f(0)$ $P(0,-f(0))$ and using above fact gives that $f(0)=0$ . $\blacksquare$ <span style="color:#f00">Claim 2</span> : $\forall x$ , $f(x)=0$ or $x^2$
Proof: $P\left(x,\frac{x^2-f(x)}{2}\right)$ gives $f(x) \left(x^2-f(x)\right) =0 $
As we can vary $x$ over all real numbers, we get that $\forall x$ , $f(x)=0$ or $f(x)=x^2$ . $\blacksquare$ Hence, $f(1)=0$ or $f(1)=1$ .
<span style="color:#f00">Case 1</span> : $f(1)=0$ We prove that $f(x)=0 \forall x$ is the only function in this case.
Assume that there exists some $m$ such that $f(m) \neq 0 $
Using claim 2, we get that $f(m)=m^2$ $P(1,m)$ gives $m^2=f(1-m)$ Using claim 2, $m^2= \left(1-m \right)^2 $ Solving, we get that $m=\frac{1}{2}$ Hence, $f(m) \neq 0 \Rightarrow m=\frac{1}{2} ....(1)$ $P\left(\frac{1}{2},0\right)$ gives $f\left(\frac{1}{4}\right)=0$ contradicting (1).
Hence, no such $m$ exists.
Hence, $\boxed{f(x)=0 \forall x}$ which indeed works.
<span style="color:#f00">Case 2</span> : $f(1)=1$ We prove that $f(x)=x^2 \forall x$ is the only function in this case.
Assume that there exists some $m \neq 0$ such that $f(m) \neq m^2 $ Using claim 2, we get that $f(m)=0$ $P(m,1)$ gives $1=f(m^2-1)$ .
Using claim 2, we get that $1=\left(m^2-1\right)^2$ Solving we get that $m=0$ or $+-\sqrt2$ .
As $m \neq 0$ , we get that $f(m) \neq m^2 \Rightarrow m=+-\sqrt2 ....(2)$ But both $P(\sqrt2,0)$ and $P(-\sqrt2,0)$ give $P(2)=0$ Contradicting (2).
Hence, no such $m$ exists.
Using claim 2, we get that $\boxed{f(x)=x^2 \forall x}$ ,which indeed works | [
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"<blockquote>Determine all functions $f:\\mathbb{R}\\to\\mathbb{R}$ such that all real numbers $x$ and $y$ satisfy \\[f(f(x)+y)=f(x^2-y)+4f(x)y.\\]</blockquote>\n\nCool!\n\nLet \\(P(x,y)\\) denote the assertion of the given functi... | [
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} |
On a board there is a regular polygon $A_1A_2\ldots A_{99}.$ Ana and Barbu alternatively occupy empty vertices of the polygon and write down triangles on a list: Ana only writes obtuse triangles, while Barbu only writes acute ones.
At the first turn, Ana chooses three vertices $X,Y$ and $Z$ and writes down $\triangle XYZ.$ Then, Barbu chooses two of $X,Y$ and $Z,$ for example $X$ and $Y$ , and an unchosen vertex $T$ , and writes down $\triangle XYT.$ The game goes on and at each turn, the player must choose a new vertex $R$ and write down $\triangle PQR$ , where $P$ is the last vertex chosen by the other player, and $Q$ is one of the other vertices of the last triangle written down by the other player.
If one player cannot perform a move, then the other one wins. If both people play optimally, determine who has a winning strategy. | <blockquote><blockquote>Ana always wins.
Let's say $A$ and $B$ to the Ana and Barbu,respectively. Since $99$ is odd, $A$ will play the last move, so after any move of $B$ , there is at least $1$ unchosen vertex of polygon. Let's show that if $B$ chooses triangle $XYZ$ and $L$ is unchosen point, then at least $1$ of triangles $ZLX$ and $ZLY$ is obtuse. Obviously $XYZ$ is acute. If $L$ lies on arc $XZ$ that doesn't contain $Y$ , then $\angle XLZ=180-\angle XYZ>90$ ,so $XLZ$ is obtuse. Similarly if $L$ lies on arc $YZ$ that doesn't contain $X$ ,then $YLZ$ is obtuse. If $L$ lies on arc $XY$ that doesn't contain $Z$ , then since $\angle LXZ+\angle LYZ=180$ at least $1$ of $\angle LXZ$ and $\angle LYZ$ should be obtuse, because they can't be $90 ($ if a regular polygon has odd number of vertices, then center of this polygon doesn't lie on any of its diogonals $)$ . So at least one of $XLZ$ and $YLZ$ is obtuse, and $A$ can capture $L$ and write down $ZSL$ such that $ZSL$ is obtuse and $S\in\{X,Y\}$ .
Note: 300th posts :)</blockquote>
You have to take in account that even if XLZ is obtuse, it still doesn’t mean you can always take it. (Because if the last vertex chosen by Barbu was Y, your triangle must have Y as well.</blockquote>
No, since I assumed that $Z$ is the last vertex chosen by $B$ (I wrote that $B$ chooses $XYZ$ ), a triangle that $A$ will choose should contain point $Z$ , not $Y$ . | [
"Ana wins. At first turn Ana takes A1, A2, A3. Now pair up the other vertices (A4, A5), (A6, A7)…. (A98, A99). I claim that every time Barbu takes some vertice, Ana can take its pair. Wlog assume Barbu makes a triangle XYZ with Y being the new vertice. Let G be the pair of Y. And say Y and G are on different sides ... | [
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"path": "Contest Collections/2022 Contests/2022 Romania EGMO TST/2781229.json"
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