id stringlengths 16 16 | problem stringlengths 0 13.6k | answer stringlengths 0 1.87k | solution stringlengths 0 14.4k | source stringlengths 3 73 | origin stringclasses 2
values | domain stringclasses 683
values | category stringclasses 9
values | difficulty float64 -1 10 ⌀ | extra stringlengths 124 397 |
|---|---|---|---|---|---|---|---|---|---|
1fdfb68b7173f701 | For positive integers $m$ and $n$, let $d(m, n)$ be the number of distinct primes that divide both $m$ and $n$. For instance, $d(60,126)=d\left(2^{2} \times 3 \times 5,2 \times 3^{2} \times 7\right)=2$. Does there exist a sequence $\left(a_{n}\right)$ of positive integers such that:
(i) $a_{1} \geqslant 2018^{2018}$;
(... | proof | Such a sequence does exist.
Let $p_{1}a_{n-1}$. Thus $d\left(a_{m}, a_{n}\right)=d\left(b_{m}, b_{n}\right)=$ $d(m, n)$, and so all three requirements are satisfied. | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
8928a4545e51b438 | Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that
$$
n!+f(m)!\mid f(n)!+f(m!)
$$
for all $m, n \in \mathbb{N}$. | f(n)=n | Answer: $f(n)=n$ for all $n \in \mathbb{N}$.
Taking $m=n=1$ in $(*)$ yields $1+f(1)!\mid f(1)!+f(1)$ and hence $1+f(1)!\mid f(1)-1$. Since $|f(1)-1|<f(1)!+1$, this implies $f(1)=1$.
For $m=1$ in $(*)$ we have $n!+1 \mid f(n)!+1$, which implies $n!\leqslant f(n)$, i.e. $f(n) \geqslant n$.
On the other hand, taking $(m, ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
db8c5d45391a55d1 | Find all primes $p$ and $q$ such that $3 p^{q-1}+1$ divides $11^{p}+17^{p}$. | (p, q)=(3,3) | Answer: $(p, q)=(3,3)$.
For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$.
Observe that $N=11^{p}+17^{p} \equiv 4(\bmod 8)$, so $8 \nmid 3 p^{q-1}+1>4$. Consider an odd prime divisor $r$ of $3 p^{q-1}+1$. Obviously, $r \notin\{3,11,17\}$. There exists $b$ such that $17 b \equiv 1$ $(\bmod ... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
4b449e356f0463d6 | Let $P(x)=a_{d} x^{d}+\cdots+a_{1} x+a_{0}$ be a non-constant polynomial with nonnegative integer coefficients having $d$ rational roots. Prove that
$$
\operatorname{lcm}(P(m), P(m+1), \ldots, P(n)) \geqslant m\binom{n}{m}
$$
for all positive integers $n>m$. | proof | Let $x_{i}=-\frac{p_{i}}{q_{i}}(1 \leqslant i \leqslant d)$ be the roots of $P(x)$, where $p_{i}, q_{i} \in \mathbb{N}$ and $\operatorname{gcd}\left(p_{i}, q_{i}\right)=1$. By Gauss' lemma, we have $P(x)=c\left(q_{1} x+p_{1}\right)\left(q_{2} x+p_{2}\right) \cdots\left(q_{d} x+p_{d}\right)$ for some $c \in \mathbb{N}$,... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
9bc0fac2f57f9f2a | Let $x$ and $y$ be positive integers. If for each positive integer $n$ we have that
$$
(n y)^{2}+1 \mid x^{\varphi(n)}-1
$$
prove that $x=1$. | proof | Let us take $n=3^{k}$ and suppose that $p$ is a prime divisor of $\left(3^{k} y\right)^{2}+1$ such that $p \equiv 2$ $(\bmod 3)$.
Since $p$ divides $x^{\varphi(n)}-1=x^{2 \cdot 3^{k-1}}-1$, the order of $x$ modulo $p$ divides both $p-1$ and $2 \cdot 3^{k-1}$, but $\operatorname{gcd}\left(p-1,2 \cdot 3^{k-1}\right) \mid... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
db0f2cc420fd35d3 | Let $a_{0}$ be an arbitrary positive integer. Consider the infinite sequence $\left(a_{n}\right)_{n \geq 1}$, defined inductively as follows: given $a_{0}, a_{1}, \ldots, a_{n-1}$ define the term $a_{n}$ as the smallest positive integer such that $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by $n$. Prove that there exists a... | proof | Define $b_{n}=\frac{a_{0}+a_{1}+\ldots+a_{n}}{n}$ for every positive integer $n$. According to the condition, $b_{n}$ is a positive integer for every positive integer $n$.
Since $a_{n+1}$ is the smallest positive integer such that $\frac{a_{0}+a_{1}+\ldots+a_{n}+a_{n+1}}{n+1}$ is a positive integer and
$$
\frac{a_{0}... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
40749e0abfff2c7a | Let $a, b, c$ be real numbers such that $0 \leq a \leq b \leq c$. Prove that if
$$
a+b+c=a b+b c+c a>0,
$$
then $\sqrt{b c}(a+1) \geq 2$. When does the equality hold? | proof | Let $a+b+c=ab+bc+ca=k$. Since $(a+b+c)^{2} \geq 3(ab+bc+ca)$, we get that $k^{2} \geq 3k$. Since $k>0$, we obtain that $k \geq 3$.
We have $bc \geq ca \geq ab$, so from the above relation we deduce that $bc \geq 1$.
By AM-GM, $b+c \geq 2 \sqrt{bc}$ and consequently $b+c \geq 2$. The equality holds iff $b=c$.
The const... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
79023d7d7a6190eb | Let $a_{i j}, i=1,2, \ldots, m$ and $j=1,2, \ldots, n$, be positive real numbers. Prove that
$$
\sum_{i=1}^{m}\left(\sum_{j=1}^{n} \frac{1}{a_{i j}}\right)^{-1} \leq\left(\sum_{j=1}^{n}\left(\sum_{i=1}^{m} a_{i j}\right)^{-1}\right)^{-1} .
$$
When does the equality hold? | proof | We will use the following
Lemma. If $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ are positive real numbers then
$$
\frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}}}+\frac{1}{\sum_{j=1}^{n} \frac{1}{b_{j}}} \leq \frac{1}{\sum_{j=1}^{n} \frac{1}{a_{j}+b_{j}}}
$$
The equality holds when $\frac{a_{1}}{b_{1}}=\frac{a... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
097cab4716e4920f | Let $a, b, c$ be positive real numbers, such that $(a b)^{2}+(b c)^{2}+(c a)^{2}=3$. Prove that
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 .
$$ | proof | The inequality is equivalent with
$$
\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right) \geq 1 \Leftrightarrow\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right) \geq(a+1)(b+1)(c+1) .
$$
Thus:
$$
\begin{gathered}
\prod_{c y c}\left(a^{3}+1\right)-\prod_{c y c}(a+1)=\sum_{c y c} a^{3}+\sum_{c... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
65a62f885e01c7ad | Let $A B C D$ be a square of center $O$ and let $M$ be the symmetric of the point $B$ with respect to the point $A$. Let $E$ be the intersection of $C M$ and $B D$, and let $S$ be the intersection of $M O$ and $A E$. Show that $S O$ is the angle bisector of $\angle E S B$. | proof | We have
$$
\left\{\begin{array}{l}
D C \equiv D A \\
\angle E D C \equiv \angle E D A \quad \Rightarrow \triangle D E C \equiv \triangle D E A \Rightarrow \angle D A E \equiv \angle D C E(*) . \\
D E \equiv D E
\end{array}\right.
$$
Let $C M \cap A D=\{P\}$, then follows $\triangle C D P \equiv \triangle B A P$ and $... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
7e6bf932c4593ce6 | Let be a triangle $\triangle A B C$ with $m(\angle A B C)=75^{\circ}$ and $m(\angle A C B)=45^{\circ}$. The angle bisector of $\angle C A B$ intersects $C B$ at the point $D$. We consider the point $E \in(A B)$, such that $D E=D C$. Let $P$ be the intersection of the lines $A D$ and $C E$. Prove that $P$ is the midpoin... | proof | Let $P^{\prime}$ be the midpoint of the segment $A D$. We will prove that $P^{\prime}=P$. Let $F \in A C$ such that $D F \perp A C$. The triangle $C D F$ is isosceles with $F D=F C$ and the triangle $D P^{\prime} F$ is equilateral as $m(\angle A D F)=60^{\circ}$. Thus, the triangle $F C P^{\prime}$ is isosceles $\left(... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
6cfaf25c5bc615e8 | Let $A B C$ be a scalene and acute triangle, with circumcentre $O$. Let $\omega$ be the circle with centre $A$, tangent to $B C$ at $D$. Suppose there are two points $F$ and $G$ on $\omega$ such that $F G \perp A O, \angle B F D=\angle D G C$ and the couples of points $(B, F)$ and $(C, G)$ are in different halfplanes w... | proof | Consider any two points $F, G$ on $\omega$ such that $\angle B F D=\angle D G C$. Exploiting the isosceles triangles $\triangle A F G, \triangle A F D$, and $\triangle A D G$, we deduce (using directed angles throughout):
$$
\begin{gathered}
\angle D B F - \angle G C D = 180^{\circ} - \angle B F D - \angle B D F - \le... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
843546198af0ee75 | Given an acute triangle $A B C$, let $M$ be the midpoint of $B C$ and $H$ the orthocentre. Let $\Gamma$ be the circle with diameter $H M$, and let $X, Y$ be distinct points on $\Gamma$ such that $A X, A Y$ are tangent to $\Gamma$. Prove that $B X Y C$ is cyclic. | proof | Let $D$ be the foot of the altitude from $A$ to $B C$, which also lies on $\Gamma$. Let $O$ be the circumcentre of $\triangle A B C$. Since $\angle H D M=90^{\circ}$, note that rays $H D$ and $H M$ meet the circumcircle at points which are reflections in $O M$. Then, since $\angle B A D=\angle O A C$, we recover the we... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
74a2ecc261f1c3d3 | Let $A B C(B C>A C)$ be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $A B$ at the point $D$. The circumcircles of triangles $B C D, O C D$ and $A O B$ intersect the ray $C A$ (beyond $A$ ) at the points $Q, P$ and $K$, respectively, such that $P \in(A K)$ and $K... | proof | As $D C$ is tangent to $k$ at $C$ then $\angle O C D=90^{\circ}$. Denote by $X$ the midpoint of $A B$. Then $\angle O X A=90^{\circ}$ because of $O X$ is the perpendicular bisector of the side $A B$. The pentagon $P X O C D$ is inscribed in the circle with diameter $O D$, hence $\angle P X A=$ $\angle P X D=\angle P C ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
05481f6330a5659b | Let $A B C$ be an acute triangle, and $A X, A Y$ two isogonal lines. Also, suppose that $K, S$ are the feet of perpendiculars from $B$ to $A X, A Y$, and $T, L$ are the feet of perpendiculars from $C$ to $A X, A Y$ respectively. Prove that $K L$ and $S T$ intersect on $B C$. | proof | Denote $\phi=\widehat{X A B}=\widehat{Y A C}, \alpha=\widehat{C A X}=\widehat{B A Y}$. Then, because the quadrilaterals ABSK and ACTL are cyclic, we have
$$
\widehat{B S K}+\widehat{B A K}=180^{\circ}=\widehat{B S K}+\phi=\widehat{L A C}+\widehat{L T C}=\widehat{L T C}+\phi,
$$
so, due to the 90-degree angles formed,... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
79910f53d6076bdc | Let $A D, B E$, and $C F$ denote the altitudes of triangle $\triangle A B C$. Points $E^{\prime}$ and $F^{\prime}$ are the reflections of $E$ and $F$ over $A D$, respectively. The lines $B F^{\prime}$ and $C E^{\prime}$ intersect at $X$, while the lines $B E^{\prime}$ and $C F^{\prime}$ intersect at the point $Y$. Prov... | proof | We will prove that the desired point of concurrency is the midpoint of $B C$. Assume that $\triangle A B C$ is acute. Let $(A B C)^{5}$ intersect $(A E F)$ at the point $Y^{\prime}$; we will prove that $Y=Y^{\prime}$.
![](https://cdn.mathpix.com/cropped/2024_12_07_82afe765b60b274413c4g-24.jpg?height=1623&width=1529&top... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
8008c0e8418dd2ee | Given an acute triangle $ABC$, $(c)$ is the circumcircle with center $O$ and $H$ the orthocenter of the triangle $ABC$. The line $AO$ intersects $(c)$ at the point $D$. Let $D_{1}, D_{2}$ and $H_{2}, H_{3}$ be the symmetrical points of the points $D$ and $H$ with respect to the lines $AB, AC$ respectively. Let $\left(c... | proof | It is well known that the symmetrical points $H_{1}, H_{2}, H_{3}$ of $H$ with respect to the sides $B C, A B, A C$ of the triangle $A B C$ respectively lie on the circle (c).
Figure 8: G8... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
2a817d9cc2054339 | Given semicircle (c) with diameter $A B$ and center $O$. On the (c) we take point $C$ such that the tangent at the $C$ intersects the line $A B$ at the point $E$. The perpendicular line from $C$ to $A B$ intersects the diameter $A B$ at the point $D$. On the (c) we get the points $H, Z$ such that $C D=C H=C Z$. The lin... | proof | Since $C H = C Z$ we have $O C \perp H Z$. So from the cyclic quadrilateral $S O D I$ we get
$$
C S \cdot C O = C I \cdot C D.
$$
Figure 9: G9
We draw the perpendicular line $(v)$ to $H ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
76c2beea99bd52f1 | Let $\mathbb{P}$ be the set of all prime numbers. Find all functions $f: \mathbb{P} \rightarrow \mathbb{P}$ such that
$$
f(p)^{f(q)} + q^p = f(q)^{f(p)} + p^q
$$
holds for all $p, q \in \mathbb{P}$. | f(p)=p | Obviously, the identical function $f(p)=p$ for all $p \in \mathbb{P}$ is a solution. We will show that this is the only one.
First we will show that $f(2)=2$. Taking $q=2$ and $p$ any odd prime number, we have
$$
f(p)^{f(2)}+2^{p}=f(2)^{f(p)}+p^{2} .
$$
Assume that $f(2) \neq 2$. It follows that $f(2)$ is odd and so... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
b9bf75fea698162b | Let $S \subset\{1, \ldots, n\}$ be a nonempty set, where $n$ is a positive integer. We denote by $s$ the greatest common divisor of the elements of the set $S$. We assume that $s \neq 1$ and let $d$ be its smallest divisor greater than 1. Let $T \subset\{1, \ldots, n\}$ be a set such that $S \subset T$ and $|T| \geq 1+... | proof | Let $t$ be the greatest common divisor of the elements in $T$. Due to the fact that $S \subset T$, we immediately get that $t \mid s$. Let us assume for the sake of contradiction that $t \neq 1$. From the previous observation we get that $t \geq d$.
By taking into account that $|T| \geq 1+\left[\frac{n}{d}\right]$, we... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
e16362d33fc50cd7 | Let $n(n \geq 1)$ be a positive integer and $U=\{1, \ldots, n\}$. Let $S$ be a nonempty subset of $U$ and let $d(d \neq 1)$ be the smallest common divisor of all elements of the set $S$. Find the smallest positive integer $k$ such that for any subset $T$ of $U$, consisting of $k$ elements, with $S \subset T$, the great... | 1+\left[\frac{n}{d}\right] | We will show that $k_{\min }=1+\left[\frac{n}{d}\right]$ (here [.] denotes the integer part).
Obviously, the number of elements of $S$ is not greater than $\left[\frac{n}{d}\right]$, i.e. $|S| \leq\left[\frac{n}{d}\right]$, and $S \neq U$.
If $S \subset T$ and the greatest common divisor of elements of $T$ is equal to... | olympiads_ref | yoonholee/math-corpus-combined | 8 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
6d487b635bb2389b | 100 couples are invited to a traditional Moldovan dance. The 200 people stand in a line, and then in a step, two of them (not necessarily adjacent) may swap positions. Find the least $C$ such that whatever the initial order, they can arrive at an ordering where everyone is dancing next to their partner in at most $C$ s... | N-1 | With 100 replaced by $N$, the answer is $C=C(N)=N-1$. Throughout, we will say that the members of a couple have the same.
$N=2$ : We use this as a base case for induction for both bounds. Up to labelling, there is one trivial initial order, and two non-trivial ones, namely
$$
1,1,2,2 ; \quad 1, \sqrt{2,2,1} ; \quad 1,... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
cb87482785bfe7e4 | An $5 \times 5$ array must be completed with all numbers $\{1,2, \ldots, 25\}$, one number in each cell. Find the maximal positive integer $k$, such that for any completion of the array there is a $2 \times 2$ square (subarray), whose numbers have a sum not less than $k$. | 45 | We will prove that $k_{\max }=45$.
We number the columns and the rows and we select all possible $3^{2}=9$ choices of an odd column with an odd row.
Collecting all such pairs of an odd column with an odd row, we double count some squares. Indeed, we take some $3^{2}$ squares 5 times, some 12 squares 3 times and there ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
8c3ebd860c0e6421 | A town-planner has built an isolated city whose road network consists of 2 N roundabouts, each connecting exactly three roads. A series of tunnels and bridges ensure that all roads in the town meet only at roundabouts. All roads are two-way, and each roundabout is oriented clockwise.
Vlad has recently passed his drivi... | N \text{ odd} | $N$ odd. In fact, the number of trajectories has the same parity as $N$.
The setting is a (multi)graph where every vertex has degree three. Each vertex has an orientation, an ordering of its incident edges. We call Vlad's possible paths trajectories, and a complete trajectory if he traverses every edge in both directio... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
83850912f4f435d9 | Find all functions \( f: \mathbb{R}^{+} \rightarrow \mathbb{R} \) and \( g: \mathbb{R}^{+} \rightarrow \mathbb{R} \) such that
\[
f\left(x^{2}+y^{2}\right)=g(x y)
\]
holds for all \( x, y \in \mathbb{R}^{+} \).
## Proposed by Greece | proof | Given any $u \geqslant 2$, take $a, b \in \mathbb{R}^{+}$ such that $a+b=u$ and $a b=1$. This is possible as the equation $x^{2}-u x+1$ for $u \geqslant 2$ has two positive real solutions. (Discriminant is $u^{2}-4 \geqslant 0$, sum and product of solutions are positive.) Now taking $x=\sqrt{a}, y=\sqrt{b}$ we get $f(u... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
7488354e399baaee | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f\left(x^{2}+y\right) \geqslant\left(\frac{1}{x}+1\right) f(y)
$$
holds for all $x \in \mathbb{R} \backslash\{0\}$ and all $y \in \mathbb{R}$.
## Proposed by Uzbekistan | f(x)=0 | We will show that $f(x)=0$ for all $x \in \mathbb{R}$ which obviously satisfies the equation.
For $x=-1$ and $y=t+1$ we get $f(t) \geqslant 0$ for every $t \in \mathbb{R}$.
For $x=\frac{1}{n}$, we get that
$$
f\left(y+\frac{1}{n^{2}}\right) \geqslant(n+1) f(y) .
$$
Therefore
$$
f\left(y+\frac{2}{n^{2}}\right) \geqsl... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
743f5af17b50e8b0 | Find all functions \( f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+} \) such that
\[
f(x+f(x)+f(y))=2 f(x)+y
\]
holds for all \( x, y \in \mathbb{R}^{+} \).
## Proposed by Greece | proof | 1. We will show that \( f(x) = x \) for every \( x \in \mathbb{R}^{+} \). It is easy to check that this function satisfies the equation.
We write \( P(x, y) \) for the assertion that \( f(x + f(x) + f(y)) = 2 f(x) + y \).
We first show that \( f \) is injective. So assume \( f(a) = f(b) \). Now \( P(1, a) \) and \( P(... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
6560601f8df3a5f1 | Let $f, g$ be functions from the positive integers to the integers. Vlad the impala is jumping around the integer grid. His initial position is $\mathbf{x}_{0}=(0,0)$, and for every $n \geqslant 1$, his jump is
$$
\mathbf{x}_{n}-\mathbf{x}_{n-1}=( \pm f(n), \pm g(n)) \text { or }( \pm g(n), \pm f(n))
$$
with eight po... | proof | 1.
(a) Yes it is always possible. The key idea is the following: Let $b(n)$ be the number of 1's in the binary expansion of $n=0,1,2, \ldots$.
Lemma: Given a polynomial $f$ with integer coefficients and degree at most $d$, then
$$
\sum_{k=0}^{2^{d+1}-1}(-1)^{b(k)} f(n+k)=f(n)-f(n+1)-f(n+2)+\cdots \pm f\left(n+\left(... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
80a85c97f049a0f0 | Find all functions \( f: \mathbb{R}^{+} \longrightarrow \mathbb{R}^{+} \) such that
\[
f(x f(x+y))=y f(x)+1
\]
holds for all \( x, y \in \mathbb{R}^{+} \).
## Proposed by North Macedonia | f(x)=\frac{1}{x} | 1. We will show that $f(x)=\frac{1}{x}$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.
We write $P(x, y)$ for the assertion that $f(x f(x+y))=y f(x)+1$.
We first show that $f$ is injective. So assume $f\left(x_{1}\right)=f\left(x_{2}\right)$ and take any $x1$ there is ... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
f5e9b7ba215eb441 | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$$
f(x y)=f(x) f(y)+f(f(x+y))
$$
holds for all $x, y \in \mathbb{R}$.
## Proposed by Romania | proof | 1. We will show that $f(x)=0$ for every $x \in \mathbb{R}$ or $f(x)=x-1$ for every $x \in \mathbb{R}$. It is easy to check that both of these functions work.
We write $P(x, y)$ for the assertion that $f(x y)=f(x) f(y)+f(f(x+y))$. For later use we write $Q(x, y)$ for the assertion that $f(x y)=f(x) f(y)$ and $R(x, y)$ ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
eaf8518f0810a4a4 | Let $\mathcal{A}_{n}$ be the set of $n$-tuples $x=\left(x_{1}, \ldots, x_{n}\right)$ with $x_{i} \in\{0,1,2\}$. A triple $x, y, z$ of distinct elements of $\mathcal{A}_{n}$ is called good if there is some $i$ such that $\left\{x_{i}, y_{i}, z_{i}\right\}=\{0,1,2\}$. A subset $A$ of $\mathcal{A}_{n}$ is called good if e... | 2\left(\frac{3}{2}\right)^{n} | 1. We proceed by induction on $n$, the case $n=1$ being trivial. Let
$$
A_{0}=\left\{\left(x_{1}, \ldots, x_{n}\right) \in A: x_{n} \neq 0\right\}
$$
and define $A_{1}$ and $A_{2}$ similarly.
Since $A$ is good and $A_{0}$ is a subset of $A$, then $A_{0}$ is also good. Therefore, any three of its elements have a coord... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
195e08971068030b | Let $K$ and $N>K$ be fixed positive integers. Let $n$ be a positive integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be distinct integers. Suppose that whenever $m_{1}, m_{2}, \ldots, m_{n}$ are integers, not all equal to 0, such that $\left|m_{i}\right| \leqslant K$ for each $i$, then the sum
$$
\sum_{i=1}^{n} m_{i} a_{... | n=\left\lfloor\log _{K+1} N\right\rfloor | The answer is $n=\left\lfloor\log _{K+1} N\right\rfloor$.
Note first that for $n \leqslant\left\lfloor\log _{K+1} N\right\rfloor$, taking $a_{i}=(K+1)^{i-1}$ works. Indeed let $r$ be maximal such that $m_{r} \neq 0$. Then on the one hand we have
$$
\left|\sum_{i=1}^{n} m_{i} a_{i}\right| \leqslant \sum_{i=1}^{n} K(K+1... | olympiads_ref | yoonholee/math-corpus-combined | 8 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
9aec473e413c101a | In an exotic country, the National Bank issues coins that can take any value in the interval $[0,1]$. Find the smallest constant $c>0$ such that the following holds, no matter the situation in that country:
Any citizen of the exotic country that has a finite number of coins, with a total value of no more than 1000, ca... | 11-\frac{11}{1001} | 1. The answer is $c=\frac{1000}{91}=11-\frac{11}{1001}$. Clearly, if $c^{\prime}$ works, so does any $c>c^{\prime}$. First we prove that $c=11-\frac{11}{1001}$ is good.
We start with 100 empty boxes. First, we consider only the coins that individually value more than $\frac{1000}{1001}$. As their sum cannot overpass 1... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
ac2df18e3a79d9f6 | A sequence of $2 n+1$ non-negative integers $a_{1}, a_{2}, \ldots, a_{2 n+1}$ is given. There's also a sequence of $2 n+1$ consecutive cells enumerated from 1 to $2 n+1$ from left to right, such that initially the number $a_{i}$ is written on the $i$-th cell, for $i=1,2, \ldots 2 n+1$. Starting from this initial positi... | C_{n} \cdot C_{n} | The answer is: $C_{n} \cdot C_{n}$, where $C_{n}=\frac{1}{n+1}\binom{2 n}{n}$ is the $n$-th Catalan number.
We divide the proof into several steps. First, some terminology: the last (rightmost) $n$ cells will be called the back cells and the front (leftmost) $n$ cells will be called the front cells. The central, $(n+1)... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
b004076f33362c42 | Angel has a warehouse, which initially contains 100 piles of 100 pieces of rubbish each. Each morning, Angel either clears every piece of rubbish from a single pile, or one piece of rubbish from each pile. However, every evening, a demon sneaks into the warehouse and adds one piece of rubbish to each non-empty pile, or... | 199 | 1. We will show that he can do so by the morning of day 199 but not earlier.
If we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be
$$
V= \begin{cases}n & m=0 \\ n+\frac{1}{2} & m=1 \\ n+1 & m \geqslant 2\end{cases}
$$
We al... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
b3485132fd6ea025 | Let $A B C$ be a triangle with $A B < A C < B C$. On the side $B C$ we consider points $D$ and $E$ such that $B A = B D$ and $C E = C A$. Let $K$ be the circumcenter of triangle $A D E$ and let $F, G$ be the points of intersection of the lines $A D, K C$ and $A E, K B$ respectively. Let $\omega_{1}$ be the circumcircle... | proof | 1. Since the triangles $B A D, K A D$ and $K D E$ are isosceles, then $\angle B A D=\angle B D A$ and $\angle K A D=\angle K D A$ and $\angle K D E=\angle K E D$. Therefore,
$$
\angle B A K=\angle B A D-\angle K A D=\angle B D A-\angle K D A=\angle K D E=\angle K E D=180^{\circ}-\angle B E K .
$$
So the points $B, E,... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
0d5375dff8a00012 | Let $I$ and $O$ be the incenter and the circumcenter of a triangle $ABC$, respectively, and let $s_{a}$ be the exterior bisector of angle $\angle BAC$. The line through $I$ perpendicular to $IO$ meets the lines $BC$ and $s_{a}$ at points $P$ and $Q$, respectively. Prove that $IQ=2IP$.
## Proposed by Serbia | I Q=2 I P | Denote by $I_{b}$ and $I_{c}$ the respective excenters opposite to $B$ and $C$. Also denote the midpoint of side $B C$ by $D$, the midpoint of the arc $B A C$ by $M$, and the midpoint of segment $A M$ by $N$. Recall that $M$ is on the perpendicular bisector of $B C$, i.e. on line $O D$. Points $I, O, D, P$ lie on the c... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
3e27f6bdd19403cf | Let $A B C$ be a triangle with $A B < A C$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle B X A = \angle A Y C$ and $X$ lies on the opposite side of $A B$ to $C$ while $Y$ lies on the opposite side of $A C$ to $B$.
Show that, as ... | proof | 1. Extend $X A$ and $Y A$ to meet $\omega$ again at $X^{\prime}$ and $Y^{\prime}$ respectively. We then have that:
$$
\angle Y^{\prime} Y C=\angle A Y C=\angle B X A=\angle B X X^{\prime} .
$$
so $B C X^{\prime} Y^{\prime}$ is an isosceles trapezium and hence $X^{\prime} Y^{\prime} \| B C$.
![](https://cdn.mathpix.co... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
897d45c8ff3103d1 | Let $A B C$ be a right-angled triangle with $\angle B A C=90^{\circ}$. Let the height from $A$ cut its side $B C$ at $D$. Let $I, I_{B}, I_{C}$ be the incenters of triangles $A B C, A B D, A C D$ respectively. Let also $E_{B}, E_{C}$ be the excenters of $A B C$ with respect to vertices $B$ and $C$ respectively. If $K$ ... | proof | Since $\angle E_{C} B I=90^{\circ}=\angle I C E_{B}$, we conclude that $E_{C} B C E_{B}$ is cyclic. Moreover, we have that
$$
\angle B A I_{B}=\frac{1}{2} \angle B A D=\frac{1}{2} \widehat{C},
$$
so $A I_{B} \perp C I$. Similarly $A I_{C} \perp B I$. Therefore $I$ is the orthocenter of triangle $A I_{B} I_{C}$. It fo... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
8e2a3609f78d45f3 | Let $A B C$ be an acute triangle with $A C > A B$ and circumcircle $\Gamma$. The tangent from $A$ to $\Gamma$ intersects $B C$ at $T$. Let $M$ be the midpoint of $B C$ and let $R$ be the reflection of $A$ in $B$. Let $S$ be a point so that $S A B T$ is a parallelogram and finally let $P$ be a point on line $S B$ such t... | proof | 1. Let $N$ be the midpoint of $B S$ which, as $S A B T$ is a parallelogram, is also the midpoint of $T A$. Using $S T\|A B\| M P$ we get:
$$
\frac{N B}{B P}=\frac{1}{2} \cdot \frac{S B}{B P}=\frac{T B}{2 \cdot B M}=\frac{T B}{B C}
$$
which shows that $T A \| C P$.
![](https://cdn.mathpix.com/cropped/2024_12_07_557efa... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
cfade955ad05150b | Let $A B C$ be an acute triangle such that $A B < A C$. Let $\omega$ be the circumcircle of $A B C$ and assume that the tangent to $\omega$ at $A$ intersects the line $B C$ at $D$. Let $\Omega$ be the circle with center $D$ and radius $A D$. Denote by $E$ the second intersection point of $\omega$ and $\Omega$. Let $M$ ... | proof | 1. Denote by $S$ the intersection point of $\Omega$ and the segment $B C$. Because $D A=D S$, we have $\angle D S A=\angle D A S$. Now using that $D A$ is tangent to $\omega$ we obtain:
$$
\angle B A S=\angle D A S-\angle D A B=\angle D S A-\angle D C A=\angle C A S .
$$
This means that the line $A S$ is the angle bi... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
a29db9eb2223eecd | Let $A B C$ be an acute scalene triangle. Its $C$-excircle tangent to the segment $A B$ meets $A B$ at point $M$ and the extension of $B C$ beyond $B$ at point $N$. Analogously, its $B$-excircle tangent to the segment $A C$ meets $A C$ at point $P$ and the extension of $B C$ beyond $C$ at point $Q$. Denote by $A_{1}$ t... | proof | 1. We shall use the standard notations for $ABC$, i.e., $\angle ABC = \beta$, $BC = a$, etc. We also write $s = \frac{a+b+c}{2}$ for the semiperimeter and $r$ for the inradius.
Let $MN$ intersect the altitude $AD$ (where $D$ lies on $BC$) at the point $L$. We have that $\angle BAD = 90^\circ - \beta$ and $\angle AML =... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
514b5e77a0b2ad10 | Let $A B C$ be a scalene triangle and let $I$ be its incenter. The projections of $I$ on $B C, C A$ and $A B$ are $D, E$ and $F$ respectively. Let $K$ be the reflection of $D$ over the line $A I$, and let $L$ be the second point of intersection of the circumcircles of the triangles $B F K$ and $C E K$. If $\frac{1}{3} ... | D E=2 K L | Writing $A E=A F=x, B F=B D=y$ and $C E=C D=z$, the condition $\frac{1}{3} B C=A C-A B$ translates to $y+z=3(z-y)$ giving $z=2 y$, i.e. $C D=2 B D$.
Letting $B^{\prime}$ be the reflection of $B$ on $A I$ we have that $B^{\prime}$ belongs on $A C$ with $B^{\prime} E=B F=$ $B D=\frac{1}{2} C D=\frac{1}{2} C E$ therefore... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
5f645d0acbb078ac | Let $n \geqslant 3$ be an integer and let
$$
M=\left\{\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}: 1 \leqslant k \leqslant n \text { and } 1 \leqslant a_{1}<\cdots<a_{k} \leqslant n\right\}
$$
be the set of the arithmetic means of the elements of all non-empty subsets of $\{1,2, \ldots, n\}$. Find $\min \{|a-b|: a, b \in M$ w... | \frac{1}{(n-1)(n-2)} | We observe that $M$ is composed of rational numbers of the form $a=\frac{x}{k}$, where $1 \leqslant k \leqslant n$. As the arithmetic mean of $1, \ldots, n$ is $\frac{n+1}{2}$, if we look at these rational numbers in their irreducible form, we can say that $1 \leqslant k \leqslant n-1$.
A non-zero difference $|a-b|$ w... | olympiads_ref | yoonholee/math-corpus-combined | 8 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
5cc454892c0a3493 | Denote by $\ell(n)$ the largest prime divisor of $n$. Let $a_{n+1}=a_{n}+\ell\left(a_{n}\right)$ be a recursively defined sequence of integers with $a_{1}=2$. Determine all natural numbers $m$ such that there exists some $i \in \mathbb{N}$ with $a_{i}=m^{2}$. | proof | We will show that all such numbers are exactly the prime numbers.
Let $p_{1}, p_{2}, \ldots$ be the sequence of prime numbers. We will prove the following:
Claim: Assume $a_{n}=p_{i} p_{i+1}$. Then for each $k=1,2, \ldots, p_{i+2}-p_{i}$ we have that $a_{n+k}=$ $\left(p_{i}+k\right) p_{i+1}$.
Proof. By induction on $k$... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
6875e3b74b7ddce3 | Let $n$ be a positive integer. Determine, in terms of $n$, the greatest integer which divides every number of the form $p+1$, where $p \equiv 2 \bmod 3$ is a prime number which does not divide $n$.
## Proposed by Bulgaria | 3 \text{ when } n \text{ is odd, and } 6 \text{ when } n \text{ is even} | Let $k$ be the greatest such integer. We will show that $k=3$ when $n$ is odd and $k=6$ when $n$ is even.
We will say that a number $p$ is nice if $p$ is a prime number of the form $2 \bmod 3$ which does not divide $N$.
Note first that if $3 \mid p+1$ for every nice number $p$ and so $k$ is a multiple of 3.
If $n$ is... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
f16b37471ef2522f | Can every positive rational number $q$ be written as
$$
\frac{a^{2021}+b^{2023}}{c^{2022}+d^{2024}}
$$
where $a, b, c, d$ are all positive integers?
Proposed by United Kingdom | proof | The answer is yes. Set \(a = x^{2023}, b = x^{2021}\) and \(c = y^{2024}, d = y^{2022}\) for some integers \(x, y\) and let \(q = \frac{m}{n}\) in lowest terms. Then we could try to solve
\[
\frac{a^{2021} + b^{2023}}{c^{2022} + d^{2024}} = \frac{2 x^{2021 \times 2023}}{2 y^{2022 \times 2024}} = \frac{x^{2021 \times 2... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
716a724eb154be2a | A natural number $n$ is given. Determine all $(n-1)$-tuples of nonnegative integers $a_{1}, a_{2}, \ldots, a_{n-1}$ such that
$$
\left[\frac{m}{2^{n}-1}\right]+\left[\frac{2 m+a_{1}}{2^{n}-1}\right]+\left[\frac{2^{2} m+a_{2}}{2^{n}-1}\right]+\left[\frac{2^{3} m+a_{3}}{2^{n}-1}\right]+\cdots+\left[\frac{2^{n-1} m+a_{n-... | a_{k}=2^{n-1}+2^{k-1}-1 | 1. We will show that there is a unique such $n$-tuple: $a_{k}=2^{n-1}+2^{k-1}-1$ for $k=1, \ldots, n-1$.
Write $N=2^{n}-1$ and $f_{k}(x)=\left[\frac{2^{k} x+a_{k}}{N}\right]$ for $k=0,1, \ldots, n-1$, where $a_{0}=0$. Since
$$
\sum_{k=0}^{n-1} f_{k}(m)-\sum_{k=0}^{n-1} f_{k}(m-1)=1
$$
for each $m \in \mathbb{Z}$, the... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
705fa29ea57db81f | Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b)+[a, b]=2021^{c}$. If $|a-b|$ is a prime number, prove that the number $(a+b)^{2}+4$ is composite.
## Proposed by Serbia | proof | We write $p=|a-b|$ and assume for contradiction that $q=(a+b)^{2}+4$ is a prime number.
Since $(a, b) \mid [a, b]$, we have that $(a, b) \mid 2021^{c}$. As $(a, b)$ also divides $p=|a-b|$, it follows that $(a, b) \in \{1,43,47\}$. We will consider all 3 cases separately:
(1) If $(a, b)=1$, then $1+a b=2021^{c}$, and t... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
f55506c63d46eb64 | A super-integer triangle is defined to be a triangle whose lengths of all sides and at least one height are positive integers. We will deem certain positive integer numbers to be good with the condition that if the lengths of two sides of a super-integer triangle are two (not necessarily different) good numbers, then t... | proof | Evidently, all right-angle triangles with integer sides are super-integer triangles. We will use the following notation $(a, b, c\{h\})$ to denote a super-integer triangle whose sides are $a$, $b$ and $c$ and the height of integer length is $h$. The height will be written in curly brackets next to the corresponding sid... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
9552c65d32b344cb | Fourteen friends met at a party. One of them, Fredek, wanted to go to bed early. He said goodbye to 10 of his friends, forgot about the remaining 3, and went to bed. After a while he returned to the party, said goodbye to 10 of his friends (not necessarily the same as before), and went to bed. Later Fredek came back a ... | 32 | Answer: Fredek returned at least 32 times.
Assume Fredek returned $k$ times, i.e., he was saying good-bye $k+1$ times to his friends. There exists a friend of Fredek, call him $X_{13}$, about whom Fredek forgot $k$ times in a row, starting from the very first time - otherwise Fredek would have come back less than $k$ ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
b5746ec6acc52c83 | Two positive integers are written on the blackboard. Initially, one of them is 2000 and the other is smaller than 2000. If the arithmetic mean \( m \) of the two numbers on the blackboard is an integer, the following operation is allowed: one of the two numbers is erased and replaced by \( m \). Prove that this operati... | 10 | Each time the operation is performed, the difference between the two numbers on the blackboard will become one half of its previous value (regardless of which number was erased). The mean value of two integers is an integer if and only if their difference is an even number. Suppose the initial numbers were \(a=2000\) a... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
db135446940b29a9 | Let \( n \) be a positive integer not divisible by 2 or 3. Prove that for all integers \( k \), the number \((k+1)^{n}-k^{n}-1\) is divisible by \( k^{2}+k+1 \). | proof | Note that $n$ must be congruent to 1 or 5 modulo 6, and proceed by induction on $\lfloor n / 6\rfloor$. It can easily be checked that the assertion holds for $n \in\{1,5\}$. Let $n>6$, and put $t=k^{2}+k+1$. The claim follows by:
$$
\begin{aligned}
(k+1)^{n}-k^{n}-1 & =(t+k)(k+1)^{n-2}-(t-(k+1)) k^{n-2}-1 \\
& \equiv ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
d075cf4aa0c09cdc | Prove that for all positive real numbers $a, b, c$ we have
$$
\sqrt{a^{2}-a b+b^{2}}+\sqrt{b^{2}-b c+c^{2}} \geqslant \sqrt{a^{2}+a c+c^{2}}.
$$ | proof | If $|O A|=a,|O B|=b,|O C|=c$ (see Figure 7), then the inequality follows from $|A C| \leqslant|A B|+|B C|$ by applying the cosine theorem to triangles $A O B$,
$B O C$ and $A O C$. The same argument holds if the quadrangle $O A B C$ is concave. | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
78321674a15b759b | Find all real solutions to the following system of equations:
$$
\left\{\begin{aligned}
x+y+z+t & =5 \\
xy+yz+zt+tx & =4 \\
xyz+yz t+zt x+tx y & =3 \\
xyzt & =-1
\end{aligned}\right.
$$ | x=\frac{1 \pm \sqrt{2}}{2}, y=2, z=\frac{1 \mp \sqrt{2}}{2}, t=2 \text{ or } x=2, y=\frac{1 \pm \sqrt{2}}{2}, z=2, t=\frac{1 \mp \sqrt{2}}{2} | Answer: $x=\frac{1 \pm \sqrt{2}}{2}, y=2, z=\frac{1 \mp \sqrt{2}}{2}, t=2$ or $x=2, y=\frac{1 \pm \sqrt{2}}{2}$, $z=2, t=\frac{1 \mp \sqrt{2}}{2}$.
Let $A=x+z$ and $B=y+t$. Then the system of equations is equivalent to
$$
\left\{\begin{aligned}
A+B & =5 \\
A B & =4 \\
B x z+A y t & =3 \\
(B x z) \cdot(A y t) & =-4 .
... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
8eabf240c09b1a22 | For every positive integer $n$, let
$$
x_{n}=\frac{(2 n+1) \cdot(2 n+3) \cdots \cdots \cdot(4 n-1) \cdot(4 n+1)}{2 n \cdot(2 n+2) \cdots \cdots \cdot(4 n-2) \cdot 4 n}
$$
Prove that $\frac{1}{4 n}<x_{n}-\sqrt{2}<\frac{2}{n}$.
## Solutions
Translate the above text into English, keeping the original text's line break... | proof | Squaring both sides of the given equality and applying \(x(x+2) \leqslant (x+1)^{2}\) to the numerator of the obtained fraction and cancelling we have
\[
x_{n}^{2} \leqslant \frac{(2 n+1) \cdot (4 n+1)}{(2 n)^{2}}2 + \frac{1}{n}
\]
Hence
\[
\frac{1}{n}\sqrt{2} \text{ and } x_{n} < 2. \text{ The result then follows f... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
e8beb3762bdd2092 | The numbers $1,2, \ldots, 49$ are placed in a $7 \times 7$ array, and the sum of the numbers in each row and in each column is computed. Some of these 14 sums are odd while others are even. Let $A$ denote the sum of all the odd sums and $B$ the sum of all even sums. Is it possible that the numbers were placed in the ar... | proof | Answer: no.
If this were possible, then $2 \cdot(1+\ldots+49)=A+B=2 B$. But $B$ is even since it is the sum of even numbers, whereas $1+\ldots+49=25 \cdot 49$ is odd. This is a contradiction. | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
db88bdc6dbbe0f19 | Let \( p \) and \( q \) be two different primes. Prove that
\[
\left\lfloor\frac{p}{q}\right\rfloor+\left\lfloor\frac{2 p}{q}\right\rfloor+\left\lfloor\frac{3 p}{q}\right\rfloor+\ldots+\left\lfloor\frac{(q-1) p}{q}\right\rfloor=\frac{1}{2}(p-1)(q-1) .
\]
(Here \(\lfloor x\rfloor\) denotes the largest integer not grea... | \frac{1}{2}(p-1)(q-1) | The line $y=\frac{p}{q} x$ contains the diagonal of the rectangle with vertices $(0,0)$, $(q, 0)$, $(q, p)$, and $(0, p)$ and passes through no points with integer coordinates in the interior of that rectangle. For $k=1,2, \ldots, q-1$ the summand $\left\lfloor\frac{k p}{q}\right\rfloor$ counts the number of interior p... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
f2df1fc361ae67d8 | Given a rhombus $A B C D$, find the locus of the points $P$ lying inside the rhombus and satisfying $\angle A P D + \angle B P C = 180^{\circ}$. | proof | Answer: the locus of the points $P$ is the union of the diagonals $A C$ and $B D$.
Let $Q$ be a point such that $P Q C D$ is a parallelogram (see Figure 4). Then $A B Q P$ is also a parallelogram. From the equality $\angle A P D + \angle B P C = 180^{\circ}$ it follows that $\angle B Q C + \angle B P C = 180^{\circ}$,... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
3980ecb30aa52228 | Let $f$ be a real-valued function defined on the positive integers satisfying the following condition: For all $n>1$ there exists a prime divisor $p$ of $n$ such that
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)
$$
Given that $f(2001)=1$, what is the value of $f(2002)$? | 2 | Answer: 2.
For any prime $p$ we have $f(p)=f(1)-f(p)$ and thus $f(p)=\frac{f(1)}{2}$. If $n$ is a product of two primes $p$ and $q$, then $f(n)=f(p)-f(q)$ or $f(n)=f(q)-f(p)$, so $f(n)=0$. By the same reasoning we find that if $n$ is a product of three primes, then there is a prime $p$ such that
$$
f(n)=f\left(\frac{... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
d96044f13647714e | What is the smallest positive odd integer having the same number of positive divisors as 360? | 31185 | Answer: 31185.
An integer with the prime factorization $p_{1}^{r_{1}} \cdot p_{2}^{r_{2}} \cdot \ldots \cdot p_{k}^{r_{k}}$ (where $p_{1}, p_{2}, \ldots$, $p_{k}$ are distinct primes) has precisely $\left(r_{1}+1\right) \cdot\left(r_{2}+1\right) \cdot \ldots \cdot\left(r_{k}+1\right)$ distinct positive divisors. Since... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
a6dff317fb2cffac | From a sequence of integers $(a, b, c, d)$ each of the sequences
$$
(c, d, a, b),(b, a, d, c),(a+n c, b+n d, c, d),(a+n b, b, c+n d, d),
$$
for arbitrary integer $n$ can be obtained by one step. Is it possible to obtain $(3,4,5,7)$ from $(1,2,3,4)$ through a sequence of such steps? | no | Answer: no.
Under all transformations $(a, b, c, d) \rightarrow\left(a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}\right)$ allowed in the problem we have $|a d-b c|=\left|a^{\prime} d^{\prime}-b^{\prime} c^{\prime}\right|$, but $|1 \cdot 4-2 \cdot 3|=2 \neq 1=|3 \cdot 7-4 \cdot 5|$.
Remark. The transformations allow... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
f8674f0624b92aab | Let \(a, b, c, d\) be real numbers such that
\[
\begin{aligned}
a+b+c+d & =-2 \\
a b+a c+a d+b c+b d+c d & =0
\end{aligned}
\]
Prove that at least one of the numbers \(a, b, c, d\) is not greater than -1. | proof | We can assume that $a$ is the least among $a, b, c, d$ (or one of the least, if some of them are equal), there are $n>0$ negative numbers among $a, b, c, d$, and the sum of the positive ones is $x$.
Then we obtain
$$
-2=a+b+c+d \geqslant n a+x.
$$
Squaring we get
$$
4=a^{2}+b^{2}+c^{2}+d^{2}
$$
which implies
$$
4... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
339e5c1cae97c550 | Let \( n \) be a positive integer. Prove that
\[
\sum_{i=1}^{n} x_{i}\left(1-x_{i}\right)^{2} \leqslant\left(1-\frac{1}{n}\right)^{2}
\]
for all nonnegative real numbers \( x_{1}, x_{2}, \ldots, x_{n} \) such that \( x_{1}+x_{2}+\cdots+x_{n}=1 \). | proof | Expanding the expressions at both sides we obtain the equivalent inequality
$$
-\sum_{i} x_{i}^{3}+2 \sum_{i} x_{i}^{2}-\frac{2}{n}+\frac{1}{n^{2}} \geqslant 0
$$
It is easy to check that the left hand side is equal to
$$
\sum_{i}\left(2-\frac{2}{n}-x_{i}\right)\left(x_{i}-\frac{1}{n}\right)^{2}
$$
and hence is non... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
e149ccda40c236b0 | Let $P$ be a set of $n \geqslant 3$ points in the plane, no three of which are on a line. How many possibilities are there to choose a set $T$ of $\left(\begin{array}{c}n-1 \\ 2\end{array}\right)$ triangles, whose vertices are all in $P$, such that each triangle in $T$ has a side that is not a side of any other triangl... | proof | For a fixed point $x \in P$, let $T_{x}$ be the set of all triangles with vertices in $P$ which have $x$ as a vertex. Clearly, $\left|T_{x}\right|=\left(\begin{array}{c}n-1 \\ 2\end{array}\right)$, and each triangle in $T_{x}$ has a side which is not a side of any other triangle in $T_{x}$. For any $x, y \in P$ such th... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
f883d5fa448fd6b1 | Let $N$ be a positive integer. Two persons play the following game. The first player writes a list of positive integers not greater than 25, not necessarily different, such that their sum is at least 200. The second player wins if he can select some of these numbers so that their sum $S$ satisfies the condition $200-N ... | 11 | If $N=11$, then the second player can simply remove numbers from the list, starting with the smallest number, until the sum of the remaining numbers is less than 212. If the last number removed was not 24 or 25, then the sum of the remaining numbers is at least $212-23=189$. If the last number removed was 24 or 25, the... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
17888a02b56e4267 | Let \( n \) be a positive integer. Prove that the equation
\[
x + y + \frac{1}{x} + \frac{1}{y} = 3n
\]
does not have solutions in positive rational numbers. | proof | Suppose $x=\frac{p}{q}$ and $y=\frac{r}{s}$ satisfy the given equation, where $p, q, r, s$ are positive integers and $\operatorname{gcd}(p, q)=1, \operatorname{gcd}(r, s)=1$. We have
$$
\frac{p}{q}+\frac{r}{s}+\frac{q}{p}+\frac{s}{r}=3 n
$$
or
$$
\left(p^{2}+q^{2}\right) r s+\left(r^{2}+s^{2}\right) p q=3 n p q r s,... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
14f05f4a45065572 | Does there exist an infinite non-constant arithmetic progression, each term of which is of the form $a^{b}$, where $a$ and $b$ are positive integers with $b \geqslant 2$?
Answer: no. | proof | For an arithmetic progression $a_{1}, a_{2}, \ldots$ with difference $d$ the following holds:
$$
\begin{aligned}
S_{n} & =\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n+1}}=\frac{1}{a_{1}}+\frac{1}{a_{1}+d}+\ldots+\frac{1}{a_{1}+n d} \geqslant \\
& \geqslant \frac{1}{m}\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
d3062937e9654702 | Let $\mathbb{Q}_{+}$ be the set of positive rational numbers. Find all functions $f: \mathbb{Q}_{+} \rightarrow \mathbb{Q}_{+}$ which for all $x \in \mathbb{Q}_{+}$ fulfill
(1) $f\left(\frac{1}{x}\right)=f(x)$
(2) $\left(1+\frac{1}{x}\right) f(x)=f(x+1)$ | f\left(\frac{p}{q}\right)=a p q | Set $g(x)=\frac{f(x)}{f(1)}$. Function $g$ fulfils (1), (2) and $g(1)=1$. First we prove that if $g$ exists then it is unique. We prove that $g$ is uniquely defined on $x=\frac{p}{q}$ by induction on $\max (p, q)$. If $\max (p, q)=1$ then $x=1$ and $g(1)=1$. If $p=q$ then $x=1$ and $g(x)$ is unique. If $p \neq q$ then ... | olympiads_ref | yoonholee/math-corpus-combined | 8 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
9318fa8585293015 | Prove that any real solution of
$$
x^{3}+p x+q=0
$$
satisfies the inequality \(4 q x \leq p^{2}\). | proof | Let $x_{0}$ be a root of the cubic, then $x^{3}+p x+q=\left(x-x_{0}\right)\left(x^{2}+a x+b\right)=$ $x^{3}+\left(a-x_{0}\right) x^{2}+\left(b-a x_{0}\right) x-b x_{0}$. So $a=x_{0}, p=b-a x_{0}=b-x_{0}^{2},-q=b x_{0}$. Hence $p^{2}=b^{2}-2 b x_{0}^{2}+x_{0}^{4}$. Also $4 x_{0} q=-4 x_{0}^{2} b$. So $p^{2}-4 x_{0} q=b^... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
5f20b7b3baaf2f0b | Let \( x, y \) and \( z \) be positive real numbers such that \( x y z = 1 \). Prove that
\[
(1+x)(1+y)(1+z) \geq 2\left(1+\sqrt[3]{\frac{y}{x}}+\sqrt[3]{\frac{z}{y}}+\sqrt[3]{\frac{x}{z}}\right)
\] | proof | Put \( a = b x, b = c y \) and \( c = a z \). The given inequality then takes the form
\[
\begin{aligned}
\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right) & \geq 2\left(1+\sqrt[3]{\frac{b^{2}}{a c}}+\sqrt[3]{\frac{c^{2}}{a b}}+\sqrt[3]{\frac{a^{2}}{b c}}\right) \\
& =2\left(1+\frac{a+b+c}... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
37b64cdeb895e0b8 | Let \( a, b, c \) be positive real numbers. Prove that
\[
\frac{2 a}{a^{2}+b c}+\frac{2 b}{b^{2}+c a}+\frac{2 c}{c^{2}+a b} \leq \frac{a}{b c}+\frac{b}{c a}+\frac{c}{a b} .
\] | proof | First we prove that
$$
\frac{2 a}{a^{2}+b c} \leq \frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\right)
$$
which is equivalent to \(0 \leq b(a-c)^{2}+c(a-b)^{2}\), and therefore holds true. Now we turn to the inequality
$$
\frac{1}{b}+\frac{1}{c} \leq \frac{1}{2}\left(\frac{2 a}{b c}+\frac{b}{c a}+\frac{c}{a b}\right),
$$... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
505181316cb2759e | Let $X$ be a subset of $\{1,2,3, \ldots, 10000\}$ with the following property: If $a, b \in X, a \neq b$, then $a \cdot b \notin X$. What is the maximal number of elements in $X$?
Answer: 9901. | 9901 | If $X=\{100,101,102, \ldots, 9999,10000\}$, then for any two selected $a$ and $b, a \neq b$, $a \cdot b \geq 100 \cdot 101>10000$, so $a \cdot b \notin X$. So $X$ may have 9901 elements.
Suppose that $x_{1}<100$ and consider the pairs
$$
\begin{gathered}
200-x_{1},\left(200-x_{1}\right) \cdot x_{1} \\
200-x_{2},\left... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
33fd3e1704aed57d | A lattice point in the plane is a point whose coordinates are both integers. The centroid of four points $\left(x_{i}, y_{i}\right), i=1,2,3,4$, is the point $\left(\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4}, \frac{y_{1}+y_{2}+y_{3}+y_{4}}{4}\right)$. Let $n$ be the largest natural number with the following property: There are ... | 12 | To prove \( n \geq 12 \), we have to show that there are 12 lattice points \(\left(x_{i}, y_{i}\right)\), \(i=1,2, \ldots, 12\), such that no four determine a lattice point centroid. This is guaranteed if we just choose the points such that \(x_{i} \equiv 0(\bmod 4)\) for \(i=1, \ldots, 6\), \(x_{i} \equiv 1(\bmod 4)\)... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
5bab829b47b0c3ed | Let $A B C$ be an arbitrary triangle and $A M B, B N C, C K A$ regular triangles outward of $A B C$. Through the midpoint of $M N$ a perpendicular to $A C$ is constructed; similarly through the midpoints of $N K$ resp. $K M$ perpendiculars to $A B$ resp. $B C$ are constructed. Prove that these three perpendiculars inte... | proof | Let $O$ be the midpoint of $M N$, and let $E$ and $F$ be the midpoints of $A B$ and $B C$, respectively. As triangle $M B C$ transforms into triangle $A B N$ when rotated $60^{\circ}$ around $B$ we get $M C=A N$ (it is also a well-known fact). Considering now the quadrangles $A M B N$ and $C M B N$ we get $O E=O F$ (fr... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
b66ee758c40e1e69 | Find all pairs of positive integers $(a, b)$ such that $a-b$ is a prime and $ab$ is a perfect square. Answer: Pairs $(a, b)=\left(\left(\frac{p+1}{2}\right)^{2},\left(\frac{p-1}{2}\right)^{2}\right)$, where $p$ is a prime greater than 2. | (a, b)=\left(\left(\frac{p+1}{2}\right)^{2},\left(\frac{p-1}{2}\right)^{2}\right) | Let $p$ be a prime such that $a-b=p$ and let $a b=k^{2}$. Insert $a=b+p$ in the equation $a b=k^{2}$. Then
$$
k^{2}=(b+p) b=\left(b+\frac{p}{2}\right)^{2}-\frac{p^{2}}{4}
$$
which is equivalent to
$$
p^{2}=(2 b+p)^{2}-4 k^{2}=(2 b+p+2 k)(2 b+p-2 k) .
$$
Since $2 b+p+2 k>2 b+p-2 k$ and $p$ is a prime, we conclude $2... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
d32fec5838b42cf9 | All the positive divisors of a positive integer $n$ are stored into an array in increasing order. Mary has to write a program which decides for an arbitrarily chosen divisor $d>1$ whether it is a prime. Let $n$ have $k$ divisors not greater than $d$. Mary claims that it suffices to check divisibility of $d$ by the firs... | proof | Let $d>1$ be a divisor of $n$. Suppose Mary's program outputs "composite" for $d$. That means it has found a divisor of $d$ greater than 1. Since $d>1$, the array contains at least 2 divisors of $d$, namely 1 and $d$. Thus Mary's program does not check divisibility of $d$ by $d$ (the first half gets complete before rea... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
43fd99e6c7e561d7 | Every integer is coloured with exactly one of the colours BLUE, GREEN, RED, YELLOW. Can this be done in such a way that if $a, b, c, d$ are not all 0 and have the same colour, then $3 a-2 b \neq 2 c-3 d$?
Answer: Yes. | proof | A colouring with the required property can be defined as follows. For a non-zero integer $k$ let $k^{*}$ be the integer uniquely defined by $k=5^{m} \cdot k^{*}$, where $m$ is a nonnegative integer and $5 \nmid k^{*}$. We also define $0^{*}=0$. Two non-zero integers $k_{1}, k_{2}$ receive the same colour if and only if... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
dbd12f5ad9ac86d5 | Let $n$ be a positive integer such that the sum of all the positive divisors of $n$ (except $n$) plus the number of these divisors is equal to $n$. Prove that $n=2 m^{2}$ for some integer $m$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation resu... | proof | Let $t_{1}<t_{2}<\cdots<t_{s}$ be all positive odd divisors of $n$, and let $2^{k}$ be the maximal power of 2 that divides $n$. Then the full list of divisors of $n$ is the following:
$$
t_{1}, \ldots, t_{s}, 2 t_{1}, \ldots, 2 t_{s}, \ldots, 2^{k} t_{1}, \ldots, 2^{k} t_{s} .
$$
Hence,
$$
2 n=\left(2^{k+1}-1\right)... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
b47c46a784f4baa1 | An $m \times n$ table is given, in each cell of which a number +1 or -1 is written. It is known that initially exactly one -1 is in the table, all the other numbers being +1. During a move, it is allowed to choose any cell containing -1, replace this -1 by 0, and simultaneously multiply all the numbers in the neighbori... | Those (m, n) for which at least one of m, n is odd. | Let us erase a unit segment which is the common side of any two cells in which two zeroes appear. If the final table consists of zeroes only, all the unit segments (except those which belong to the boundary of the table) are erased. We must erase a total of
$$
m(n-1)+n(m-1)=2 m n-m-n
$$
such unit segments.
On the ot... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
f001f4d2568fe09a | A circle is divided into 13 segments, numbered consecutively from 1 to 13. Five fleas called $A, B, C, D$ and $E$ are sitting in the segments 1, 2, 3, 4 and 5. A flea is allowed to jump to an empty segment five positions away in either direction around the circle. Only one flea jumps at the same time, and two fleas can... | not found | Write the numbers from 1 to 13 in the order $\mathbf{1}, 6,11, \mathbf{3}, 8,13,5,10,2,7,12,4$, 9. Then each time a flea jumps it moves between two adjacent numbers or between the first and the last number in this row. Since a flea can never move past another flea, the possible permutations are
| 3 | 5 | 2 | 4 | | 1 ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
07ee2834bddbc2a0 | Through a point \( P \) exterior to a given circle pass a secant and a tangent to the circle. The secant intersects the circle at \( A \) and \( B \), and the tangent touches the circle at \( C \) on the same side of the diameter through \( P \) as \( A \) and \( B \). The projection of \( C \) on the diameter is \( Q ... | proof | Denoting the centre of the circle by $O$, we have $O Q \cdot O P=O A^{2}=O B^{2}$. Hence $\triangle O A Q \sim \triangle O P A$ and $\triangle O B Q \sim \triangle O P B$. Since $\triangle A O B$ is isosceles, we have
$\angle O A P+\angle O B P=180^{\circ}$, and therefore
$$
\begin{aligned}
\angle A Q P+\angle B Q P &... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
5945d136777430a4 | Consider a rectangle with side lengths 3 and 4, and pick an arbitrary inner point on each side. Let \( x, y, z \) and \( u \) denote the side lengths of the quadrilateral spanned by these points. Prove that \( 25 \leq x^{2}+y^{2}+z^{2}+u^{2} \leq 50 \). | 25 \leq x^{2}+y^{2}+z^{2}+u^{2} \leq 50 | Let \(a, b, c\) and \(d\) be the distances of the chosen points from the midpoints of the sides of the rectangle (with \(a\) and \(c\) on the sides of length 3). Then
\[
\begin{aligned}
x^{2}+y^{2}+z^{2}+u^{2}= & \left(\frac{3}{2}+a\right)^{2}+\left(\frac{3}{2}-a\right)^{2}+\left(\frac{3}{2}+c\right)^{2}+\left(\frac{3... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
6720eab45ac31a35 | Consider the sequence \(a_{k}\) defined by \(a_{1}=1, a_{2}=\frac{1}{2}\),
\[
a_{k+2}=a_{k}+\frac{1}{2} a_{k+1}+\frac{1}{4 a_{k} a_{k+1}} \quad \text{for } k \geq 1
\]
Prove that
\[
\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}}+\frac{1}{a_{3} a_{5}}+\cdots+\frac{1}{a_{98} a_{100}}<4
\] | 4 | Note that
$$
\frac{1}{a_{k} a_{k+2}}a_{k}+\frac{1}{2} a_{k+1}
$$
which is evident for the given sequence. Now we have
$$
\begin{aligned}
\frac{1}{a_{1} a_{3}}+\frac{1}{a_{2} a_{4}} & +\frac{1}{a_{3} a_{5}}+\cdots+\frac{1}{a_{98} a_{100}} \\
& <\frac{2}{a_{1} a_{2}}-\frac{2}{a_{2} a_{3}}+\frac{2}{a_{2} a_{3}}-\frac{2... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
f5f7e3993990678d | Let \( a, b, c \) be positive real numbers with \( abc = 1 \). Prove that
$$
\frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2} \leq 1
$$ | proof | For any positive real $x$ we have $x^{2}+1 \geq 2 x$. Hence
$$
\begin{aligned}
\frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2} & \leq \frac{a}{2 a+1}+\frac{b}{2 b+1}+\frac{c}{2 c+1} \\
& =\frac{1}{2+1 / a}+\frac{1}{2+1 / b}+\frac{1}{2+1 / c}=: R .
\end{aligned}
$$
$R \leq 1$ is equivalent to
$$
\left(2+\frac{1... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
e1e957c90c2379a5 | A rectangular array has $n$ rows and six columns, where $n>2$. In each cell there is written either 0 or 1. All rows in the array are different from each other. For each pair of rows $\left(x_{1}, x_{2}, \ldots, x_{6}\right)$ and $\left(y_{1}, y_{2}, \ldots, y_{6}\right)$, the row $\left(x_{1} y_{1}, x_{2} y_{2}, \ldot... | proof | Clearly there must be rows with some zeroes. Consider the case when there is a row with just one zero; we can assume it is $(0,1,1,1,1,1)$. Then for each row $\left(1, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}\right)$ there is also a row $\left(0, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}\right)$; the conclusion follows. Consider the ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
d6b2143123eb802c | A rectangle is divided into $200 \times 3$ unit squares. Prove that the number of ways of splitting this rectangle into rectangles of size $1 \times 2$ is divisible by 3. | N_{6 k+2} \equiv 0(\bmod 3) | Let us denote the number of ways to split some figure into dominos by a small picture of this figure with a sign \#. For example, $\# \boxplus=2$.
Let $N_{n}=\#$ ( $n$ rows) and $\gamma_{n}=\#$ ( $n-2$ full rows and one row with two cells).
We are going to find a recurrence relation for the numbers $N_{n}$.
Observe ... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Combinatorics", "problem_type": "Combinatorics", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
c81e69cd16758cbf | What is the smallest number of circles of radius $\sqrt{2}$ that are needed to cover a rectangle
(a) of size $6 \times 3$?
(b) of size $5 \times 3$?
Answer: (a) Six circles, (b) five circles. | 6 | (a) Consider the four corners and the two midpoints of the sides of length 6. The distance between any two of these six points is 3 or more, so one circle cannot cover two of these points, and at least six circles are needed.
On the other hand one circle will cover a $2 \times 2$ square, and it is easy to see that six... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
fc7466ce67baddde | Let the lines $e$ and $f$ be perpendicular and intersect each other at $O$. Let $A$ and $B$ lie on $e$ and $C$ and $D$ lie on $f$, such that all the five points $A, B, C, D$ and $O$ are distinct. Let the lines $b$ and $d$ pass through $B$ and $D$ respectively, perpendicularly to $A C$; let the lines $a$ and $c$ pass th... | proof | Let $A_{1}$ be the intersection of $a$ with $BD$, $B_{1}$ the intersection of $b$ with $AC$, $C_{1}$ the intersection of $c$ with $BD$ and $D_{1}$ the intersection of $d$ with $AC$. It follows easily by the given right angles that the following three sets each are concyclic:
- $A, A_{1}, D, D_{1}, O$ lie on a circle $... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
33fb9b461581ea6a | A sequence $\left(x_{n}\right), n \geq 0$, is defined as follows: $x_{0}=a, x_{1}=2$ and $x_{n}=2 x_{n-1} x_{n-2}-$ $x_{n-1}-x_{n-2}+1$ for $n>1$. Find all integers $a$ such that $2 x_{3 n}-1$ is a perfect square for all $n \geq 1$.
Answer: $a=\frac{(2 m-1)^{2}+1}{2}$ where $m$ is an arbitrary positive integer. | a=\frac{(2 m-1)^{2}+1}{2} | Let $y_{n}=2 x_{n}-1$. Then
$$
\begin{aligned}
y_{n} & =2\left(2 x_{n-1} x_{n-2}-x_{n-1}-x_{n-2}+1\right)-1 \\
& =4 x_{n-1} x_{n-2}-2 x_{n-1}-2 x_{n-2}+1 \\
& =\left(2 x_{n-1}-1\right)\left(2 x_{n-2}-1\right)=y_{n-1} y_{n-2}
\end{aligned}
$$
when $n>1$. Notice that $y_{n+3}=y_{n+2} y_{n+1}=y_{n+1}^{2} y_{n}$. We see ... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
5d86e06712845aed | Let \( x \) and \( y \) be positive integers and assume that \( z = \frac{4xy}{x+y} \) is an odd integer. Prove that at least one divisor of \( z \) can be expressed in the form \( 4n-1 \) where \( n \) is a positive integer. | proof | Let $x=2^{s} x_{1}$ and $y=2^{t} y_{1}$ where $x_{1}$ and $y_{1}$ are odd integers. Without loss of generality we can assume that $s \geq t$. We have
$$
z=\frac{2^{s+t+2} x_{1} y_{1}}{2^{t}\left(2^{s-t} x_{1}+y_{1}\right)}=\frac{2^{s+2} x_{1} y_{1}}{2^{s-t} x_{1}+y_{1}}
$$
If $s \neq t$, then the denominator is odd a... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
a5e92f1536ec24dd | Let $a, b, c, d, e, f$ be non-negative real numbers satisfying $a+b+c+d+e+f=6$. Find the maximal possible value of
$$
a b c+b c d+c d e+d e f+e f a+f a b
$$
and determine all 6-tuples $(a, b, c, d, e, f)$ for which this maximal value is achieved.
Answer: 8. | 8 | If we set $a=b=c=2, d=e=f=0$, then the given expression is equal to 8. We will show that this is the maximal value. Applying the inequality between arithmetic and geometric mean we obtain
$$
\begin{aligned}
8 & =\left(\frac{(a+d)+(b+e)+(c+f)}{3}\right)^{3} \geq(a+d)(b+e)(c+f) \\
& =(a b c+b c d+c d e+d e f+e f a+f a b... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
11ee050a5213426e | The altitudes of a triangle are 12, 15 and 20. What is the area of the triangle?
Answer: 150. | 150 | Denote the sides of the triangle by \(a, b\) and \(c\) and its altitudes by \(h_{a}, h_{b}\) and \(h_{c}\). Then we know that \(h_{a}=12, h_{b}=15\) and \(h_{c}=20\). By the well known relation \(a: b = h_{b}: h_{a}\) it follows \(b = \frac{h_{a}}{h_{b}} a = \frac{12}{15} a = \frac{4}{5} a\). Analogously, \(c = \frac{h... | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
b94125cb45cfede7 | In a triangle $ABC$, points $D, E$ lie on sides $AB, AC$ respectively. The lines $BE$ and $CD$ intersect at $F$. Prove that if
$$
BC^2 = BD \cdot BA + CE \cdot CA,
$$
then the points $A, D, F, E$ lie on a circle. | proof | Let $G$ be a point on the segment $B C$ determined by the condition $B G \cdot B C = B D \cdot B A$. (Such a point exists because $B D \cdot B A < B C^2$.) Then the points $A, D, G, C$ lie on a circle. Moreover, we have
$$
C E \cdot C A = B C^2 - B D \cdot B A = B C \cdot (B G + C G) - B C \cdot B G = C B \cdot C G,
$... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
6c6f421fb6188740 | Let the medians of the triangle \(ABC\) intersect at the point \(M\). A line \(t\) through \(M\) intersects the circumcircle of \(ABC\) at \(X\) and \(Y\) so that \(A\) and \(C\) lie on the same side of \(t\). Prove that \(BX \cdot BY = AX \cdot AY + CX \cdot CY\). | proof | Let us start with a lemma: If the diagonals of an inscribed quadrilateral $A B C D$ intersect at $O$, then $\frac{A B \cdot B C}{A D \cdot D C}=\frac{B O}{O D}$. Indeed,
$$
\frac{A B \cdot B C}{A D \cdot D C}=\frac{\frac{1}{2} A B \cdot B C \cdot \sin B}{\frac{1}{2} A D \cdot D C \cdot \sin D}=\frac{\operatorname{area... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Geometry", "problem_type": "Geometry", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
c75f0aa6b1971a9a | Does there exist a sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive integers such that the sum of every $n$ consecutive elements is divisible by $n^{2}$ for every positive integer $n$ ?
Answer: Yes. One such sequence begins 1, 3, 5, 55, 561, 851, 63253, 110055, ... | proof | We will show that whenever we have positive integers $a_{1}, \ldots, a_{k}$ such that $n^{2} \mid a_{i+1}+\cdots+a_{i+n}$ for every $n \leq k$ and $i \leq k-n$, then it is possible to choose $a_{k+1}$ such that $n^{2} \mid a_{i+1}+\cdots+a_{i+n}$ for every $n \leq k+1$ and $i \leq k+1-n$. This directly implies the posi... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
1546350fdf4194a9 | A 12-digit positive integer consisting only of digits 1, 5, and 9 is divisible by 37. Prove that the sum of its digits is not equal to 76. | proof | Let $N$ be the initial number. Assume that its digit sum is equal to 76.
The key observation is that $3 \cdot 37=111$, and therefore $27 \cdot 37=999$. Thus we have a divisibility test similar to the one for divisibility by 9: for $x=a_{n} 10^{3 n}+a_{n-1} 10^{3(n-1)}+$ $\cdots+a_{1} 10^{3}+a_{0}$, we have $x \equiv a... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Number Theory", "problem_type": "Number Theory", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
b212e621d0200d12 | Determine all polynomials $p(x)$ with real coefficients such that
$$
p\left((x+1)^{3}\right)=(p(x)+1)^{3}
$$
and
$$
p(0)=0
$$
Answer: $p(x)=x$. | p(x)=x | Consider the sequence defined by
$$
\left\{\begin{array}{l}
a_{0}=0 \\
a_{n+1}=\left(a_{n}+1\right)^{3}
\end{array}\right.
$$
It follows inductively that \( p(a_{n}) = a_{n} \). Since the polynomials \( p \) and \( x \) agree on infinitely many points, they must be equal, so \( p(x) = x \). | olympiads_ref | yoonholee/math-corpus-combined | 6.8 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
766509ecbc65839f | Prove that if the real numbers $a, b$ and $c$ satisfy $a^{2}+b^{2}+c^{2}=3$ then
$$
\frac{a^{2}}{2+b+c^{2}}+\frac{b^{2}}{2+c+a^{2}}+\frac{c^{2}}{2+a+b^{2}} \geq \frac{(a+b+c)^{2}}{12} .
$$
When does equality hold? | proof | Let $2+b+c^{2}=u, 2+c+a^{2}=v, 2+a+b^{2}=w$. We note that it follows from $a^{2}+b^{2}+c^{2}=3$ that $a, b, c \geq-\sqrt{3}>-2$. Therefore, $u, v$ and $w$ are positive. From the Cauchy-Schwartz inequality we get then
$$
\begin{aligned}
(a+b+c)^{2} & =\left(\frac{a}{\sqrt{u}} \sqrt{u}+\frac{b}{\sqrt{v}} \sqrt{v}+\frac{... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Inequalities", "competition": null, "grade": null, "yoonholee_split": "train"} | ||
4ec30c4edc0cb999 | Does there exist an angle $\alpha \in(0, \pi / 2)$ such that $\sin \alpha, \cos \alpha, \tan \alpha$ and $\cot \alpha$, taken in some order, are consecutive terms of an arithmetic progression?
Answer: No. | proof | Suppose that there is an $x$ such that $0 < \sin x$, we can reduce by $\cos x - \sin x$ and get
$$
1 = \frac{\cos x + \sin x}{\cos x \sin x} = \frac{1}{\sin x} + \frac{1}{\cos x}.
$$
But $0 < \sin x < \cos x$ implies $\frac{1}{\sin x} > \frac{1}{\cos x}$, so $\frac{1}{\sin x} + \frac{1}{\cos x} > 2$, which is a contr... | olympiads_ref | yoonholee/math-corpus-combined | 6 | {"dataset": "NuminaMath", "topic": "Algebra", "problem_type": "Algebra", "competition": null, "grade": null, "yoonholee_split": "train"} |
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