problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
G2 For a fixed triangle $A B C$ we choose a point $M$ on the ray $C A$ (after $A$ ), a point $N$ on the ray $A B$ (after $B$ ) and a point $P$ on the ray $B C$ (after $C$ ) in a way such that $A M-B C=B N-A C=C P-A B$. Prove that the angles of triangle $M N P$ do not depend on the choice of $M, N, P$.
| ## Solution
Consider the points $M^{\prime}$ on the ray $B A$ (after $A$ ), $N^{\prime}$ on the ray $C B$ (after $B$ ) and $P^{\prime}$ on the ray $A C$ (after $C$ ), so that $A M=A M^{\prime}, B N=B N^{\prime}, C P=C P^{\prime}$. Since $A M-B C=B N-A C=B N^{\prime}-A C$, we get $C M=A C+A M=B C+B N^{\prime}=C N^{\pri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 321 |
G3 The vertices $A$ and $B$ of an equilateral $\triangle A B C$ lie on a circle $k$ of radius 1 , and the vertex $C$ is inside $k$. The point $D \neq B$ lies on $k, A D=A B$ and the line $D C$ intersects $k$ for the second time in point $E$. Find the length of the segment $C E$.
| ## Solution
As $A D=A C, \triangle C D A$ is isosceles. If $\varangle A D C=\varangle A C D=\alpha$ and $\varangle B C E=\beta$, then $\beta=120^{\circ}-\alpha$. The quadrilateral $A B E D$ is cyclic, so $\varangle A B E=180^{\circ}-\alpha$. Then $\varangle C B E=$ $120^{\circ}-\alpha$ so $\varangle C B E=\beta$. Thus... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 322 |
G4 Let $A B C$ be a triangle, $(B C<A B)$. The line $\ell$ passing trough the vertices $C$ and orthogonal to the angle bisector $B E$ of $\angle B$, meets $B E$ and the median $B D$ of the side $A C$ at points $F$ and $G$, respectively. Prove that segment $D F$ bisect the segment $E G$.
| ## Solution
We will prove that desired covering is impossible.
Let assume the opposite i.e. a square with side length $a$, can be tiled with $k$ congruent right angled triangles, whose sides are of lengths $b, b \sqrt{3}$ and $2 b$.
Then the area of such a triangle is $\frac{b^{2} \sqrt{3}}{2}$.
And the area of the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 324 |
G6 Let $A B C$ be a triangle with $A<90^{\circ}$. Outside of a triangle we consider isosceles triangles $A B E$ and $A C Z$ with bases $A B$ and $A C$, respectively. If the midpoint $D$ of the side $B C$ is such that $D E \perp D Z$ and $E Z=2 \cdot E D$, prove that $\widehat{A E B}=2 \cdot \widehat{A Z C}$.
| ## Solution
Since $D$ is the midpoint of the side $B C$, in the extension of the line segment $Z D$ we take a point $H$ such that $Z D=D H$. Then the quadrilateral $B H C Z$ is parallelogram and therefore we have
$$
B H=Z C=Z A
$$
=a b \sin A$.
But it is also true that $(A B C D)=4(A O D)=4 \cdot \frac{O A \cdot O D}{2} \sin \theta=2 O A \cdot O D \sin \theta=$ $=2 \cdot \frac{x}{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 327 |
G9 Let $O$ be a point inside the parallelogram $A B C D$ such that
$$
\angle A O B+\angle C O D=\angle B O C+\angle C O D
$$
Prove that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\triangle A O B, \triangle B O C, \triangle C O D$ and $\triangle D O A$.
 point on $\Gamma$. Denote by $\gamma$ the circle of diameter $A M$, by $X$ the (other than $M$ ) intersection... | ## Solution
Consider the line $\rho$ tangent to $\gamma$ at $A$, and take the points $\{K\}=A M \cap X Y,\{L\}=$ $\rho \cap X M$, and $\{F\}=O A \cap X Y$.
(Remark: Moving $M$ into its reflection with respect to the line $O A$ will move $X Y$ into its reflection with respect to $O A$. These old and the new $X Y$ meet... | proof | Geometry | proof | Yes | Yes | olympiads | false | 329 |
G11 Consider $A B C$ an acute-angled triangle with $A B \neq A C$. Denote by $M$ the midpoint of $B C$, by $D, E$ the feet of the altitudes from $B, C$ respectively and let $P$ be the intersection point of the lines $D E$ and $B C$. The perpendicular from $M$ to $A C$ meets the perpendicular from $C$ to $B C$ at point... |
Solution
Let $F$ be the foot of the altitude from $A$ and let $S$ be the intersection point of $A M$ and $R C$. As $P C$ is an altitude of the triangle $P R S$, the claim is equivalent to $R M \perp P S$, since the latter implies that $M$ is the orthocenter of $P R S$. Due to $R M \perp A C$, we need to prove that $A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 330 |
NT1 Find all the positive integers $x$ and $y$ that satisfy the equation
$$
x(x-y)=8 y-7
$$
| ## Solution 1:
The given equation can be written as:
$$
\begin{aligned}
& x(x-y)=8 y-7 \\
& x^{2}+7=y(x+8)
\end{aligned}
$$
Let $x+8=m, m \in \mathbb{N}$. Then we have: $x^{2}+7 \equiv 0(\bmod m)$, and $x^{2}+8 x \equiv 0(\bmod m)$. So we obtain that $8 x-7 \equiv 0(\bmod m) \quad(1)$.
Also we obtain $8 x+8^{2}=8(x... | (x,y)=(63,56) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 331 |
NT2 Let $n \geq 2$ be a fixed positive integer. An integer will be called " $n$-free" if it is not a multiple of an $n$-th power of a prime. Let $M$ be an infinite set of rational numbers, such that the product of every $n$ elements of $M$ is an $n$-free integer. Prove that $M$ contains only integers.
| ## Solution
We first prove that $M$ can contain only a finite number of non-integers. Suppose that there are infinitely many of them: $\frac{p_{1}}{q_{1}}, \frac{p_{2}}{q_{2}}, \ldots, \frac{p_{k}}{q_{k}}, \ldots$, with $\left(p_{k}, q_{k}\right)=1$ and $q_{k}>1$ for each $k$. Let $\frac{p}{q}=\frac{p_{1} p_{2} \ldots... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 332 |
NT3 Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots a_{n}, \ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\left(a_{n}\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.
| ## Solution
Since $a_{n-1} \equiv s\left(a_{n-1}\right)$ (all congruences are modulo 9 ), we have $2 a_{n-1} \equiv a_{n} \equiv 2008 \equiv 10$, so $a_{n-1} \equiv 5$. But $a_{n-1}<2008$, so $s\left(a_{n-1}\right) \leq 28$ and thus $s\left(a_{n-1}\right)$ can equal 5,14 or 23 . We check $s(2008-5)=s(2003)=5, s(2008-1... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 333 |
NT4 Find all integers $n$ such that $n^{4}+8 n+11$ is a product of two or more consecutive integers.
|
Solution
We will prove that $n^{4}+8 n+11$ is never a multiple of 3 . This is clear if $n$ is a multiple of 3 . If
$n$ is not a multiple of 3 , then $n^{4}+8 n+11=\left(n^{4}-1\right)+12+8 n=(n-1)(n+1)\left(n^{2}+1\right)+12+8 n$, where $8 n$ is the only term not divisible by 3 . Thus $n^{4}+8 n+11$ is never the prod... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 334 |
NT5 Is it possible to arrange the numbers $1^{1}, 2^{2}, \ldots, 2008^{2008}$ one after the other, in such a way that the obtained number is a perfect square? (Explain your answer.)
| ## Solution
We will use the following lemmas.
Lemma 1. If $x \in \mathbb{N}$, then $x^{2} \equiv 0$ or $1(\bmod 3)$.
Proof: Let $x \in \mathbb{N}$, then $x=3 k, x=3 k+1$ or $x=3 k+2$, hence
$$
\begin{aligned}
& x^{2}=9 k^{2} \equiv 0(\bmod 3) \\
& x^{2}=9 k^{2}+6 k+1 \equiv 1(\bmod 3), \\
& x^{2}=9 k^{2}+12 k+4 \eq... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 335 |
NT6 Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function, satisfying the following condition:
for every integer $n>1$, there exists a prime divisor $p$ of $n$ such that $f(n)=f\left(\frac{n}{p}\right)-f(p)$. If
$$
f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right)=2006
$$
determine the value o... | ## Solution
If $n=p$ is prime number, we have
$$
f(p)=f\left(\frac{p}{p}\right)-f(p)=f(1)-f(p)
$$
i.e.
$$
f(p)=\frac{f(1)}{2}
$$
If $n=p q$, where $p$ and $q$ are prime numbers, then
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)=f(q)-f(p)=\frac{f(1)}{2}-\frac{f(1)}{2}=0
$$
If $n$ is a product of three prime numbers, we... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 336 |
NT7 Determine the minimal prime number $p>3$ for which no natural number $n$ satisfies
$$
2^{n}+3^{n} \equiv 0(\bmod p)
$$
| ## Solution
We put $A(n)=2^{n}+3^{n}$. From Fermat's little theorem, we have $2^{p-1} \equiv 1(\bmod p)$ and $3^{p-1} \equiv 1(\bmod p)$ from which we conclude $A(n) \equiv 2(\bmod p)$. Therefore, after $p-1$ steps
at most, we will have repetition of the power. It means that in order to determine the minimal prime num... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 337 |
NT8 Let $a, b, c, d, e, f$ are nonzero digits such that the natural numbers $\overline{a b c}, \overline{d e f}$ and $\overline{a b c d e f}$ are squares.
a) Prove that $\overline{a b c d e f}$ can be represented in two different ways as a sum of three squares of natural numbers.
b) Give an example of such a number.... |
Solution
a) Let $\overline{a b c}=m^{2}, \overline{d e f}=n^{2}$ and $\overline{a b c d e f}=p^{2}$, where $11 \leq m \leq 31,11 \leq n \leq 31$ are natural numbers. So, $p^{2}=1000 \cdot m^{2}+n^{2}$. But $1000=30^{2}+10^{2}=18^{2}+26^{2}$. We obtain the following relations
$$
\begin{gathered}
p^{2}=\left(30^{2}+10... | 225625=475^{2}=450^{2}+150^{2}+25^{2}=270^{2}+390^{2}+25^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 338 |
NT9 Let $p$ be a prime number. Find all positive integers $a$ and $b$ such that:
$$
\frac{4 a+p}{b}+\frac{4 b+p}{a}
$$
and
$$
\frac{a^{2}}{b}+\frac{b^{2}}{a}
$$
are integers.
| ## Solution
Since $a$ and $b$ are symmetric we can assume that $a \leq b$. Let $d=(a, b), a=d u, b=d v$ and $(u, v)=1$. Then we have:
$$
\frac{a^{2}}{b}+\frac{b^{2}}{a}=\frac{d\left(u^{3}+v^{3}\right)}{u v}
$$
Since,
$$
\left(u^{3}+v^{3}, u\right)=\left(u^{3}+v^{3}, v\right)=1
$$
we deduce that $u \mid d$ and $v \... | (,b)={(1,1),(2,2),(p,p),(2p,2p),(5,25),(6,18),(18,6),(25,5),(30,150),(150,30)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 339 |
NT10 Prove that $2^{n}+3^{n}$ is not a perfect cube for any positive integer $n$.
| ## Solution
If $n=1$ then $2^{1}+3^{1}=5$ is not perfect cube.
Perfect cube gives residues $-1,0$ and 1 modulo 9 . If $2^{n}+3^{n}$ is a perfect cube, then $n$ must be divisible with 3 (congruence $2^{n}+3^{n}=x^{3}$ modulo 9 ).
If $n=3 k$ then $2^{3 k}+3^{2 k}>\left(3^{k}\right)^{3}$. Also, $\left(3^{k}+1\right)^{3... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 340 |
NT11 Determine the greatest number with $n$ digits in the decimal representation which is divisible by 429 and has the sum of all digits less than or equal to 11 .
| ## Solution
Let $A=\overline{a_{n} a_{n-1} \ldots a_{1}}$ and notice that $429=3 \cdot 11 \cdot 13$.
Since the sum of the digits $\sum a_{i} \leq 11$ and $\sum a_{i}$ is divisible by 3 , we get $\sum a_{i}=3,6$ or 9. As 11 divides $A$, we have
$$
11 \mid a_{n}-a_{n-1}+a_{n-2}-a_{n-3}+\ldots
$$
in other words $11 \m... | 30030000\ldots | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 341 |
NT12 Solve the equation $\frac{p}{q}-\frac{4}{r+1}=1$ in prime numbers.
| ## Solution
We can rewrite the equation in the form
$$
\begin{gathered}
\frac{p r+p-4 q}{q(r+1)}=1 \Rightarrow p r+p-4 q=q r+q \\
p r-q r=5 q-p \Rightarrow r(p-q)=5 q-p
\end{gathered}
$$
It follows that $p \neq q$ and
$$
\begin{gathered}
r=\frac{5 q-p}{p-q}=\frac{4 q+q-p}{p-q} \\
r=\frac{4 q}{p-q}-1
\end{gathered}
... | (p,q,r)=(3,2,7), | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 342 |
ALG 1. A number $A$ is written with $2 n$ digits, each of whish is 4 , and a number $B$ is written with $n$ digits, each of which is 8 . Prove that for each $n, A+2 B+4$ is a total square.
| ## Solution.
$$
\begin{aligned}
A & =\underbrace{44 \ldots 44}_{2 n}=\underbrace{44 \ldots 4}_{n} \underbrace{44 \ldots 4}_{n}=\underbrace{44 \ldots 4}_{n} \underbrace{400 \ldots 0}_{n}-\underbrace{44 \ldots 4}_{n}+\underbrace{88 \ldots 8}_{n}=\underbrace{44 \ldots 4}_{n} \cdot\left(10^{n}-1\right)+B \\
& =4 \cdot \un... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 343 |
ALG 2. Let $a, b, c$ be lengths of triangle sides, $p=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $q=\frac{a}{c}+\frac{c}{b}+\frac{b}{a}$.
Prove that $|p-q|<1$.
|
Solution: One has
$$
\begin{aligned}
a b c|p-q| & =a b c\left|\frac{c-b}{a}+\frac{a-c}{b}+\frac{b-a}{c}\right| \\
& =\left|b c^{2}-b^{2} c+a^{2} c-a c^{2}+a b^{2}-a^{2} b\right|= \\
& =\left|a b c-a c^{2}-a^{2} b+a^{2} c-b^{2} c+b c^{2}+a b^{2}-a b c\right|= \\
& =\left|(b-c)\left(a c-a^{2}-b c+a b\right)\right|= \\
... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 344 |
ALG 3: Let $a, b, c$ be real numbers such that $a^{2}+b^{2}+c^{2}=1$. Prove that. $P=a b+b c+c a-2(a+b+c) \geq-\frac{5}{2}$. Are there values of $a, b, c$, such that $P=-\frac{5}{2}$.
|
Solution: We have $a b+b c+c a=\frac{(a+b+c)^{2}-c^{2}-b^{2}-a^{2}}{2}=\frac{(a+b+c)^{2}-1}{2}$.
If put $t=a+b+c$ we obtain
$$
P=\frac{t^{2}-1}{2}-2 t=\frac{t^{2}-4 t-1}{2}=\frac{(t-2)^{2}-5}{2} \geq-\frac{5}{2}
$$
Obviously $P=-\frac{5}{2}$ when $t=2$, i.e. $a+b+c=2$, or $c=2-a-b$. Substitute in $a^{2}+b^{2}+c^{2}... | P\geq-\frac{5}{2} | Inequalities | proof | Yes | Yes | olympiads | false | 345 |
## ALG 4.
Let $a, b, c$ be rational numbers such that
$$
\frac{1}{a+b c}+\frac{1}{b+a c}=\frac{1}{a+b}
$$
Prove that $\sqrt{\frac{c-3}{c+1}}$ is also a rational number
|
Solution. By cancelling the denominators
$$
(a+b)^{2}(1+c)=a b+c\left(a^{2}+b^{2}\right)+a b c^{2}
$$
and
$$
a b(c-1)^{2}=(a+b)^{2}
$$
If $c=-1$, we obtrin the contradiction
$$
\frac{1}{a-b}+\frac{1}{b-a}=\frac{1}{a+b}
$$
Furtherrdore,
$$
\begin{aligned}
(c-3)(c+1) & =(c-1)^{2}-4=\frac{(a+b)^{2}}{a b}-4 \\
& =\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 346 |
ALG 5. Let $A B C$ be a scalene triangle with $B C=a, A C=b$ and $A B=c$, where $a_{r} b, c$ are positive integers. Prove that
$$
\left|a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a\right| \geq 2
$$
|
Solution. Denote $E=a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a$. We have
$$
\begin{aligned}
E= & \left(a b c-c^{2} a\right)+\left(c a^{2}-a^{2} b\right)+\left(b c^{2}-b^{2} c\right)+\left(a b^{2}-a b c\right)= \\
& (b-c)\left(a c-a^{2}-b c+a b\right)=(b-c)\left(a a^{2}-b\right)(c-a)
\end{aligned}
$$
So, $|E|=|a... | 2 | Algebra | proof | Yes | Yes | olympiads | false | 347 |
ALG 6'. Let $a, b, c$ be positive numbers such that $a b+b c+c a=3$. Prove that
$$
a+b+c \geq a b c+2
$$
|
Solution. Eliminating $c$ gives
$$
a+b+c-a b c=a+b+(1-a b) c=a+b+\frac{(1-a b)(3-a b)}{a+b}
$$
Put $x=\sqrt{a b}$. Then $a+b \geq 2 x$, and since $1<x^{2}<3, \frac{(1-a b)(3-a b)}{a+b} \geq \frac{\left(1-x^{2}\right)\left(3-x^{2}\right)}{2 x}$.
It then suffices to prove that
$$
2 x+\frac{\left(1-x^{2}\right)\left(... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 349 |
ALG 7 .
Let $x, y, z$ be real numbers greater than -1 . Prove that
$$
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2
$$
|
Solution. We have $y \leq \frac{1+y^{2}}{2}$, hence $\quad$
$$
\frac{1+x^{2}}{1+y+z^{2}} \geq \frac{1+x^{2}}{1+z^{2}+\frac{1+\dot{y}^{2}}{2}}
$$
and the similar inequalities.
Setting $a=1+x^{2}, b=1+y^{2}, c=1+z^{2}$, it sufices to prove that
$$
\frac{a}{2 c+b}+\frac{b}{2 a+c}+\frac{c}{2 b+a} \geq 1
$$
for all $a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 350 |
ALG 8. Prove that there exist two sets $A=\{x, y, z\}$ and $B=\{m, n, p\}$ of positive integers greater than 2003 such that the sets have no common elements and the equalities $x+y+z=m+n+p$ and $x^{2}+y^{2}+z^{2}=m^{2}+n^{2}+p^{2}$ hold.
|
Solution. Let $A B C$ be a triangle with $B C=a, A C=b, A B=c$ and $ak+3=c
$$
a triangle with such length sides there exist. After the simple calculations we have
$$
\begin{gathered}
A=\left\{3(k+1)^{2}-2,3(k+2)^{2}+4,3(k+3)^{2}-2\right\} \\
B=\left\{3(k+1)^{2}, 3(k+2)^{2}, 3(k+3)^{2}\right\}
\end{gathered}
$$
It e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 351 |
COM 2 Natural numbers 1,2,3, .., 2003 are written in an arbitrary sequence $a_{1}, a_{2}, a_{3}, \ldots a_{2003}$. Let $b_{1}=1 a_{1}, b_{2}=2 a_{2}, b_{3}=3 a_{3}, \ldots, b_{2003}=2003 a_{2003}$, and $B$ be the maximum of the numbers $b_{1}, b_{2}, b_{3}, \ldots, b_{2003}$.
a) If $a_{1}=2003, a_{2}=2002, a_{3}=2001... |
Solution: a) Using the inequality between the arithmetical and geometrical mean, we obtain that $b_{n}=n(2004-n) \leq\left(\frac{n+(2004-n)}{2}\right)^{2}=1002^{2}$ for $n=1,2,3, \ldots, 2003$. The equality holds if and only if $n=2004-n$, i.e. $n=1002$. Therefore, $B=b_{1002}=1002 \times(2004-1002)=1002^{2}$. b) Let ... | 1002^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 353 |
COM 3. Prove that amongst any 29 natural numbers there are 15 such that sum of them is divisible by 15 .
|
Solution: Amongst any 5 natural numbers there are 3 such that sum of them is divisible by 3 . Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisiblc by 3. It is easy to see ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 354 |
COM 5. If $m$ is a number from the set $\{1,2,3,4\}$ and each point of the plane is painted in red or blue, prove that in the plane there exists at least an equilateral triangle with the vertices of the same colour and with length side $m$.
|
Solution. Suppose that in the plane there no exists an equilateral triangle with the vertices of the same colour and length side $m=1,2,3,4$.
First assertion: we shall prove that in the plane there no exists a segment with the length 2 such that the ends and the midpint of this segment have the same colour. Suppose t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 355 |
GEO 1. Is there a convex quadrilateral, whose diagonals divide it into four triangles, such that their areas are four distinct prime integers.
|
Solution. No. Let the areas of those triangles be the prime numbers $p, q, r$ and $t$. But for the areas of the triangles we have $\mathrm{pq}=\mathrm{rt}$, where the triangles with areas $\mathrm{p}$ and $\mathrm{q}$ have only a common vertex. This is not possible for distinct primes.
| proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 356 |
GEO 2. Is there a triangle whose area is $12 \mathrm{~cm}^{2}$ and whose perimeter is $12 \mathrm{~cm}$.
|
Solution. No. Let $\mathrm{r}$ be the radius of the inscribed circle. Then $12=6 \mathrm{r}$, i.e. $\mathrm{r}=2 \mathrm{~cm}$. But the area of the inscribed circle is $4 \pi>12$, and it is known that the area of any triangle is bigger than the area of its inscribed circle.
| proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 357 |
## GEO 3.
Let $G$ be the centroid of the the triangle $A B C$. Reflect point $A$ across $C$ at $A^{\prime}$. Prove that $G, B, C, A^{\prime}$ are on the same circle if and only if $G A$ is perpendicular to $G C$.
|
Solution. Observe first that $G A \perp G C$ if and only if $5 A C^{2}=A B^{2}+B C^{2}$. Indeed,
$$
G A \perp G C \Leftrightarrow \frac{4}{9} m_{a}^{2}+\frac{4}{9} m_{c}^{2}=b^{2} \Leftrightarrow 5 b^{2}=a^{2}+c^{2}
$$
Moreover,
$$
G B^{2}=\frac{4}{9} m_{b}^{2}=\frac{2 a^{2}+2 c^{2}-b^{2}}{9}=\frac{9 b^{2}}{9}=b^{2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 358 |
GEO 5. Let three congruent circles intersect in one point $M$ and $A_{1}, A_{2}$ and $A_{3}$ be the other intersection points for those circles. Prove that $M$ is a.orthocenter for a triangle $A_{1} A_{2} A_{3}$.
|
Solution: The quadrilaterals $\mathrm{O}_{3} M O_{2} A_{1}, \mathrm{O}_{3} M O_{1} A_{2}$ and $O_{1} M O_{2} A_{3}$ are rombes. Therefore, $O_{2} A_{1} \| M O_{3}$ and $M O_{3} \| O_{1} A_{2}$, which imply $O_{2} A_{1} \| O_{1} A_{2}$. Because $O_{2} A_{1}=O_{3}{ }^{*} M=O_{1} A_{2}$ the quadrilateral $O_{2} A_{1} A_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 359 |
GEO 7. Through a interior point of a triangle, three lines parallel to the sides of the triangle are constructed. In that way the triangle is divided on six figures, areas equal $a, b, c, \alpha, \beta, \gamma$ (see the picture).
 Let $F$ and $G$ be the centers of the two circles of radius $R$ passing through $A$ and $B$; and $B$ and $C$, respectively. Let $O$ be the point for which the the rectangle $A B G O$ is a parallelogram. Then $\angle O A D=\angle G B C$, and the triangles $O A D$ and $G B C$ are congruent (sas... | proof | Geometry | proof | Yes | Yes | olympiads | false | 363 |
87.3. Let $f$ be a strictly increasing function defined in the set of natural numbers satisfying the conditions $f(2)=a>2$ and $f(m n)=f(m) f(n)$ for all natural numbers $m$ and $n$. Determine the smallest possible value of $a$.
|
Solution. Since $f(n)=n^{2}$ is a function satisfying the conditions of the problem, the smallest posiible $a$ is at most 4. Assume $a=3$. It is easy to prove by induction that $f\left(n^{k}\right)=f(n)^{k}$ for all $k \geq 1$. So, taking into account that $f$ is strictly increasing, we get
$$
\begin{gathered}
f(3)^{... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 364 |
87.4. Let $a, b$, and $c$ be positive real numbers. Prove:
$$
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
|
Solution. The arithmetic-geometric inequality yields
$$
3=3 \sqrt[3]{\frac{a^{2}}{b^{2}} \cdot \frac{b^{2}}{c^{2}} \cdot \frac{c^{2}}{a^{2}}} \leq \frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}
$$
or
$$
\sqrt{3} \leq \sqrt{\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}}
$$
On the other ha... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 365 |
88.1. The positive integer $n$ has the following property: if the three last digits of $n$ are removed, the number $\sqrt[3]{n}$ remains. Find $n$.
|
Solution. If $x=\sqrt[3]{n}$, and $y, 0 \leq y1000$, and $x>31$. On the other hand, $x^{3}<1000 x+1000$, or $x\left(x^{2}-1000\right)<1000$. The left hand side of this inequality is an increasing function of $x$, and $x=33$ does not satisfy the inequality. So $x<33$. Since $x$ is an integer, $x=32$ and $n=32^{3}=32768... | 32768 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 366 |
88.2. Let $a, b$, and $c$ be non-zero real numbers and let $a \geq b \geq c$. Prove the inequality
$$
\frac{a^{3}-c^{3}}{3} \geq a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$$
When does equality hold?
|
Solution. Since $c-b \leq 0 \leq a-b$, we have $(a-b)^{3} \geq(c-b)^{3}$, or
$$
a^{3}-3 a^{2} b+3 a b^{2}-b^{3} \geq c^{3}-3 b c^{2}+3 b^{2} c-b^{3}
$$
On simplifying this, we immediately have
$$
\frac{1}{3}\left(a^{3}-c^{3}\right) \geq a^{2} b-a b^{2}+b^{2} c-b c^{2}=a b c\left(\frac{a-b}{c}+\frac{b-c}{a}\right)
$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 367 |
88.3. Two concentric spheres have radii $r$ and $R, r<R$. We try to select points $A, B$ and $C$ on the surface of the larger sphere such that all sides of the triangle $A B C$ would be tangent to the surface of the smaller sphere. Show that the points can be selected if and only if $R \leq 2 r$.
|
Solution. Assume $A, B$, and $C$ lie on the surface $\Gamma$ of a sphere of radius $R$ and center $O$, and $A B, B C$, and $C A$ touch the surface $\gamma$ of a sphere of radius $r$ and center $O$. The circumscribed and inscribed circles of $A B C$ then are intersections of the plane $A B C$ with $\Gamma$ and $\gamma$... | R\leq2r | Geometry | proof | Yes | Yes | olympiads | false | 368 |
88.4. Let $m_{n}$ be the smallest value of the function
$$
f_{n}(x)=\sum_{k=0}^{2 n} x^{k}
$$
Show that $m_{n} \rightarrow \frac{1}{2}$, as $n \rightarrow \infty$.
|
Solution. For $n>1$,
$$
\begin{gathered}
f_{n}(x)=1+x+x^{2}+\cdots \\
=1+x\left(1+x^{2}+x^{4}+\cdots\right)+x^{2}\left(1+x^{2}+x^{4} \cdots\right) \\
=1+x(1+x) \sum_{k=0}^{n-1} x^{2 k}
\end{gathered}
$$
From this we see that $f_{n}(x) \geq 1$, for $x \leq-1$ and $x \geq 0$. Consequently, $f_{n}$ attains its minimum ... | \frac{1}{2} | Algebra | proof | Yes | Yes | olympiads | false | 369 |
89.2. Three sides of a tetrahedron are right-angled triangles having the right angle at their common vertex. The areas of these sides are $A, B$, and $C$. Find the total surface area of the tetrahedron.
|
Solution 1. Let $P Q R S$ be the tetrahedron of the problem and let $S$ be the vertex common to the three sides which are right-angled triangles. Let the areas of $P Q S, Q R S$, and $R P S$ be $A, B$, and $C$, respectively. Denote the area of $Q R S$ by $X$. If $S S^{\prime}$ is the altitude from $S$ (onto $P Q R$ ) ... | A+B+C+\sqrt{A^{2}+B^{2}+C^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 370 |
89.3. Let $S$ be the set of all points $t$ in the closed interval $[-1,1]$ such that for the sequence $x_{0}, x_{1}, x_{2}, \ldots$ defined by the equations $x_{0}=t, x_{n+1}=2 x_{n}^{2}-1$, there exists a positive integer $N$ such that $x_{n}=1$ for all $n \geq N$. Show that the set $S$ has infinitely many elements.
|
Solution. All numbers in the sequence $\left\{x_{n}\right\}$ lie in the interval $[-1,1]$. For each $n$ we can pick an $\alpha_{n}$ such that $x_{n}=\cos \alpha_{n}$. If $x_{n}=\cos \alpha_{n}$, then $x_{n+1}=2 \cos ^{2} \alpha_{n}-1=\cos \left(2 \alpha_{n}\right)$. The nuber $\alpha_{n+1}$ can be chosen as $2 \alpha_... | notfound | Algebra | proof | Yes | Yes | olympiads | false | 371 |
90.1. Let $m, n$, and $p$ be odd positive integers. Prove that the number
$$
\sum_{k=1}^{(n-1)^{p}} k^{m}
$$
is divisible by $n$.
|
Solution. Since $n$ is odd, the sum has an even number of terms. So we can write it as
$$
\sum_{k=1}^{\frac{1}{2}(n-1)^{p}}\left(k^{m}+\left((n-1)^{p}-k+1\right)^{m}\right)
$$
Because $m$ is odd, each term in the sum has $k+(n-1)^{p}-k+1=(n-1)^{p}+1$ as a factor. As $p$ is odd, too, $(n-1)^{p}+1=(n-1)^{p}+1^{p}$ has... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 372 |
90.2. Let $a_{1}, a_{2}, \ldots, a_{n}$ be real numbers. Prove
$$
\sqrt[3]{a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}} \leq \sqrt{a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}}
$$
When does equality hold in (1)?
|
Solution. If $0 \leq x \leq 1$, then $x^{3 / 2} \leq x$, and equality holds if and only if $x=0$ or $x=1$. - The inequality is true as an equality, if all the $a_{k}$ 's are zeroes. Assume that at least one of the numbers $a_{k}$ is non-zero. Set
$$
x_{k}=\frac{a_{k}^{2}}{\sum_{j=1}^{n} a_{j}^{2}}
$$
Then $0 \leq x_... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 373 |
90.3. Let $A B C$ be a triangle and let $P$ be an interior point of $A B C$. We assume that a line $l$, which passes through $P$, but not through $A$, intersects $A B$ and $A C$ (or their extensions over $B$ or $C$ ) at $Q$ and $R$, respectively. Find $l$ such that the perimeter of the triangle $A Q R$ is as small as ... |
Solution. (See Figure 2.) Let
$$
s=\frac{1}{2}(A R+R Q+Q A)
$$
Let $\mathcal{C}$ be the excircle of $A Q R$ tangent to $Q R$, i.e. the circle tangent to $Q R$ and the extensions of $A R$ and $A Q$. Denote the center of $\mathcal{C}$ by $I$ and the measure of $\angle Q A R$ by $\alpha . I$ is on the bisector of $\ang... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 374 |
90.4. It is possible to perform three operations $f, g$, and $h$ for positive integers: $f(n)=$ $10 n, g(n)=10 n+4$, and $h(2 n)=n$; in other words, one may write 0 or 4 in the end of the number and one may divide an even number by 2. Prove: every positive integer can be constructed starting from 4 and performing a fi... |
Solution. All odd numbers $n$ are of the form $h(2 n)$. All we need is to show that every even number can be obtained fron 4 by using the operations $f, g$, and $h$. To this end, we show that a suitably chosen sequence of inverse operations $F=f^{-1}, G=g^{-1}$, and $H=h^{-1}$ produces a smaller even number or the num... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 375 |
91.1. Determine the last two digits of the number
$$
2^{5}+2^{5^{2}}+2^{5^{3}}+\cdots+2^{5^{1991}}
$$
written in decimal notation.
|
Solution. We first show that all numbers $2^{5^{k}}$ are of the form $100 p+32$. This can be shown by induction. The case $k=1$ is clear $\left(2^{5}=32\right)$. Assume $2^{5^{k}}=100 p+32$. Then, by the binomial formula,
$$
2^{5^{k+1}}=\left(2^{5^{k}}\right)^{5}=(100 p+32)^{5}=100 q+32^{5}
$$
and
$$
\begin{gathere... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 376 |
91.3. Show that
$$
\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}<\frac{2}{3}
$$
for all $n \geq 2$.
|
Solution. Since
$$
\frac{1}{j^{2}}<\frac{1}{j(j-1)}=\frac{1}{j-1}-\frac{1}{j}
$$
we have
$$
\begin{gathered}
\sum_{j=k}^{n} \frac{1}{j^{2}}<\left(\frac{1}{k-1}-\frac{1}{k}\right)+\left(\frac{1}{k}-\frac{1}{k+1}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right) \\
=\frac{1}{k-1}-\frac{1}{n}<\frac{1}{k-1}
\end{gat... | \frac{2389}{3600}<\frac{2}{3} | Inequalities | proof | Yes | Yes | olympiads | false | 378 |
91.4. Let $f(x)$ be a polynomial with integer coefficients. We assume that there exists a positive integer $k$ and $k$ consecutive integers $n, n+1, \ldots, n+k-1$ so that none of the numbers $f(n), f(n+1), \ldots, f(n+k-1)$ is divisible by $k$. Show that the zeroes of $f(x)$ are not integers.
|
Solution. Let $f(x)=a_{0} x^{d}+a_{1} x^{d-1}+\cdots+a_{d}$. Assume that $f$ has a zero $m$ which is an integer. Then $f(x)=(x-m) g(x)$, where $g$ is a polynomial. If $g(x)=b_{0} x^{d-1}+b_{1} x^{d-2}+$ $\cdots+b_{d-1}$, then $a_{0}=b_{0}$, and $a_{k}=b_{k}-m b_{k-1}, 1 \leq k \leq d-1$. So $b_{0}$ is an integer, and ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 379 |
92.1. Determine all real numbers $x>1, y>1$, and $z>1$, satisfying the equation
$$
\begin{aligned}
x+y+z+\frac{3}{x-1} & +\frac{3}{y-1}+\frac{3}{z-1} \\
& =2(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2})
\end{aligned}
$$
|
Solution. Consider the function $f$,
$$
f(t)=t+\frac{3}{t-1}-2 \sqrt{t+2}
$$
defined for $t>1$. The equation of the problem can be written as
$$
f(x)+f(y)+f(z)=0
$$
We reformulate the formula for $f$ :
$$
\begin{aligned}
f(t) & =\frac{1}{t-1}\left(t^{2}-t+3-2(t-1) \sqrt{t+2}\right) \\
& =\frac{1}{t-1}\left(t^{2}-... | \frac{3+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 380 |
92.2. Let $n>1$ be an integer and let $a_{1}, a_{2}, \ldots, a_{n}$ be $n$ different integers. Show that the polynomial
$$
f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{n}\right)-1
$$
is not divisible by any polynomial with integer coefficients and of degree greater than zero but less than $n$ and s... |
Solution. Suppose $g(x)$ is a polynomial of degree $m$, where $1 \leq m<n$, with integer coefficients and leading coefficient 1 , such that
$$
f(x)=g(x) h(x)
$$
whre $h(x)$ is a polynomial. Let
$$
\begin{aligned}
& g(x)=x^{m}+b_{m-1} x^{m-1}+\cdots+b_{1} x+b_{0} \\
& h(x)=x^{n-m}+c_{n-m-1} x^{n-m-1}+\cdots+c_{1} x+... | proof | Algebra | proof | Yes | Yes | olympiads | false | 381 |
92.4. Peter has many squares of equal side. Some of the squares are black, some are white. Peter wants to assemble a big square, with side equal to $n$ sides of the small squares, so that the big square has no rectangle formed by the small squares such that all the squares in the vertices of the rectangle are of equal... |
Solution. We show that Peter only can make a $4 \times 4$ square. The construction is possible, if $n=4$ :
Now consider the case $n=5$. We may assume that at least 13 of the 25 squares are ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 382 |
93.1. Let $F$ be an increasing real function defined for all $x, 0 \leq x \leq 1$, satisfying the conditions
$$
\begin{aligned}
& F\left(\frac{x}{3}\right)=\frac{F(x)}{2} \\
& F(1-x)=1-F(x)
\end{aligned}
$$
Determine $F\left(\frac{173}{1993}\right)$ and $F\left(\frac{1}{13}\right)$.
|
Solution. Condition (i) implies $F(0)=\frac{1}{2} F(0)$, so $F(0)=0$. Because of condition (ii), $F(1)=1-F(0)=1$. Also $F\left(\frac{1}{3}\right)=\frac{1}{2}$ and $F\left(\frac{2}{3}\right)=1-F\left(\frac{1}{3}\right)=\frac{1}{2}$. Since $F$ is an increasing function, this is possible only if $F(x)=\frac{1}{2}$ for al... | \frac{3}{16},\frac{1}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 383 |
93.2. A hexagon is inscribed in a circle of radius $r$. Two of the sides of the hexagon have length 1, two have length 2 and two have length 3. Show that $r$ satisfies the equation
$$
2 r^{3}-7 r-3=0
$$
 We join the vertices of the hexagon to the center $O$ of its circumcircle. We denote by $\alpha$ the central angles corresponding the chords of length 1 , by $\beta$ those corresponding the chords of length 2 , and by $\gamma$ those corresponding the chords of length 3. Clearly $\alpha+\beta+... | 2r^{3}-7r-3=0 | Geometry | proof | Yes | Yes | olympiads | false | 384 |
93.3. Find all solutions of the system of equations
$$
\left\{\begin{aligned}
s(x)+s(y) & =x \\
x+y+s(z) & =z \\
s(x)+s(y)+s(z) & =y-4
\end{aligned}\right.
$$
where $x, y$, and $z$ are positive integers, and $s(x), s(y)$, and $s(z)$ are the numbers of digits in the decimal representations of $x, y$, and $z$, respect... |
Solution. The first equation implies $x \geq 2$ and the first and third equation together imply
$$
s(z)=y-x-4
$$
So $y \geq x+5 \geq 7$. From (1) and the second equation we obtain $z=2 y-4$. Translated to the values of $s$, these equation imply $s(x) \leq s(2 y) \leq s(y)+1$ and $s(x) \leq s(y)$. We insert these ine... | (2,8,12) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 385 |
93.4. Denote by $T(n)$ the sum of the digits of the decimal representation of a positive integer $n$.
a) Find an integer $N$, for which $T(k \cdot N)$ is even for all $k, 1 \leq k \leq 1992$, but $T(1993 \cdot N)$ is odd.
b) Show that no positive integer $N$ exists such that $T(k \cdot N)$ is even for all positive in... |
Solution. a) If $s$ has $n$ decimal digits and $m=10^{n+r} s+s$, then $T(k m)$ is even at least as long as $k s<10^{n+r}$, because all non-zero digits appear in pairs in $k m$. Choose $N=5018300050183$ or $s=50183, n=5, r=3$. Now $1992 \cdot s=99964536<10^{8}$, so $T(k N)$ is even for all $k \leq 1992$. But $1993 \cdo... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 386 |
94.1. Let $O$ be an interior point in the equilateral triangle $A B C$, of side length $a$. The lines $A O, B O$, and $C O$ intersect the sides of the triangle in the points $A_{1}, B_{1}$, and $C_{1}$. Show that
$$
\left|O A_{1}\right|+\left|O B_{1}\right|+\left|O C_{1}\right|<a
$$
|
Solution. Let $H_{A}, H_{B}$, and $H_{C}$ be the orthogonal projections of $O$ on $B C, C A$, and $A B$, respectively. Because $60^{\circ}\left|O A_{1}\right| \frac{\sqrt{3}}{2}
$$
In the same way,
$$
\left|O H_{B}\right|>\left|O B_{1}\right| \frac{\sqrt{3}}{2} \quad \text { and } \quad\left|O H_{C}\right|>\left|O C... | proof | Geometry | proof | Yes | Yes | olympiads | false | 387 |
94.2. We call a finite plane set $S$ consisting of points with integer coefficients a twoneighbour set, if for each point $(p, q)$ of $S$ exactly two of the points $(p+1, q),(p, q+1)$, $(p-1, q),(p, q-1)$ belong to $S$. For which integers $n$ there exists a two-neighbour set which contains exactly $n$ points?
|
Solution. The points $(0,0),(1,0),(1,1),(0,1)$ clearly form a two-neighbour set (which we abbreviate as $2 \mathrm{NS})$. For every even number $n=2 k \geq 8$, the set $S=\{(0,0), \ldots$, $(k-2,0),(k-2,1),(k-2,2), \ldots,(0,2),(0,1)\}$ is a $2 \mathrm{NS}$. We show that there is no $2 \mathrm{NS}$ with $n$ elements f... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 388 |
94.3. A piece of paper is the square $A B C D$. We fold it by placing the vertex $D$ on the point $H$ of the side $B C$. We assume that $A D$ moves onto the segment $G H$ and that $H G$ intersects $A B$ at $E$. Prove that the perimeter of the triangle $E B H$ is one half of the perimeter of the square.
 The fold gives rise to an isosceles trapezium $A D H G$. Because of symmetry, the distance of the vertex $D$ from the side $G H$ equals the distance of the vertex
$H$ from side $A D$; the latter distance is the side length $a$ of the square. The line $G H$ thus is tangent to the circle with c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 389 |
94.4. Determine all positive integers $n<200$, such that $n^{2}+(n+1)^{2}$ is the square of an integer.
|
Solution. We determine the integral solutions of
$$
n^{2}+(n+1)^{2}=(n+p)^{2}, \quad p \geq 2
$$
The root formula for quadratic equations yields
$$
n=p-1+\sqrt{2 p(p-1)} \geq 2(p-1)
$$
Because $n<200$, we have $p \leq 100$. Moreover, the number $2 p(p-1)$ has to be the square of an integer. If $p$ is odd, $p$ and ... | 20,3,119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 390 |
95.1. Let $A B$ be a diameter of a circle with centre $O$. We choose a point $C$ on the circumference of the circle such that $O C$ and $A B$ are perpendicular to each other. Let $P$ be an arbitrary point on the (smaller) arc $B C$ and let the lines $C P$ and $A B$ meet at $Q$. We choose $R$ on $A P$ so that $R Q$ and... |
Solution 1. (See Figure 7.) Draw $P B$. By the Theorem of Thales, $\angle R P B=\angle A P B=$ $90^{\circ}$. So $P$ and $Q$ both lie on the circle with diameter $R B$. Because $\angle A O C=90^{\circ}$, $\angle R P Q=\angle C P A=45^{\circ}$. Then $\angle R B Q=45^{\circ}$, too, and $R B Q$ is an isosceles right trian... | proof | Geometry | proof | Yes | Yes | olympiads | false | 391 |
95.2. Messages are coded using sequences consisting of zeroes and ones only. Only sequences with at most two consecutive ones or zeroes are allowed. (For instance the sequence 011001 is allowed, but 011101 is not.) Determine the number of sequences consisting of exactly 12 numbers.
|
Solution 1. Let $S_{n}$ be the set of acceptable sequences consisting of $2 n$ digits. We partition $S_{n}$ in subsets $A_{n}, B_{n}, C_{n}$, and $D_{n}$, on the basis of the two last digits of the sequence. Sequences ending in 00 are in $A_{n}$, those ending in 01 are in $B_{n}$, those ending in 10 are in $C_{n}$, an... | 466 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 392 |
95.3. Let $n \geq 2$ and let $x_{1}, x_{2}, \ldots x_{n}$ be real numbers satisfying $x_{1}+x_{2}+\ldots+x_{n} \geq 0$ and $x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=1$. Let $M=\max \left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$. Show that
$$
M \geq \frac{1}{\sqrt{n(n-1)}}
$$
When does equality hold in (1)?
|
Solution. Denote by $I$ the set of indices $i$ for which $x_{i} \geq 0$, and by $J$ the set of indices $j$ for which $x_{j}<0$. Let us assume $M<\frac{1}{\sqrt{n(n-1)}}$. Then $I \neq\{1,2, \ldots, n\}$, since otherwise we would have $\left|x_{i}\right|=x_{i} \leq \frac{1}{\sqrt{n(n-1)}}$ for every $i$, and $\sum_{i=1... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 393 |
95.4. Show that there exist infinitely many mutually non-congruent triangles $T$, satisfying
(i) The side lengths of $T$ are consecutive integers.
(ii) The area of $T$ is an integer.
|
Solution. Let $n \geq 3$, and let $n-1, n, n+1$ be the side lengths of the triangle. The semiperimeter of the triangle then equals on $\frac{3 n}{2}$. By Heron's formula, the area of the triangle is
$$
\begin{gathered}
T=\sqrt{\frac{3 n}{2} \cdot\left(\frac{3 n}{2}-n+1\right)\left(\frac{3 n}{2}-n\right)\left(\frac{3 ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 394 |
96.1. Show that there exists an integer divisible by 1996 such that the sum of the its decimal digits is 1996 .
|
Solution. The sum of the digits of 1996 is 25 and the sum of the digits of $2 \cdot 1996=3992$ is 23 . Because $1996=78 \cdot 25+46$, the number obtained by writing 781996 's and two 3992 in succession satisfies the condition of the problem. - As $3 \cdot 1996=5998$, the sum of the digits of 5988 is 30 , and $1996=65 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 395 |
96.2. Determine all real numbers $x$, such that
$$
x^{n}+x^{-n}
$$
is an integer for all integers $n$.
|
Solution. Set $f_{n}(x)=x^{n}+x^{-n}$. $f_{n}(0)$ is not defined for any $n$, so we must have $x \neq 0$. Since $f_{0}(x)=2$ for all $x \neq 0$, we have to find out those $x \neq 0$ for which $f_{n}(x)$ is an integer foe every $n>0$. We note that
$$
x^{n}+x^{-n}=\left(x+x^{-1}\right)\left(x^{n-1}+x^{1-n}\right)-\left... | \frac{}{2}\ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 396 |
96.3. The circle whose diameter is the altitude dropped from the vertex $A$ of the triangle $A B C$ intersects the sides $A B$ and $A C$ at $D$ and $E$, respectively $(A \neq D, A \neq E)$. Show that the circumcentre of $A B C$ lies on the altitude dropped from the vertex $A$ of the triangle $A D E$, or on its extensi... |
Solution. (See Figure 8.) Let $A F$ be the altitude of $A B C$. We may assume that $\angle A C B$ is sharp. From the right triangles $A C F$ and $A F E$ we obtain $\angle A F E=\angle A C F . \angle A D E$ and $\angle A F E$ subtend the same arc, so they are equal. Thus $\angle A C B=\angle A D E$, and the triangles $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 397 |
96.4. The real-valued function $f$ is defined for positive integers, and the positive integer a satisfies
$$
\begin{gathered}
f(a)=f(1995), \quad f(a+1)=f(1996), \quad f(a+2)=f(1997) \\
f(n+a)=\frac{f(n)-1}{f(n)+1} \quad \text { for all positive integers } n
\end{gathered}
$$
(i) Show that $f(n+4 a)=f(n)$ for all po... |
Solution. To prove (i), we the formula $f(n+a)=\frac{f(n)-1}{f(n)+1}$ repeatedly:
$$
\begin{gathered}
f(n+2 a)=f((n+a)+a)=\frac{\frac{f(n)-1}{f(n)+1}-1}{\frac{f(n)-1}{f(n)+1}+1}=-\frac{1}{f(n)} \\
f(n+4 a)=f((n+2 a)+2 a)=-\frac{1}{-\frac{1}{f(n)}}=f(n)
\end{gathered}
$$
(ii) If $a=1$, then $f(1)=f(a)=f(1995)=f(3+498... | 3 | Algebra | proof | Yes | Yes | olympiads | false | 398 |
97.1. Let A be a set of seven positive numbers. Determine the maximal number of triples $(x, y, z)$ of elements of A satisfying $x<y$ and $x+y=z$.
|
Solution. Let $0<a_{1}<a_{2}<\ldots<a_{7}$ be the elements of the set $A$. If $\left(a_{i}, a_{j}, a_{k}\right)$ is a triple of the kind required in the problem, then $a_{i}<a_{j}<a_{i}+a_{j}=a_{k}$. There are at most $k-1$ pairs $\left(a_{i}, a_{j}\right)$ such that $a_{i}+a_{j}=a_{k}$. The number of pairs satisfying... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 399 |
97.2. Let $A B C D$ be a convex quadrilateral. We assume that there exists a point $P$ inside the quadrilateral such that the areas of the triangles $A B P, B C P, C D P$, and $D A P$ are equal. Show that at least one of the diagonals of the quadrilateral bisects the other diagonal.
 We first assume that $P$ does not lie on the diagonal $A C$ and the line $B P$ meets the diagonal $A C$ at $M$. Let $S$ and $T$ be the feet of the perpendiculars from $A$ and $C$ on the line $B P$. The triangles $A P B$ and $C B P$ have equal area. Thus $A S=C T$. If $S \neq T$, then the righ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 400 |
97.3. Let $A, B, C$, and $D$ be four different points in the plane. Three of the line segments $A B, A C, A D, B C, B D$, and $C D$ have length a. The other three have length $b$, where $b>a$. Determine all possible values of the quotient $\frac{b}{a}$.
|
Solution. If the three segments of length $a$ share a common endpoint, say $A$, then the other three points are on a circle of radius $a$, centered at $A$, and they are the vertices of an equilateral triangle of side length $b$. But this means that $A$ is the center of the triangle $B C D$, and
$$
\frac{b}{a}=\frac{b... | \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 401 |
97.4. Let $f$ be a function defined in the set $\{0,1,2, \ldots\}$ of non-negative integers, satisfying $f(2 x)=2 f(x), f(4 x+1)=4 f(x)+3$, and $f(4 x-1)=2 f(2 x-1)-1$. Show that $f$ is an injection, i.e. if $f(x)=f(y)$, then $x=y$.
|
Solution. If $x$ is even, then $f(x)$ is even, and if $x$ is odd, then $f(x)$ is odd. Moreover, if $x \equiv 1 \bmod 4$, then $f(x) \equiv 3 \bmod 4$, and if $x \equiv 3 \bmod 4$, then $f(x) \equiv 1 \bmod 4$. Clearly $f(0)=0, f(1)=3, f(2)=6$, and $f(3)=5$. So at least $f$ restricted to the set $\{0,1,2,3\}$ ia an inj... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 402 |
98.1. Determine all functions $f$ defined in the set of rational numbers and taking their values in the same set such that the equation $f(x+y)+f(x-y)=2 f(x)+2 f(y)$ holds for all rational numbers $x$ and $y$.
|
Solution. Insert $x=y=0$ in the equation to obtain $2 f(0)=4 f(0)$, which implies $f(0)=0$. Setting $x=0$, one obtains $f(y)+f(-y)=2 f(y)$ of $f(-y)=f(y)$. Then assume $y=n x$, where $n$ is a positive integer. We obtain
$$
f((n+1) x)=2 f(x)+2 f(n x)-f((n-1) x)
$$
In particular, $f(2 x)=2 f(x)+2 f(x)-f(0)=4 f(x)$ and... | f(x)=^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 403 |
98.2. Let $C_{1}$ and $C_{2}$ be two circles intersecting at $A$ and $B$. Let $S$ and $T$ be the centres of $C_{1}$ and $C_{2}$, respectively. Let $P$ be a point on the segment $A B$ such that $|A P| \neq|B P|$ and $P \neq A, P \neq B$. We draw a line perpendicular to $S P$ through $P$ and denote by $C$ and $D$ the po... |
Solution. (See Figure 10.) The power of the point $P$ with respect to the circles $C_{1}$ and $C_{2}$ is $P A \cdot P B=P C \cdot P D=P E \cdot P F$. Since $S P$ is perpendicular to the chord $C D, P$
 For which positive numbers $n$ does there exist a sequence $x_{1}, x_{2}, \ldots, x_{n}$, which contains each of the numbers 1, 2, ..., $n$ exactly once and for which $x_{1}+x_{2}+\cdots+x_{k}$ is divisible by $k$ for each $k=1,2, \ldots, n$ ?
(b) Does there exist an infinite sequence $x_{1}, x_{2}, x_{3}, ... |
Solution. (a) We assume that $x_{1}, \ldots, x_{n}$ is the sequence required in the problem. Then $x_{1}+x_{2}+\cdots+x_{n}=\frac{n(n+1)}{2}$. This sum should be divisible by $n$. If $n$ is odd, this is possible, since $\frac{(n+1)}{2}$ is an integer. If, on the other hand, $n=2 m$, then $\frac{n(n+1)}{2}=m(2 m+1)=$ $... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 405 |
98.4. Let $n$ be a positive integer. Count the number of numbers $k \in\{0,1,2, \ldots, n\}$ such that $\binom{n}{k}$ is odd. Show that this number is a power of two, i.e. of the form $2^{p}$ for some nonnegative integer $p$.
|
Solution. The number of odd binomial coefficients $\binom{n}{k}$ equals the number of ones on the $n$ :th line of the Pascal Triangle $\bmod 2$ :
(We count the lines so that the uppermost l... | 2^{e(n)} | Combinatorics | proof | Yes | Yes | olympiads | false | 406 |
99.1. The function $f$ is defined for non-negative integers and satisfies the condition
$$
f(n)= \begin{cases}f(f(n+11)), & \text { if } n \leq 1999 \\ n-5, & \text { if } n>1999\end{cases}
$$
Find all solutions of the equation $f(n)=1999$.
|
Solution. If $n \geq 2005$, then $f(n)=n-5 \geq 2000$, and the equation $f(n)=1999$ has no solutions. Let $1 \leq k \leq 4$. Then
$$
\begin{gathered}
2000-k=f(2005-k)=f(f(2010-k)) \\
=f(1999-k)=f(f(2004-k))=f(1993-k)
\end{gathered}
$$
Let $k=1$. We obtain three solutions $1999=f(2004)=f(1998)=f(1992)$. Moreover, $19... | 1999=f(6n),ifonlyifn=1,2,\ldots,334 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 407 |
99.2. Consider 7-gons inscribed in a circle such that all sides of the 7-gon are of different length. Determine the maximal number of $120^{\circ}$ angles in this kind of a 7-gon.
|
Solution. It is easy to give examples of heptagons $A B C D E F G$ inscribed in a circle with all sides unequal and two angles equal to $120^{\circ}$. These angles cannot lie on adjacent vertices of the heptagon. In fact, if $\angle A B C=\angle B C D=120^{\circ}$, and arc $B C$ equals $b^{\circ}$, then arcs $A B$ and... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 408 |
99.3. The infinite integer plane $\mathbb{Z} \times \mathbb{Z}=\mathbb{Z}^{2}$ consists of all number pairs $(x, y)$, where $x$ and $y$ are integers. Let $a$ and $b$ be non-negative integers. We call any move from a point $(x, y)$ to any of the points $(x \pm a, y \pm b)$ or $(x \pm b, y \pm a) a(a, b)$-knight move. D... |
Solution. If the greatest common divisor of $a$ and $b$ is $d$, only points whose coordinates are multiples of $d$ can be reached by a sequence of $(a, b)$-knight moves starting from the origin. So $d=1$ is a necessary condition for the possibility of reaching every point in the integer plane. In any $(a, b)$-knight m... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 409 |
99.4. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers and $n \geq 1$. Show that
$$
\begin{aligned}
& n\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right) \\
& \quad \geq\left(\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}}\right)\left(n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)
\end{aligned}
$$
When does e... |
Solution. The inequality of the problem can be written as
$$
\frac{1}{1+a_{1}}+\cdots+\frac{1}{1+a_{n}} \leq \frac{n\left(\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}\right)}{n+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}}
$$
A small manipulation of the right hand side brings the inequality to the equivalent form
$$
\frac{1}{\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 410 |
00.1. In how many ways can the number 2000 be written as a sum of three positive, not necessarily different integers? (Sums like $1+2+3$ and $3+1+2$ etc. are the same.)
|
Solution. Since 3 is not a factor of 2000 , there has to be at least two different numbers among any three summing up to 2000 . Denote by $x$ the number of such sums with three different summands and by $y$ the number of sums with two different summands. Consider 3999 boxes consequtively numbered fron 1 to 3999 such t... | 333333 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 411 |
00.2. The persons $P_{1}, P_{1}, \ldots, P_{n-1}, P_{n}$ sit around a table, in this order, and each one of them has a number of coins. In the start, $P_{1}$ has one coin more than $P_{2}, P_{2}$ has one coin more than $P_{3}$, etc., up to $P_{n-1}$ who has one coin more than $P_{n}$. Now $P_{1}$ gives one coin to $P_... |
Solution. Assume that $P_{n}$ has $m$ coins in the start. Then $P_{n-1}$ has $m+1$ coins, ... and $P_{1}$ has $m+n-1$ coins. In every move a player receives $k$ coins and gives $k+1$ coins away, so her net loss is one coin. After the first round, when $P_{n}$ has given $n$ coins to $P_{1}$, $P_{n}$ has $m-1$ coins, $P... | 6or63 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 412 |
00.3. In the triangle $A B C$, the bisector of angle $B$ meets $A C$ at $D$ and the bisector of angle $C$ meets $A B$ at $E$. The bisectors meet each other at $O$. Furthermore, $O D=O E$. Prove that either $A B C$ is isosceles or $\angle B A C=60^{\circ}$.
|
Solution. (See Figure 11.) Consider the triangles $A O E$ and $A O D$. They have two equal pairs of sides and the angles facing one of these pairs are equal. Then either $A O E$ and $A O D$ are congruent or $\angle A E O=180^{\circ}-\angle A D O$. In the first case, $\angle B E O=\angle C D O$, and
=0, f(1)=1$, and
$$
\frac{1}{2} \leq \frac{f(z)-f(y)}{f(y)-f(x)} \leq 2
$$
for all $0 \leq x<y<z \leq 1$ with $z-y=y-x$. Prove that
$$
\frac{1}{7} \leq f\left(\frac{1}{3}\right) \leq \frac{4}{7}
$$
|
Solution. We set $f\left(\frac{1}{3}\right)=a$ and $f\left(\frac{2}{3}\right)=b$. Applying the inequality of the problem for $x=\frac{1}{3}, y=\frac{2}{3}$ and $z=1$, as well as for $x=0, y=\frac{1}{3}$, and $z=\frac{2}{3}$, we obtain
$$
\frac{1}{2} \leq \frac{1-b}{b-a} \leq 2, \quad \frac{1}{2} \leq \frac{b-a}{a} \l... | \frac{1}{7}\leq\leq\frac{4}{7} | Inequalities | proof | Yes | Yes | olympiads | false | 414 |
01.1. Let $A$ be a finite collection of squares in the coordinate plane such that the vertices of all squares that belong to $A$ are $(m, n),(m+1, n),(m, n+1)$, and $(m+1, n+1)$ for some integers $m$ and $n$. Show that there exists a subcollection $B$ of $A$ such that $B$ contains at least $25 \%$ of the squares in $A... |
Solution. Divide the plane into two sets by painting the strips of squares parallel to the $y$ axis alternately red and green. Denote the sets of red and green squares by $R$ and $G$, respectively. Of the sets $A \cap R$ and $A \cap G$ at least one contains at least one half of the squares in $A$. Denote this set by $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 415 |
01.2. Let $f$ be a bounded real function defined for all real numbers and satisfying for all real numbers $x$ the condition
$$
f\left(x+\frac{1}{3}\right)+f\left(x+\frac{1}{2}\right)=f(x)+f\left(x+\frac{5}{6}\right)
$$
Show that $f$ is periodic. (A function $f$ is bounded, if there exists a number $L$ such that $|f(... |
Solution. Let $g(6 x)=f(x)$. Then $g$ is bounded, and
$$
\begin{gathered}
g(t+2)=f\left(\frac{t}{6}+\frac{1}{3}\right), \quad g(t+3)=f\left(\frac{t}{6}+\frac{1}{2}\right) \\
g(t+5)=f\left(\frac{t}{6}+\frac{5}{6}\right), \quad g(t+2)+g(t+3)=g(t)+g(t+5) \\
g(t+5)-g(t+3)=g(t+2)-g(t)
\end{gathered}
$$
for all real numbe... | proof | Algebra | proof | Yes | Yes | olympiads | false | 416 |
01.3. Determine the number of real roots of the equation
$$
x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2}=0
$$
|
Solution. Write
$$
\begin{gathered}
x^{8}-x^{7}+2 x^{6}-2 x^{5}+3 x^{4}-3 x^{3}+4 x^{2}-4 x+\frac{5}{2} \\
=x(x-1)\left(x^{6}+2 x^{4}+3 x^{2}+4\right)+\frac{5}{2}
\end{gathered}
$$
If $x(x-1) \geq 0$, i.e. $x \leq 0$ or $x \geq 1$, the equation has no roots. If $0x(x-1)=\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4} \ge... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 417 |
01.4. Let $A B C D E F$ be a convex hexagon, in which each of the diagonals $A D, B E$, and $C F$ divides the hexagon in two quadrilaterals of equal area. Show that $A D, B E$, and $C F$ are concurrent.
 Denote the area of a figure by $|\cdot|$. Let $A D$ and $B E$ intersect at $P, A D$ and $C F$ at $Q$, and $B E$ and $C F$ at $R$. Assume that $P, Q$, and $R$ are different. We may assume that $P$ lies between $B$ and $R$, and $Q$ lies between $C$ and $R$. Both $|A B P|$ and $|D E P|$ differ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 418 |
02.1. The trapezium $A B C D$, where $A B$ and $C D$ are parallel and $A D<C D$, is inscribed in the circle c. Let DP be a chord of the circle, parallel to AC. Assume that the tangent to $c$ at $D$ meets the line $A B$ at $E$ and that $P B$ and $D C$ meet at $Q$. Show that $E Q=A C$.
|
Solution. (See Figure 13.) since $A D<C D, \angle P D C=\angle D C A<\angle D A C$. This implies that arc $C P$ is smaller than arc $C D$, and $P$ lies on that arc $C D$ which does not include $A$ and $B$. We show that the triangles $A D E$ and $C B Q$ are congruent. As a trapezium inscribed in a circle, $A B C D$ is ... | AC=EQ | Geometry | proof | Yes | Yes | olympiads | false | 419 |
02.2. In two bowls there are in total $N$ balls, numbered from 1 to $N$. One ball is moved from one of the bowls to the other. The average of the numbers in the bowls is increased in both of the bowls by the same amount, $x$. Determine the largest possible value of $x$.
|
Solution. Consider the situation before the ball is moved from urn one to urn two. Let the number of balls in urn one be $n$, and let the sum of numbers in the balls in that urn be $a$. The number of balls in urn two is $m$ and the sum of numbers $b$. If $q$ is the number written in the ball which was moved, the condi... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 420 |
02.3. Let $a_{1}, a_{2}, \ldots, a_{n}$, and $b_{1}, b_{2}, \ldots, b_{n}$ be real numbers, and let $a_{1}, a_{2}, \ldots, a_{n}$ be all different.. Show that if all the products
$$
\left(a_{i}+b_{1}\right)\left(a_{i}+b_{2}\right) \cdots\left(a_{i}+b_{n}\right)
$$
$i=1,2, \ldots, n$, are equal, then the products
$$... |
Solution. Let $P(x)=\left(x+b_{1}\right)\left(x+b_{2}\right) \cdots\left(x+b_{n}\right)$. Let $P\left(a_{1}\right)=P\left(a_{2}\right)=\ldots=P\left(a_{n}\right)=d$. Thus $a_{1}, a_{2}, \ldots, a_{n}$ are the roots of the $n$ :th degree polynomial equation $P(x)-d=0$. Then $P(x)-d=c\left(x-a_{1}\right)\left(x-a_{2}\ri... | proof | Algebra | proof | Yes | Yes | olympiads | false | 421 |
02.4. Eva, Per and Anna play with their pocket calculators. They choose different integers and check, whether or not they are divisible by 11. They only look at nine-digit numbers consisting of all the digits 1, 2, .., 9. Anna claims that the probability of such a number to be a multiple of 11 is exactly 1/11. Eva has... |
Solution. We write the numbers in consideration, $n=a_{0}+10 a_{1}+10^{2} a_{2}+\cdots+10^{8} a_{8}$, in the form
$$
\begin{gathered}
a_{0}+(11-1) a_{1}+(99+1) a_{2}+(1001-1) a_{3} \\
+(9999+1) a_{4}+(100001-1) a_{5}+(999999+1) a_{6} \\
\quad+(10000001-1) a_{7}+(99999999+1) a_{8} \\
=\left(a_{0}-a_{1}+a_{2}-a_{3}+a_{... | \frac{11}{126}<\frac{1}{11} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 422 |
03.1. Stones are placed on the squares of a chessboard having 10 rows and 14 columns. There is an odd number of stones on each row and each column. The squares are coloured black and white in the usual fashion. Show that the number of stones on black squares is even. Note that there can be more than one stone on a squ... |
Solution. Changing the order of rows or columns does not influence the number of stones on a row, on a column or on black squares. Thus we can order the rows and columns in such a way that the $5 \times 7$ rectangles in the upper left and lower right corner are black and the other two $5 \times 7$ rectangles are white... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 423 |
03.2. Find all triples of integers $(x, y, z)$ satisfying
$$
x^{3}+y^{3}+z^{3}-3 x y z=2003
$$
|
Solution. It is a well-known fact (which can be rediscovered e.g. by noticing that the left hand side is a polynomial in $x$ having $-(y+z)$ as a zero) that
$$
\begin{aligned}
& x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \\
&=(x+y+z) \frac{(x-y)^{2}+(y-z)^{2}+(z-x)^{2}}{2}
\end{aligne... | (668,668,667) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 424 |
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