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 |
|---|---|---|---|---|---|---|---|---|---|
G4 Let $S$ be a point inside $\varangle p O q$, and let $k$ be a circle which contains $S$ and touches the legs $O p$ and $O q$ in points $P$ and $Q$ respectively. Straight line $s$ parallel to $O p$ from $S$ intersects $O q$ in a point $R$. Let $T$ be the point of intersection of the ray $(P S$ and circumscribed circ... | ## Solution
Let $\varangle O P S=\varphi_{1}$ and $\varangle O Q S=\varphi_{2}$. We have that $\varangle O P S=\varangle P Q S=\varphi_{1}$ and $\varangle O Q S=$ $\varangle Q P S=\varphi_{2}$ (tangents to circle $k$ ).
Because $R S \| O P$ we have $\varangle O P S=\varangle R S T=\varphi_{1}$ and $\varangle R Q T=\v... | proof | Geometry | proof | Yes | Yes | olympiads | false | 107 |
NT1 Find all the pairs positive integers $(x, y)$ such that
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{[x, y]}+\frac{1}{(x, y)}=\frac{1}{2}
$$
where $(x, y)$ is the greatest common divisor of $x, y$ and $[x, y]$ is the least common multiple of $x, y$.
| ## Solution
We put $x=d u$ and $y=d v$ where $d=(x, y)$. So we have $(u, v)=1$. From the conclusion we obtain $2(u+1)(v+1)=d u v$. Because $(v, v+1)=1, v$ divides $2(u+1)$.
Case 1. $u=v$. Hence $x=y=[x, y]=(x, y)$, which leads to the solution $x=8$ and $y=8$.
Case 2. $uv$. Because of the symmetry of $u, v$ and $x, y... | (8,8),(9,24),(24,9),(5,20),(20,5),(12,15),(15,12),(8,12),(12,8),(6,12),(12,6) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 108 |
NT2 Prove that the equation $x^{2006}-4 y^{2006}-2006=4 y^{2007}+2007 y$ has no solution in the set of the positive integers.
| ## Solution
We assume the contrary is true. So there are $x$ and $y$ that satisfy the equation. Hence we have
$$
\begin{gathered}
x^{2006}=4 y^{2007}+4 y^{2006}+2007 y+2006 \\
x^{2006}+1=4 y^{2006}(y+1)+2007(y+1) \\
x^{2006}+1=\left(4 y^{2006}+2007\right)(y+1)
\end{gathered}
$$
But $4 y^{2006}+2007 \equiv 3(\bmod 4)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 109 |
NT3 Let $n>1$ be a positive integer and $p$ a prime number such that $n \mid(p-1)$ and $p \mid\left(n^{6}-1\right)$. Prove that at least one of the numbers $p-n$ and $p+n$ is a perfect square.
| ## Solution
Since $n \mid p-1$, then $p=1+n a$, where $a \geq 1$ is an integer. From the condition $p \mid n^{6}-1$, it follows that $p|n-1, p| n+1, p \mid n^{2}+n+1$ or $p \mid n^{2}-n+1$.
- Let $p \mid n-1$. Then $n \geq p+1>n$ which is impossible.
- Let $p \mid n+1$. Then $n+1 \geq p=1+n a$ which is possible only ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 110 |
NT4 Let $a, b$ be two co-prime positive integers. A number is called good if it can be written in the form $a x+b y$ for non-negative integers $x, y$. Define the function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ as $f(n)=n-n_{a}-n_{b}$, where $s_{t}$ represents the remainder of $s$ upon division by $t$. Show that an int... | ## Solution
If $n$ is good then $n=a x+b y$ also $n_{a}=(b y)_{a}$ and $n_{b}=(a x)_{b}$ so
$$
f(n)=a x-(a x)_{b}+b y-(b y)_{a}=b y^{\prime}+a x^{\prime}
$$
is also good, thus the sequence contains only good numbers which are non-negative.
Now we have to prove that if the sequence contains only non-negative integer... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 111 |
NT5 Let $p$ be a prime number. Show that $7 p+3^{p}-4$ is not a perfect square.
| ## Solution
Assume that for a prime number $p$ greater than $3, m=7 p+3^{p}-4$ is a perfect square. Let $m=n^{2}$ for some $n \in \mathbb{Z}$. By Fermat's Little Theorem,
$$
m=7 p+3^{p}-4 \equiv 3-4 \equiv-1 \quad(\bmod p)
$$
If $p=4 k+3, k \in \mathbb{Z}$, then again by Fermat's Little Theorem
$$
-1 \equiv m^{2 k+... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 112 |
## A1
For any real number a, let $\lfloor a\rfloor$ denote the greatest integer not exceeding a. In positive real numbers solve the following equation
$$
n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor=2014
$$
|
Solution1. Obviously $n$ must be positive integer. Now note that $44^{2}=19362000$ than $2014=n+\lfloor\sqrt{n}\rfloor+\lfloor\sqrt[3]{n}\rfloor>2000+44+12=2056$, a contradiction!
So $1950 \leq n \leq 2000$, therefore $\lfloor\sqrt{n}\rfloor=44$ and $\lfloor\sqrt[3]{n}\rfloor=12$. Plugging that into the original equa... | 1956 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 113 |
## A2
Let $a, b$ and $c$ be positive real numbers such that abc $=\frac{1}{8}$. Prove the inequality
$$
a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geq \frac{15}{16}
$$
When does equality hold?
|
Solution1. By using The Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that
$$
\begin{aligned}
& a^{2}+b^{2}+c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}= \\
& \quad=\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{a^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\frac{b^{2}}{4}+\fra... | \frac{15}{16} | Inequalities | proof | Yes | Yes | olympiads | false | 114 |
## A3
Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that:
$$
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} \geq 3(a+b+c+1)
$$
When does equality hold?
|
Solution1. By using AM-GM $\left(x^{2}+y^{2}+z^{2} \geq x y+y z+z x\right)$ we have
$$
\begin{aligned}
\left(a+\frac{1}{b}\right)^{2}+\left(b+\frac{1}{c}\right)^{2}+\left(c+\frac{1}{a}\right)^{2} & \geq\left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\f... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 115 |
## A4
Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Prove that
$$
\frac{7+2 b}{1+a}+\frac{7+2 c}{1+b}+\frac{7+2 a}{1+c} \geq \frac{69}{4}
$$
When does equality hold?
|
Solution1. The inequality can be written as: $\frac{5+2(1+b)}{1+a}+\frac{5+2(1+c)}{1+b}+\frac{5+2(1+a)}{1+c} \geq \frac{69}{4}$.
We substitute $1+a=x, 1+b=y, 1+c=z$.
So, we have to prove the inequality
$$
\frac{5+2 y}{x}+\frac{5+2 z}{y}+\frac{5+2 x}{z} \geq \frac{69}{4} \Leftrightarrow 5\left(\frac{1}{x}+\frac{1}{y... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 116 |
## A6
Let $a, b, c$ be positive real numbers. Prove that
$$
\left(\left(3 a^{2}+1\right)^{2}+2\left(1+\frac{3}{b}\right)^{2}\right)\left(\left(3 b^{2}+1\right)^{2}+2\left(1+\frac{3}{c}\right)^{2}\right)\left(\left(3 c^{2}+1\right)^{2}+2\left(1+\frac{3}{a}\right)^{2}\right) \geq 48^{3}
$$
When does equality hold?
|
Solution. Let $x$ be a positive real number. By AM-GM we have $\frac{1+x+x+x}{4} \geq x^{\frac{3}{4}}$, or equivalently $1+3 x \geq 4 x^{\frac{3}{4}}$. Using this inequality we obtain:
$$
\left(3 a^{2}+1\right)^{2} \geq 16 a^{3} \text { and } 2\left(1+\frac{3}{b}\right)^{2} \geq 32 b^{-\frac{3}{2}}
$$
Moreover, by i... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 117 |
## A8
Let $x, y$ and $z$ be positive real numbers such that $x y z=1$. Prove the inequality
$$
\frac{1}{x(a y+b)}+\frac{1}{y(a z+b)}+\frac{1}{z(a x+b)} \geq 3 \text {, if: }
$$
a) $a=0$ and $b=1$;
b) $a=1$ and $b=0$;
c) $a+b=1$ for $a, b>0$
When does the equality hold true?
|
Solution. a) The inequality reduces to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq 3$, which follows directly from the AM-GM inequality.
Equality holds only when $x=y=z=1$.
b) Here the inequality reduces to $\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \geq 3$, i.e. $x+y+z \geq 3$, which also follows from the AM-GM inequ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 118 |
## A9
Let $n$ be a positive integer, and let $x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{n}$ be positive real numbers such that $x_{1}+\ldots+x_{n}=y_{1}+\ldots+y_{n}=1$. Show that
$$
\left|x_{1}-y_{1}\right|+\ldots\left|x_{n}-y_{n}\right| \leq 2-\min _{1 \leq i \leq n} \frac{x_{i}}{y_{i}}-\min _{1 \leq i \leq n} \frac{... |
Solution. Up to reordering the real numbers $x_{i}$ and $y_{i}$, we may assume that $\frac{x_{1}}{y_{1}} \leq \ldots \leq \frac{x_{n}}{y_{n}}$. Let $A=\frac{x_{1}}{y_{1}}$ and $B=\frac{x_{n}}{y_{n}}$, and $\mathrm{S}=\left|x_{1}-y_{1}\right|+\ldots\left|x_{n}-y_{n}\right|$. Our aim is to prove that $S \leq 2-A-\frac{1... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 119 |
## C1
Several (at least two) segments are drawn on a board. Select two of them, and let $a$ and $b$ be their lengths. Delete the selected segments and draw a segment of length $\frac{a b}{a+b}$. Continue this procedure until only one segment remains on the board. Prove:
a) the length of the last remaining segment doe... |
Solution. a) Observe that $\frac{1}{\frac{a b}{a+b}}=\frac{1}{a}+\frac{1}{b}$. Thus, if the lengths of the initial segments on the board were $a_{1}, a_{2}, \ldots, a_{n}$, and $c$ is the length of the last remaining segment, then $\frac{1}{c}=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}$ , proving a).
b) F... | proof | Algebra | proof | Yes | Yes | olympiads | false | 120 |
## C2
In a country with $n$ cities, all direct airlines are two-way. There are $r>2014$ routes between pairs of different cities that include no more than one intermediate stop (the direction of each route matters). Find the least possible $n$ and the least possible $r$ for that value of $n$.
|
Solution. Denote by $X_{1}, X_{2}, \ldots X_{n}$ the cities in the country and let $X_{i}$ be connected to exactly $m_{i}$ other cities by direct two-way airline. Then $X_{i}$ is a final destination of $m_{i}$ direct routes and an intermediate stop of $m_{i}\left(m_{i}-1\right)$ non-direct routes. Thus $r=m_{1}^{2}+\l... | 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 121 |
## C3
For a given positive integer n, two players $A$ and B play the following game: Given is pile of $\boldsymbol{\Omega}$ stones. The players take turn alternatively with A going first. On each turn the player is allowed to take one stone, a prime number of stones, or a multiple of $n$ stones. The winner is the one ... |
Solution. Denote by $k$ the sought number and let $\left\{a_{1}, a_{2}, \ldots, a_{k}\right\}$ be the corresponding values for $a$. We will call each $a_{i}$ a losing number and every other positive integer a winning numbers. Clearly every multiple of $n$ is a winning number.
Suppose there are two different losing nu... | n-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 122 |
## C4
Let $A=1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2014$ be the product of the numbers less or equal to 2014 that give remainder 1 when divided by 3 . Find the last non-zero digit of $A$.
|
Solution. Grouping the elements of the product by ten we get:
$$
\begin{aligned}
& (30 k+1)(30 k+4)(30 k+7)(30 k+10)(30 k+13)(30 k+16) \\
& (30 k+19)(30 k+22)(30 k+25)(30 k+28)= \\
& =(30 k+1)(15 k+2)(30 k+7)(120 k+40)(30 k+13)(15 k+8) \\
& (30 k+19)(15 k+11)(120 k+100)(15 k+14)
\end{aligned}
$$
(We divide all even ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 123 |
## G2
Let $A B C$ be an acute triangle with $\overline{A B}<\overline{A C}<\overline{B C}$ and $c(O, R)$ be its circumcircle. Denote with $D$ and $E$ be the points diametrically opposite to the points $B$ and $C$, respectively. The circle $c_{1}(A, \overline{A E})$ intersects $\overline{A C}$ at point $K$, the circle ... |
Solution. Let $\mathrm{M}$ be the point of intersection of the line $D L$ with the circle $c(O, R)$ (we choose $M \equiv D$ if $L D$ is tangent to $c$ and $M$ to be the second intersecting point otherwise). It is
, D M \perp A C(M \in A C)$ and $D N \perp B C(N \in B C)$ for an acute triangle ABC with area $S$. If $H_{1}$ and $H_{2}$ are the orthocentres of the triangles $M N C$ and MND respectively. Evaluate the area of the quadrilateral $\mathrm{AH}_{1} \mathrm{BH}_{2}$.
|
Solution1. Let $O, P, K, R$ and $T$ be the midpoints of the segments $C D, M N, C N, C H_{1}$ and $M H_{1}$, respectively. From $\triangle M N C$ we have that $\overline{P K}=\frac{1}{2} \overline{M C}$ and $P K \| M C$. Analogously, from $\Delta M H_{1} C$ we have that $\overline{T R}=\frac{1}{2} \overline{M C}$ and ... | S | Geometry | math-word-problem | Yes | Yes | olympiads | false | 125 |
## G4
Let $A B C$ be a triangle such that $\overline{A B} \neq \overline{A C}$. Let $M$ be a midpoint of $\overline{B C}, H$ the orthocenter of $A B C, O_{1}$ the midpoint of $\overline{A H}$ and $O_{2}$ the circumcenter of $B C H$. Prove that $O_{1} A M O_{2}$ is a parallelogram.
|
Solution1. Let $O_{2}^{\prime}$ be the point such that $O_{1} A M O_{2}^{\prime}$ is a parallelogram. Note that $\overrightarrow{M O_{2}}=\overrightarrow{A O_{1}}=\overrightarrow{O_{1} H}$. Therefore, $O_{1} H O_{2}^{\prime} M$ is a parallelogram and $\overrightarrow{M O_{1}}=\overrightarrow{O_{2} H}$.
Since $M$ is t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 126 |
## G5
Let $A B C$ be a triangle with $\overline{A B} \neq \overline{B C}$, and let $B D$ be the internal bisector of $\measuredangle A B C(D \in A C)$. Denote the midpoint of the arc $A C$ which contains point BbyM. The circumcircle of the triangle $B D M$ intersects the segment $A B$ at point $K \neq B$, and let $J$ ... | ## Solution1.
Let the circumcircle of the triangle $B D M$ intersect the line segment $B C$ at point $L \neq B$. From $\measuredangle C B D=\measuredangle D B A$ we have $\overline{D L}=\... | proof | Geometry | proof | Yes | Yes | olympiads | false | 127 |
## G6
Let $A B C D$ be a quadrilateral whose sides $A B$ and $C D$ are not parallel, and let $O$ be the intersection of its diagonals. Denote with $H_{1}$ and $H_{2}$ the orthocenters of the triangles $O A B$ and OCD, respectively. If $M$ and $N$ are the midpoints of the segments $\overline{A B}$ and $\overline{C D}$,... | ## Solution.
Let $A^{\prime}$ and $B^{\prime}$ be the feet of the altitudes drawn from $A$ and $B$ respectively in the triangle $A O B$, and $C^{\prime}$ and $D^{\prime}$ are the feet of ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 128 |
## N1
Each letter of the word OHRID corresponds to a different digit belonging to the set $(1,2,3,4,5)$. Decipher the equality $(O+H+R+I+D)^{2}:(O-H-R+I+D)=O^{H^{R^{I_{D}^{D}}}}$.
|
Solution. Since $O, H, R, I$ and $D$ are distinct numbers from $\{1,2,3,4,5\}$, we have $O+H+R+I+D=15$ and $O-H-R+I+D=O+H+R+I+D-2(H+R)15$ and divides 225 , which is only possible for $O^{H^{R^{I^{D}}}}=25$ (must be a power of three or five). This implies that $O=5, H=2$ and $R=1$. It's easy to check that both $I=3, D=... | O=5,H=2,R=1,I=3,D=4orO=5,H=2,R=1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 129 |
N3
Find the integer solutions of the equation
$$
x^{2}=y^{2}\left(x+y^{4}+2 y^{2}\right)
$$
|
Solution. If $x=0$, then $y=0$ and conversely, if $y=0$, then $x=0$. It follows that $(x, y)=(0,0)$ is a solution of the problem. Assume $x \neq 0$ and $y \neq 0$ satisfy the equation. The equation can be transformed in the form $x^{2}-x y^{2}=y^{6}+2 y^{4}$. Then $4 x^{2}-4 x y^{2}+y^{4}=4 y^{6}+9 y^{4}$ and conseque... | (x,y)=(0,0),(12,-2),(12,2),(-8,-2),(-8,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 130 |
N4
Prove there are no integers $a$ and $b$ satisfying the following conditions:
i) $16 a-9 b$ is a prime number
ii) $\quad a b$ is a perfect square
iii) $a+b$ is a perfect square
|
Solution. Suppose $a$ and $b$ be integers satisfying the given conditions. Let $p$ be a prime number, $n$ and $m$ be integers. Then we can write the conditions as follows:
$$
\begin{aligned}
& 16 a-9 b=p \\
& a b=n^{2} \\
& a+b=m^{2}
\end{aligned}
$$
Moreover, let $d=g d c(a, b)$ and $a=d x, b=d y$ for some relative... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 131 |
## N6
Vukasin, Dimitrije, Dusan, Stefan and Filip asked their professor to guess a three consecutive positive integer numbers after they had told him these (true) sentences:
Vukasin: "Sum of the digits of one of them is a prime number. Sum of the digits of some of the other two is an even perfect number ( $n$ is perf... |
Solution. Let the middle number be $n$, so the numbers are $n-1, n$ and $n+1$. Since 4 does not divide any of them, $n \equiv 2(\bmod 4)$. Furthermore, neither 3,5 nor 7 divides $n$. Also $n+1+11 \equiv 2(\bmod 4)$ cannot be a square. Then 3 must divide $n-1$ or $n+1$. If $n-1+11$ is a square, then $3 \mid n+1$ which ... | 2013,2014,2015 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 132 |
A2. Let $a, b, c$ be positive real numbers such that $a b c=\frac{2}{3}$. Prove that
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a} \geqslant \frac{a+b+c}{a^{3}+b^{3}+c^{3}}
$$
|
Solution. The given inequality is equivalent to
$$
\left(a^{3}+b^{3}+c^{3}\right)\left(\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}\right) \geqslant a+b+c
$$
By the AM-GM Inequality it follows that
$$
a^{3}+b^{3}=\frac{a^{3}+a^{3}+b^{3}}{3}+\frac{b^{3}+b^{3}+a^{3}}{3} \geqslant a^{2} b+b^{2} a=a b(a+b)
$$
Simil... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 135 |
A3. Let $A$ and $B$ be two non-empty subsets of $X=\{1,2, \ldots, 11\}$ with $A \cup B=X$. Let $P_{A}$ be the product of all elements of $A$ and let $P_{B}$ be the product of all elements of $B$. Find the minimum and maximum possible value of $P_{A}+P_{B}$ and find all possible equality cases.
|
Solution. For the maximum, we use the fact that $\left(P_{A}-1\right)\left(P_{B}-1\right) \geqslant 0$, to get that $P_{A}+P_{B} \leqslant P_{A} P_{B}+1=11!+1$. Equality holds if and only if $A=\{1\}$ or $B=\{1\}$.
For the minimum observe, first that $P_{A} \cdot P_{B}=11!=c$. Without loss of generality let $P_{A} \l... | 12636 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 136 |
A4. Let $a, b$ be two distinct real numbers and let $c$ be a positive real number such that
$$
a^{4}-2019 a=b^{4}-2019 b=c
$$
Prove that $-\sqrt{c}<a b<0$.
|
Solution. Firstly, we see that
$$
2019(a-b)=a^{4}-b^{4}=(a-b)(a+b)\left(a^{2}+b^{2}\right)
$$
Since $a \neq b$, we get $(a+b)\left(a^{2}+b^{2}\right)=2019$, so $a+b \neq 0$. Thus
$$
\begin{aligned}
2 c & =a^{4}-2019 a+b^{4}-2019 b \\
& =a^{4}+b^{4}-2019(a+b) \\
& =a^{4}+b^{4}-(a+b)^{2}\left(a^{2}+b^{2}\right) \\
& ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 137 |
A5. Let $a, b, c, d$ be positive real numbers such that $a b c d=1$. Prove the inequality
$$
\frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d}+\frac{1}{a+b+c+d^{3}} \leqslant \frac{a+b+c+d}{4}
$$
|
Solution. From the Cauchy-Schwarz Inequality, we obtain
$$
(a+b+c+d)^{2} \leqslant\left(a^{3}+b+c+d\right)\left(\frac{1}{a}+b+c+d\right)
$$
Using this, together with the other three analogous inequalities, we get
$$
\begin{aligned}
\frac{1}{a^{3}+b+c+d}+\frac{1}{a+b^{3}+c+d}+\frac{1}{a+b+c^{3}+d} & +\frac{1}{a+b+c+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 138 |
A6. Let $a, b, c$ be positive real numbers. Prove the inequality
$$
\left(a^{2}+a c+c^{2}\right)\left(\frac{1}{a+b+c}+\frac{1}{a+c}\right)+b^{2}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)>a+b+c
$$
|
Solution. By the Cauchy-Schwarz Inequality, we have
$$
\frac{1}{a+b+c}+\frac{1}{a+c} \geqslant \frac{4}{2 a+b+2 c}
$$
and
$$
\frac{1}{b+c}+\frac{1}{a+b} \geqslant \frac{4}{a+2 b+c}
$$
Since
$$
a^{2}+a c+c^{2}=\frac{3}{4}(a+c)^{2}+\frac{1}{4}(a-c)^{2} \geqslant \frac{3}{4}(a+c)^{2}
$$
then, writing $L$ for the Le... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 139 |
A7. Show that for any positive real numbers $a, b, c$ such that $a+b+c=a b+b c+c a$, the following inequality holds
$$
3+\sqrt[3]{\frac{a^{3}+1}{2}}+\sqrt[3]{\frac{b^{3}+1}{2}}+\sqrt[3]{\frac{c^{3}+1}{2}} \leqslant 2(a+b+c)
$$
|
Solution. Using the condition we have
$$
a^{2}-a+1=a^{2}-a+1+a b+b c+c a-a-b-c=(c+a-1)(a+b-1)
$$
Hence we have
$$
\sqrt[3]{\frac{a^{3}+1}{2}}=\sqrt[3]{\frac{(a+1)\left(a^{2}-a+1\right)}{2}}=\sqrt[3]{\left(\frac{a+1}{2}\right)(c+a-1)(a+b-1)}
$$
Using the last equality together with the AM-GM Inequality, we have
$$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 140 |
C1. Let $S$ be a set of 100 positive integers having the following property:
"Among every four numbers of $S$, there is a number which divides each of the other three or there is a number which is equal to the sum of the other three."
Prove that the set $S$ contains a number which divides each of the other 99 number... |
Solution. Let $a<b$ be the two smallest numbers of $S$ and let $d$ be the largest number of $S$. Consider any two other numbers $x<y$ of $S$. For the quadruples $(a, b, x, d)$ and $(a, b, y, d)$ we cannot get both of $d=a+b+x$ and $d=a+b+y$, since $a+b+x<a+b+y$. From here, we get $a \mid b$ and $a \mid d$.
Consider a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 141 |
C2. In a certain city there are $n$ straight streets, such that every two streets intersect, and no three streets pass through the same intersection. The City Council wants to organize the city by designating the main and the side street on every intersection. Prove that this can be done in such way that if one goes a... |
Solution. Pick any street $s$ and organize the intersections along $s$ such that the intersections of the two types alternate, as in the statement of the problem.
On every other street $s_{1}$, exactly one intersection has been organized, namely the one where $s_{1}$ intersects $s$. Call this intersection $I_{1}$. We... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 142 |
C3. In a $5 \times 100$ table we have coloured black $n$ of its cells. Each of the 500 cells has at most two adjacent (by side) cells coloured black. Find the largest possible value of $n$.
|
Solution. If we colour all the cells along all edges of the board together with the entire middle row except the second and the last-but-one cell, the condition is satisfied and there are 302 black cells. The figure below exhibits this colouring for the $5 \times 8$ case.
}{2}=7 k$. This is possible only if $n \equiv 0,1 \bmod 7$.
- In order to obtain $n=7 m+1$, arrange the kids in a circ... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 144 |
G1. Let $A B C$ be a right-angled triangle with $\hat{A}=90^{\circ}$ and $\hat{B}=30^{\circ}$. The perpendicular at the midpoint $M$ of $B C$ meets the bisector $B K$ of the angle $\hat{B}$ at the point $E$. The perpendicular bisector of $E K$ meets $A B$ at $D$. Prove that $K D$ is perpendicular to $D E$.
|
Solution. Let $I$ be the incenter of $A B C$ and let $Z$ be the foot of the perpendicular from $K$ on $E C$. Since $K B$ is the bisector of $\hat{B}$, then $\angle E B C=15^{\circ}$ and since $E M$ is the perpendicular bisector of $B C$, then $\angle E C B=\angle E B C=15^{\circ}$. Therefore $\angle K E C=30^{\circ}$.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 146 |
G2. Let $A B C$ be a triangle and let $\omega$ be its circumcircle. Let $\ell_{B}$ and $\ell_{C}$ be two parallel lines passing through $B$ and $C$ respectively. The lines $\ell_{B}$ and $\ell_{C}$ intersect with $\omega$ for the second time at the points $D$ and $E$ respectively, with $D$ belonging on the arc $A B$, ... |
Solution. We write $\omega_{1}, \omega_{2}$ and $\omega^{\prime}$ for the circumcircles of $A G D, A E F$ and $O O_{1} O_{2}$ respectively. Since $O_{1}$ and $O_{2}$ are the centers of $\omega_{1}$ and $\omega_{2}$, and because $D G$ and $E F$ are parallel, we get that
$$
\angle G A O_{1}=90^{\circ}-\frac{\angle G O_... | proof | Geometry | proof | Yes | Yes | olympiads | false | 147 |
G3. Let $A B C$ be a triangle with incenter $I$. The points $D$ and $E$ lie on the segments $C A$ and $B C$ respectively, such that $C D=C E$. Let $F$ be a point on the segment $C D$. Prove that the quadrilateral $A B E F$ is circumscribable if and only if the quadrilateral $D I E F$ is cyclic.
|
Solution. Since $C D=C E$ it means that $E$ is the reflection of $D$ on the bisector of $\angle A C B$, i.e. the line $C I$. Let $G$ be the reflection of $F$ on $C I$. Then $G$ lies on the segment $C E$, the segment $E G$ is the reflection of the segment $D F$ on the line $C I$. Also, the quadraliteral $D E G F$ is cy... | proof | Geometry | proof | Yes | Yes | olympiads | false | 148 |
G4. Let $A B C$ be a triangle such that $A B \neq A C$, and let the perpendicular bisector of the side $B C$ intersect lines $A B$ and $A C$ at points $P$ and $Q$, respectively. If $H$ is the orthocenter of the triangle $A B C$, and $M$ and $N$ are the midpoints of the segments $B C$ and $P Q$ respectively, prove that... |
Solution. We have
$$
\angle A P Q=\angle B P M=90^{\circ}-\angle M B P=90^{\circ}-\angle C B A=\angle H C B
$$
and
$$
\angle A Q P=\angle M Q C=90^{\circ}-\angle Q C M=90^{\circ}-\angle A C B=\angle C B H
$$
From these two equalities, we see that the triangles $A P Q$ and $H C B$ are similar. Moreover, since $M$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 149 |
G5. Let $P$ be a point in the interior of a triangle $A B C$. The lines $A P, B P$ and $C P$ intersect again the circumcircles of the triangles $P B C, P C A$, and $P A B$ at $D, E$ and $F$ respectively. Prove that $P$ is the orthocenter of the triangle $D E F$ if and only if $P$ is the incenter of the triangle $A B C... |
Solution. If $P$ is the incenter of $A B C$, then $\angle B P D=\angle A B P+\angle B A P=\frac{\hat{A}+\hat{B}}{2}$, and $\angle B D P=\angle B C P=\frac{\hat{C}}{2}$. From triangle $B D P$, it follows that $\angle P B D=90^{\circ}$, i.e. that $E B$ is one of the altitudes of the triangle $D E F$. Similarly, $A D$ an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 150 |
G6. Let $A B C$ be a non-isosceles triangle with incenter $I$. Let $D$ be a point on the segment $B C$ such that the circumcircle of $B I D$ intersects the segment $A B$ at $E \neq B$, and the circumcircle of $C I D$ intersects the segment $A C$ at $F \neq C$. The circumcircle of $D E F$ intersects $A B$ and $A C$ at ... |
Solution. Since $B D I E$ is cyclic, and $B I$ is the bisector of $\angle D B E$, then $I D=I E$. Similarly, $I D=I F$, so $I$ is the circumcenter of the triangle $D E F$. We also have
$$
\angle I E A=\angle I D B=\angle I F C
$$
which implies that $A E I F$ is cyclic. We can assume that $A, E, M$ and $A, N, F$ are ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 151 |
G7. Let $A B C$ be a right-angled triangle with $\hat{A}=90^{\circ}$. Let $K$ be the midpoint of $B C$, and let $A K L M$ be a parallelogram with centre $C$. Let $T$ be the intersection of the line $A C$ and the perpendicular bisector of $B M$. Let $\omega_{1}$ be the circle with centre $C$ and radius $C A$ and let $\... |
Solution. Let $M^{\prime}$ be the symmetric point of $M$ with respect to $T$. Observe that $T$ is equidistant from $B$ and $M$, therefore $M$ belongs on $\omega_{2}$ and $M^{\prime} M$ is a diameter of $\omega_{2}$. It suffices to prove that $M^{\prime} A$ is perpendicular to $L M$, or equivalently, to $A K$. To see t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 152 |
N1. Find all prime numbers $p$ for which there are non-negative integers $x, y$ and $z$ such that the number
$$
A=x^{p}+y^{p}+z^{p}-x-y-z
$$
is a product of exactly three distinct prime numbers.
|
Solution. For $p=2$, we take $x=y=4$ and $z=3$. Then $A=30=2 \cdot 3 \cdot 5$. For $p=3$ we can take $x=3$ and $y=2$ and $z=1$. Then again $A=30=2 \cdot 3 \cdot 5$. For $p=5$ we can take $x=2$ and $y=1$ and $z=1$. Again $A=30=2 \cdot 3 \cdot 5$.
Assume now that $p \geqslant 7$. Working modulo 2 and modulo 3 we see th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 153 |
N2. Find all triples $(p, q, r)$ of prime numbers such that all of the following numbers are integers
$$
\frac{p^{2}+2 q}{q+r}, \quad \frac{q^{2}+9 r}{r+p}, \quad \frac{r^{2}+3 p}{p+q}
$$
|
Solution. We consider the following cases:
1st Case: If $r=2$, then $\frac{r^{2}+3 p}{p+q}=\frac{4+3 p}{p+q}$. If $p$ is odd, then $4+3 p$ is odd and therefore $p+q$ must be odd. From here, $q=2$ and $\frac{r^{2}+3 p}{p+q}=\frac{4+3 p}{p+2}=3-\frac{2}{p+2}$ which is not an integer. Thus $p=2$ and $\frac{r^{2}+3 p}{p+... | (2,3,7) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 154 |
N3. Find all prime numbers $p$ and nonnegative integers $x \neq y$ such that $x^{4}-y^{4}=$ $p\left(x^{3}-y^{3}\right)$.
|
Solution. If $x=0$ then $y=p$ and if $y=0$ then $x=p$. We will show that there are no other solutions.
Suppose $x, y>0$. Since $x \neq y$, we have
$$
p\left(x^{2}+x y+y^{2}\right)=(x+y)\left(x^{2}+y^{2}\right)
$$
If $p$ divides $x+y$, then $x^{2}+y^{2}$ must divide $x^{2}+x y+y^{2}$ and so it must also divide $x y$... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 155 |
N4. Find all integers $x, y$ such that
$$
x^{3}(y+1)+y^{3}(x+1)=19
$$
|
Solution. Substituting $s=x+y$ and $p=x y$ we get
$$
2 p^{2}-\left(s^{2}-3 s\right) p+19-s^{3}=0
$$
This is a quadratic equation in $p$ with discriminant $D=s^{4}+2 s^{3}+9 s^{2}-152$.
For each $s$ we have $D0$.
For $s \geqslant 11$ and $s \leqslant-8$ we have $D>\left(s^{2}+s+3\right)^{2}$ as this is equivalent t... | (2,1),(1,2),(-1,-20),(-20,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 156 |
N5. Find all positive integers $x, y, z$ such that
$$
45^{x}-6^{y}=2019^{z}
$$
|
Solution. We define $v_{3}(n)$ to be the non-negative integer $k$ such that $3^{k} \mid n$ but $3^{k+1} \nmid n$. The equation is equivalent to
$$
3^{2 x} \cdot 5^{x}-3^{y} \cdot 2^{y}=3^{z} \cdot 673^{z}
$$
We will consider the cases $y \neq 2 x$ and $y=2 x$ separately.
Case 1. Suppose $y \neq 2 x$. Since $45^{x}>... | (2,1,1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 157 |
N6. Find all triples $(a, b, c)$ of nonnegative integers that satisfy
$$
a!+5^{b}=7^{c}
$$
|
Solution. We cannot have $c=0$ as $a!+5^{b} \geqslant 2>1=7^{0}$.
Assume first that $b=0$. So we are solving $a!+1=7^{c}$. If $a \geqslant 7$, then $7 \mid a!$ and so $7 \nmid a!+1$. So $7 \nmid 7^{c}$ which is impossible as $c \neq 0$. Checking $a0$. In this case, if $a \geqslant 5$, we have $5 \mid a$ !, and since ... | (,b,)\in{(3,0,1),(1,2,1),(4,2,2)} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 158 |
N7. Find all perfect squares $n$ such that if the positive integer $a \geqslant 15$ is some divisor of $n$ then $a+15$ is a prime power.
|
Solution. We call positive a integer $a$ "nice" if $a+15$ is a prime power.
From the definition, the numbers $n=1,4,9$ satisfy the required property. Suppose that for some $t \in \mathbb{Z}^{+}$, the number $n=t^{2} \geqslant 15$ also satisfies the required property. We have two cases:
1. If $n$ is a power of 2 , th... | 1,4,9,16,49,64,196 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 159 |
NT1 Solve in positive integers the equation $1005^{x}+2011^{y}=1006^{z}$.
| ## Solution
We have $1006^{z}>2011^{y}>2011$, hence $z \geq 2$. Then $1005^{x}+2011^{y} \equiv 0(\bmod 4)$.
But $1005^{x} \equiv 1(\bmod 4)$, so $2011^{y} \equiv-1(\bmod 4) \Rightarrow y$ is odd, i.e. $2011^{y} \equiv-1(\bmod 1006)$.
Since $1005^{x}+2011^{y} \equiv 0(\bmod 1006)$, we get $1005^{x} \equiv 1(\bmod 100... | (2,1,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 160 |
NT2 Find all prime numbers $p$ such that there exist positive integers $x, y$ that satisfy the relation $x\left(y^{2}-p\right)+y\left(x^{2}-p\right)=5 p$.
| ## Solution
The given equation is equivalent to $(x+y)(x y-p)=5 p$. Obviously $x+y \geq 2$.
We will consider the following three cases:
Case 1: $x+y=5$ and $x y=2 p$. The equation $x^{2}-5 x+2 p=0$ has at least a solution, so we must have $0 \leq \Delta=25-8 p$ which implies $p \in\{2,3\}$. For $p=2$ we obtain the s... | p\in{2,3,7} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 161 |
NT3 Find all positive integers $n$ such that the equation $y^{2}+x y+3 x=n\left(x^{2}+x y+3 y\right)$ has at least a solution $(x, y)$ in positive integers.
| ## Solution
Clearly for $n=1$, each pair $(x, y)$ with $x=y$ is a solution. Now, suppose that $n>1$ which implies $x \neq y$. We have $03$ and $y>x$. Take $d=\operatorname{gcd}(x+y-$ $\left.3 ; x^{2}+x y+3 y\right)$. Then $d$ divides $x^{2}+x y+3 y-x(x+y-3)=3(x+y)$. Then $d$ also divides $3(x+y)-3(x+y-3)=9$, hence $d ... | n\in{1,3,4,9} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 162 |
NT4 Find all prime positive integers $p, q$ such that $2 p^{3}-q^{2}=2(p+q)^{2}$.
| ## Solution 1
The given equation can be rewritten as $2 p^{2}(p-1)=q(3 q+4 p)$.
Hence $p\left|3 q^{2}+4 p q \Rightarrow p\right| 3 q^{2} \Rightarrow p \mid 3 q$ (since $p$ is a prime number) $\Rightarrow p \mid 3$ or $p \mid q$. If $p \mid q$, then $p=q$. The equation becomes $2 p^{3}-9 p^{2}=0$ which has no prime so... | (p,q)=(3,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 163 |
NT5 Find the least positive integer such that the sum of its digits is 2011 and the product of its digits is a power of 6 .
| ## Solution
Denote this number by $N$. Then $N$ can not contain the digits $0,5,7$ and its digits must be written in increasing order. Suppose that $N$ has $x_{1}$ ones, $x_{2}$ twos, $x_{3}$ threes, $x_{4}$ fours, $x_{6}$ sixes, $x_{8}$ eights and $x_{9}$ nines, then $x_{1}+2 x_{2}+3 x_{3}+4 x_{4}+6 x_{6}+8 x_{8}+9 x... | 34\underbrace{88\ldots8}_{93}\underbrace{99\ldots9}_{140} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 164 |
G1 Let $A B C$ be an isosceles triangle with $A B=A C$. On the extension of the side $[C A]$ we consider the point $D$ such that $A D<A C$. The perpendicular bisector of the segment $[B D]$ meets the internal and the external bisectors of the angle $\widehat{B A C}$ at the points $E$ and $Z$, respectively. Prove that ... |
Solution 1
In $\triangle A B D$ the ray $[A Z$ bisects the angle $\widehat{D A B}$ and the line $Z E$ is the perpendicular bisector of the side $[B D]$. Hence $Z$ belongs to the circumcircle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 165 |
G2 Let $A D, B F$ and $C E$ be the altitudes of $\triangle A B C$. A line passing through $D$ and parallel to $A B$ intersects the line $E F$ at the point $G$. If $H$ is the orthocenter of $\triangle A B C$, find the angle $\widehat{C G H}$.
| ## Solution 1
We can see easily that points $C, D, H, F$ lies on a circle of diameter $[C H]$.
Take $\left\{F, G^{\prime}\right\}=\odot(C H F) \cap E F$. We have $\widehat{E F H}=\widehat{B A D}=\widehat{B C E}=\widehat{D F H}$ since the quadrilaterals $A E D C, A E H F, C D H F$ are cyclic. Hence $[F B$ is the bisec... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 166 |
G3 Let $A B C$ be a triangle in which ( $B L$ is the angle bisector of $\widehat{A B C}(L \in A C), A H$ is an altitude of $\triangle A B C(H \in B C)$ and $M$ is the midpoint of the side $[A B]$. It is known that the midpoints of the segments $[B L]$ and $[M H]$ coincides. Determine the internal angles of triangle $\... | ## Solution
Let $N$ be the intersection of the segments $[B L]$ and $[M H]$. Because $N$ is the midpoint of both segments $[B L]$ and $[M H]$, it follows that $B M L H$ is a parallelogram. This implies that $M L \| B C$ and $L H \| A B$ and hence, since $M$ is the midpoint of $[A B]$, the angle bisector [ $B L$ and th... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 167 |
G4 Point $D$ lies on the side $[B C]$ of $\triangle A B C$. The circumcenters of $\triangle A D C$ and $\triangle B A D$ are $O_{1}$ and $O_{2}$, respectively and $O_{1} O_{2} \| A B$. The orthocenter of $\triangle A D C$ is $H$ and $A H=O_{1} O_{2}$. Find the angles of $\triangle A B C$ if $2 m(<C)=3 m(<B)$.
| ## Solution 1
As $A D$ is the radical axis of the circumcircles of $\triangle A D C$ and $\triangle B A D$, we have that $O_{1} O_{2} \perp A D$, therefore $\widehat{D A B}=90^{\circ}$. Let ... | \widehat{BAC}=105,\widehat{ABC}=30,\widehat{ACB}=45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 168 |
G5 Inside the square $A B C D$, the equilateral triangle $\triangle A B E$ is constructed. Let $M$ be an interior point of the triangle $\triangle A B E$ such that $M B=\sqrt{2}, M C=\sqrt{6}, M D=\sqrt{5}$ and $M E=\sqrt{3}$. Find the area of the square $A B C D$.
| ## Solution
Let $K, F, H, Z$ be the projections of point $M$ on the sides of the square.
Then by Pythagorean Theorem we can prove that $M A^{2}+M C^{2}=M B^{2}+M D^{2}$.
From the given condition we obtain $M A=1$.
With center $A$ and angle $60^{\circ}$, we rotate $\triangle A M E$, so we construct the triangle $A N... | 3+\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 169 |
G6 Let $A B C D$ be a convex quadrilateral, $E$ and $F$ points on the sides $A B$ and $C D$, respectively, such that $\frac{A B}{A E}=\frac{C D}{D F}=n$. Denote by $S$ the area of the quadrilateral $A E F D$. Prove that $S \leq \frac{A B \cdot C D+n(n-1) \cdot D A^{2}+n \cdot A D \cdot B C}{2 n^{2}}$.
| ## Solution
By Ptolemy's Inequality in $A E F D$, we get $S=\frac{A F \cdot D E \cdot \sin (\widehat{A F, D E})}{2} \leq \frac{A F \cdot D E}{2} \leq$ $\frac{A E \cdot D F+A D \cdot E F}{2}=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 170 |
NT1. Let $a, b, p, q$ be positive integers such that $a$ and $b$ are relatively prime, $a b$ is even and $p, q \geq 3$. Prove that
$$
2 a^{p} b-2 a b^{q}
$$
$$
a \text { is even }
$$
cannot be a square of an integer number. | Solution. Without loss of
Let $a=2 a^{\prime}$. If $\vdots$. Without loss of generality, assume that $a$ is even and consequently $b$ is odd.
$$
2 a^{p} b-2 a b^{q}=4 a^{\prime} b\left(a^{p-1}-b^{q-1}\right)
$$
is a square, then $a^{\prime}, b$ and $a^{p-1}-b^{q-1}$ are pairwise coprime.
On the other hand, $a^{p-1}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 171 |
NT3. Find all positive integers $n, n \geq 3$, such that $n \mid(n-2)$ !.
|
Solution. For $n=3$ and $n=4$ we easily check that $n$ does not divide $(n-2)!$.
If $n$ is prime, $n \geq 5$, then $(n-2)!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot(n-2)$ is not divided by $n$, since $n$ is a prime not included in the set of factors of ( $n-2)$ !.
If $n$ is composite, $n \geq 6$, then $n=p m$, where $p$ ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 173 |
NT4. If the positive integers $x$ and $y$ are such that both $3 x+4 y$ and $4 x+3 y$ are perfect squares, prove that both $x$ and $y$ are multiples of 7 .
|
Solution. Let
$$
3 x+4 y=m^{2}, \quad 4 x+3 y=n^{2}
$$
Then
$$
7(x+y)=m^{2}+n^{2} \Rightarrow 7 \mid m^{2}+n^{2}
$$
Considering $m=7 k+r, \quad r \in\{0,1,2,3,4,5,6\}$ we find that $m^{2} \equiv u(\bmod 7), \quad u \in$ $\{0,1,2,4\}$ and similarly $n^{2} \equiv v(\bmod 7), \quad v \in\{0,1,2,4\}$. Therefore we hav... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 174 |
A1. Prove that
$$
(1+a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 3+a+b+c
$$
for any real numbers $a, b, c \geq 1$.
|
Solution. The inequality rewrites as
$$
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+(b c+a c+a b) \geq 3+a+b+c
$$
or
$$
\left(\frac{\frac{1}{a}+\frac{1}{b}}{2}+\frac{\frac{1}{b}+\frac{1}{c}}{2}+\frac{\frac{1}{a}+\frac{1}{c}}{2}\right)+(b c+a c+a b) \geq 3+a+b+c
$$
which is equivalent to
$$
\frac{(2 a b-(a+b)... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 175 |
A2. Prove that, for all real numbers $x, y, z$ :
$$
\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leq(x+y+z)^{2}
$$
When the equality holds?
|
Solution. For $x=y=z=0$ the equality is valid.
Since $(x+y+z)^{2} \geq 0$ it is enongh to prove that
$$
\frac{x^{2}-y^{2}}{2 x^{2}+1}+\frac{y^{2}-z^{2}}{2 y^{2}+1}+\frac{z^{2}-x^{2}}{2 z^{2}+1} \leq 0
$$
which is equivalent to the inequality
$$
\frac{x^{2}-y^{2}}{x^{2}+\frac{1}{2}}+\frac{y^{2}-z^{2}}{y^{2}+\frac{1... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 176 |
A3. Prove that for all real $x, y$
$$
\frac{x+y}{x^{2}-x y+y^{2}} \leq \frac{2 \sqrt{2}}{\sqrt{x^{2}+y^{2}}}
$$
|
Solution. The inequality rewrites as
$$
\frac{x+y}{x^{2}-x y+y^{2}} \leq \frac{\sqrt{2\left(x^{2}+y^{2}\right)}}{\frac{x^{2}+y^{2}}{2}}
$$
Now it is enough to prove the next two simple inequalities:
$$
x+y \leq \sqrt{2\left(x^{2}+y^{2}\right)}, \quad x^{2}-x y+y^{2} \geq \frac{x^{2}+y^{2}}{2}
$$
| proof | Inequalities | proof | Yes | Yes | olympiads | false | 177 |
A4. Prove that if $0<\frac{a}{b}<b<2 a$ then
$$
\frac{2 a b-a^{2}}{7 a b-3 b^{2}-2 a^{2}}+\frac{2 a b-b^{2}}{7 a b-3 a^{2}-2 b^{2}} \geq 1+\frac{1}{4}\left(\frac{a}{b}-\frac{b}{a}\right)^{2}
$$
|
Solution. If we denote
$$
u=2-\frac{a}{b}, \quad v=2-\frac{b}{a}
$$
then the inequality rewrites as
$$
\begin{aligned}
& \frac{u}{v+u v}+\frac{v}{u+u v} \geq 1+\frac{1}{4}(u-u)^{2} \\
& \frac{(u-v)^{2}+u v(1-u v)}{u v(u v+u+v+1)} \geq \frac{(u-u)^{2}}{4}
\end{aligned}
$$
$\mathrm{Or}$
Since $u>0, v>0, u+v \leq 2,... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 178 |
G1. Two circles $k_{1}$ and $k_{2}$ intersect a.t points $A$ and $B$. A circle $k_{3}$ centered at $A$ meet $k_{1}$ at $M$ and $P$ and $k_{2}$ at $N$ and $Q$, such that $N$ and $Q$ are on different sides of $M P$ and $A B>A M$.
Prove rhat the angles $\angle M B Q$ and $\angle N B P$ are equal.
|
Solution As $A M=A P$, we have
$$
\angle M B A=\frac{1}{2} \operatorname{arcAM}=\frac{1}{2} \operatorname{arc} A P=\angle A B P
$$
and likewise
$$
\angle Q B A=\frac{1}{2} \operatorname{arc} A Q=\frac{1}{2} \operatorname{arc} c A N=\angle A B N
$$
Summing these equalities yields $\angle M B Q=\angle N B P$ as need... | notfound | Geometry | proof | Yes | Yes | olympiads | false | 179 |
G5. Let $A B C$ be a triangle with $\angle C=90^{\circ}$ and $D \in C .4, E \in C^{\prime} B$, and $k_{1}, k_{2}, k_{3}, k_{4}$ semicircles with diameters $C A, C B, C D, C E$ respectively, which have common part with the triangle $A B C$. Let also,
$$
k_{1} \cap k_{2}=\{C, K\}, k_{3} \cap k_{4}=\{C, M\}, k_{2} \cap ... |
Solution. The points $K, L, M, N$ belong to the segments $A B, B D, D E, E A$ respectively, where $C K \perp A B, C L \perp B D, C M \perp D E, C N \perp A E$. Then quadrilaterals $C D L M$ and $C E N M$ are inscribed. Let $\angle C A E=\varphi, \angle D C L=\theta$. Then $\angle E M N=\angle E C N=\varphi$ and $\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 181 |
C1. A polygon having $n$ sides is arbitrarily decomposed in triangles having all the vertices among the vertices of the polygon. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is side of the polygon and white those triangles that have in ... |
Solution. Denote by $b, r, w$ the number of black, red white triangles respectively.
It is easy to prove that the polygon is divided into $n-2$ triangles, hence
$$
b+r+w=n-2
$$
Each side of the polygon is a side of exactly one triangle of the decomposition, and thus
$$
2 b+r=n
$$
Subtracting the two relations yie... | b-2 | Geometry | proof | Yes | Yes | olympiads | false | 182 |
C2. Given $m \times n$ table, each cell signed with "-". The following operations are
(i) to change all the signs in entire row to the opposite, i. e. every "-" to "+", and every "+" to "-";
(ii) to change all the signs in entire column to the opposite, i. e. every "-" to "+" and every "+" to " -".
(a) Prove that i... |
Solution. If we apply (i) to $l$ rows and (ii) to $k$ columns we obtain $(m-k) l+(n-l) k$
(a) We have equation $(100-k) l+(100-l) k=2004$, or $100 l+100 k-2 l k=2004$, le
$$
50 l+50 k-1 k=1002
$$
Rewrite the lasc equation as
$$
(50-l)(50-h)=2.500-100.2=1498
$$
Since $1498=2 \cdot 7 \cdot 107$, this equation has n... | 102 | Combinatorics | proof | Yes | Yes | olympiads | false | 183 |
Problem A1. The real numbers $a, b, c, d$ satisfy simultaneously the equations
$$
a b c-d=1, b c d-a=2, c d a-b=3, d a b-c=-6
$$
Prove that $a+b+c+d \neq 0$.
|
Solution. Suppose that $a+b+c+d=0$. Then
$$
a b c+b c d+c d a+d a b=0
$$
If $a b c d=0$, then one of numbers, say $d$, must be 0 . In this case $a b c=0$, and so at least two of the numbers $a, b, c, d$ will be equal to 0 , making one of the given equations impossible. Hence $a b c d \neq 0$ and, from (1),
$$
\frac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 184 |
Problem A2. Determine all four digit numbers $\overline{a b c d}$ such that
$$
a(a+b+c+d)\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(a^{6}+2 b^{6}+3 c^{6}+4 d^{6}\right)=\overline{a b c d}
$$
|
Solution. From $\overline{a b c d}\overline{1 b c d}=(1+b+c+d)\left(1+b^{2}+c^{2}+d^{2}\right)\left(1+2 b^{6}+3 c^{6}+4 d^{6}\right) \geq$ $(b+1)\left(b^{2}+1\right)\left(2 b^{6}+1\right)$, so $b \leq 2$. Similarly one gets $c\overline{2 b c d}=2(2+b+c+d)\left(4+b^{2}+c^{2}+d^{2}\right)\left(64+2 b^{6}+3 c^{6}+4 d^{6}... | 2010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 185 |
Problem A3. Find all pairs $(x, y)$ of real numbers such that $|x|+|y|=1340$ and $x^{3}+y^{3}+2010 x y=670^{3}$.
|
Solution. Answer: $(-670 ;-670),(1005 ;-335),(-335 ; 1005)$.
To prove this, let $z=-670$. We have
$$
0=x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left((x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right)
$$
Thus either $x+y+z=0$, or $x=y=z$. In the latter case we get $x=y=-670$, which satisfies both the equations. In the former ... | (-670,-670),(1005,-335),(-335,1005) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 186 |
Problem A4. Let $a, b, c$ be positive real numbers such that $a b c(a+b+c)=3$. Prove the inequality
$$
(a+b)(b+c)(c+a) \geq 8
$$
and determine all cases when equality holds.
|
Solution. We have
$A=(a+b)(b+c)(c+a)=\left(a b+a c+b^{2}+b c\right)(c+a)=(b(a+b+c)+a c)(c+a)$,
so by the given condition
$$
A=\left(\frac{3}{a c}+a c\right)(c+a)=\left(\frac{1}{a c}+\frac{1}{a c}+\frac{1}{a c}+a c\right)(c+a)
$$
Aplying the AM-GM inequality for four and two terms respectively, we get
$$
A \geq 4 ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 187 |
Problem A5. The real positive numbers $x, y, z$ satisfy the relations $x \leq 2$, $y \leq 3, x+y+z=11$. Prove that $\sqrt{x y z} \leq 6$.
|
Solution. For $x=2, y=3$ and $z=6$ the equality holds.
After the substitutions $x=2-u, y=3-v$ with $u \in[0,2), v \in[0,3)$, we obtain that $z=6+u+v$ and the required inequality becomes
$$
(2-u)(3-v)(6+u+v) \leqslant 36
$$
We shall need the following lemma.
Lemma. If real numbers $a$ and $b$ satisfy the relations ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 188 |
Problem G1. Consider a triangle $A B C$ with $\angle A C B=90^{\circ}$. Let $F$ be the foot of the altitude from $C$. Circle $\omega$ touches the line segment $F B$ at point $P$, the altitude $C F$ at point $Q$ and the circumcircle of $A B C$ at point $R$. Prove that points $A, Q, R$ are collinear and $A P=A C$.
Let $O$ be the circumcenter of the triangle $A B M$. From $\angle B A M=105^{\circ}$ follows $\angle M B O=15^{\circ}$. Let $M^{\prime}, C^{\pri... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 190 |
Problem G3. Let $A B C$ be an acute-angled triangle. A circle $\omega_{1}\left(O_{1}, R_{1}\right)$ passes through points $B$ and $C$ and meets the sides $A B$ and $A C$ at points $D$ and $E$, respectively. Let $\omega_{2}\left(O_{2}, R_{2}\right)$ be the circumcircle of the triangle $A D E$. Prove that $O_{1} O_{2}$ ... |
Solution. Recall that, in every triangle, the altitude and the diameter of the circumcircle drawn from the same vertex are isogonal. The proof offers no difficulty, being a simple angle chasing around the circumcircle of the triangle.
Let $O$ be the circumcenter of the triangle $A B C$. From the above, one has $\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 191 |
Problem G4. Let $A L$ and $B K$ be angle bisectors in the non-isosceles triangle $A B C(L \in B C, K \in A C)$. The perpendicular bisector of $B K$ intersects the line $A L$ at point $M$. Point $N$ lies on the line $B K$ such that $L N \| M K$. Prove that $L N=N A$.
. Then $\angle C B K=\angle A B K=\angle A M K=\angle N L A$. Thus $A B L N$ is cyclic, whence $\angle N A L=$ $\angle N B L=\angle C B K=\angle N L A$. No... | LN=NA | Geometry | proof | Yes | Yes | olympiads | false | 192 |
Problem C1. There are two piles of coins, each containing 2010 pieces. Two players A and B play a game taking turns (A plays first). At each turn, the player on play has to take one or more coins from one pile or exactly one coin from each pile. Whoever takes the last coin is the winner. Which player will win if they ... |
Solution. B wins.
In fact, we will show that A will lose if the total number of coins is a multiple of 3 and the two piles differ by not more than one coin (call this a balanced position). To this end, firstly notice that it is not possible to move from one balanced position to another. The winning strategy for B con... | Bwins | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 193 |
Problem N1. Find all positive integers $n$ such that $n 2^{n+1}+1$ is a perfect square.
|
Solution. Answer: $n=0$ and $n=3$.
Clearly $n 2^{n+1}+1$ is odd, so, if this number is a perfect square, then $n 2^{n+1}+1=$ $(2 x+1)^{2}, x \in \mathbb{N}$, whence $n 2^{n-1}=x(x+1)$.
The integers $x$ and $x+1$ are coprime, so one of them must divisible by $2^{n-1}$, which means that the other must be at most $n$. ... | n=0n=3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 195 |
Problem N2. Find all positive integers $n$ such that $36^{n}-6$ is a product of two or more consecutive positive integers.
|
Solution. Answer: $n=1$.
Among each four consecutive integers there is a multiple of 4 . As $36^{n}-6$ is not a multiple of 4 , it must be the product of two or three consecutive positive integers.
Case I. If $36^{n}-6=x(x+1)$ (all letters here and below denote positive integers), then $4 \cdot 36^{n}-23=(2 x+1)^{2}... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 196 |
A1. Let $a, b, c$ be positive real numbers such that $a b c=8$. Prove that
$$
\frac{a b+4}{a+2}+\frac{b c+4}{b+2}+\frac{c a+4}{c+2} \geq 6
$$
|
Solution. We have $a b+4=\frac{8}{c}+4=\frac{4(c+2)}{c}$ and similarly $b c+4=\frac{4(a+2)}{a}$ and $c a+4=\frac{4(b+2)}{b}$. It follows that
$$
(a b+4)(b c+4)(c a+4)=\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2)
$$
so that
$$
\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8
$$
Applying AM-GM, we conclude:
$$
\... | 6 | Inequalities | proof | Yes | Yes | olympiads | false | 197 |
A2. Given positive real numbers $a, b, c$, prove that
$$
\frac{8}{(a+b)^{2}+4 a b c}+\frac{8}{(a+c)^{2}+4 a b c}+\frac{8}{(b+c)^{2}+4 a b c}+a^{2}+b^{2}+c^{2} \geq \frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}
$$
|
Solution. Since $2 a b \leq a^{2}+b^{2}$, it follows that $(a+b)^{2} \leq 2\left(a^{2}+b^{2}\right)$ and $4 a b c \leq 2 c\left(a^{2}+b^{2}\right)$, for any positive reals $a, b, c$. Adding these inequalities, we find
$$
(a+b)^{2}+4 a b c \leq 2\left(a^{2}+b^{2}\right)(c+1)
$$
so that
$$
\frac{8}{(a+b)^{2}+4 a b c}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 198 |
A3. Determine the number of pairs of integers $(m, n)$ such that
$$
\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}} \in \mathbb{Q}
$$
|
Solution. Let $r=\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}}$. Then
$$
n+m+2 \sqrt{n+\sqrt{2016}} \cdot \sqrt{m-\sqrt{2016}}=r^{2}
$$
and
$$
(m-n) \sqrt{2106}=\frac{1}{4}\left(r^{2}-m-n\right)^{2}-m n+2016 \in \mathbb{Q}
$$
Since $\sqrt{2016} \notin \mathbb{Q}$, it follows that $m=n$. Then
$$
\sqrt{n^{2}-2016}=\fra... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 199 |
A4. If $x, y, z$ are non-negative real numbers such that $x^{2}+y^{2}+z^{2}=x+y+z$, then show that:
$$
\frac{x+1}{\sqrt{x^{5}+x+1}}+\frac{y+1}{\sqrt{y^{5}+y+1}}+\frac{z+1}{\sqrt{z^{5}+z+1}} \geq 3
$$
When does the equality hold?
|
Solution. First we factor $x^{5}+x+1$ as follows:
$$
\begin{aligned}
x^{5}+x+1 & =x^{5}-x^{2}+x^{2}+x+1=x^{2}\left(x^{3}-1\right)+x^{2}+x+1=x^{2}(x-1)\left(x^{2}+x+1\right)+x^{2}+x+1 \\
& =\left(x^{2}+x+1\right)\left(x^{2}(x-1)+1\right)=\left(x^{2}+x+1\right)\left(x^{3}-x^{2}+1\right)
\end{aligned}
$$
Using the $A M... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 200 |
A5. Let $x, y, z$ be positive real numbers such that $x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$.
a) Prove the inequality
$$
x+y+z \geq \sqrt{\frac{x y+1}{2}}+\sqrt{\frac{y z+1}{2}}+\sqrt{\frac{z x+1}{2}}
$$
b) (Added by the problem selecting committee) When does the equality hold?
| ## Solution.
a) We rewrite the inequality as
$$
(\sqrt{x y+1}+\sqrt{y z+1}+\sqrt{z x+1})^{2} \leq 2 \cdot(x+y+z)^{2}
$$
and note that, from CBS,
$$
\text { LHS } \leq\left(\frac{x y+1}{x}+\frac{y z+1}{y}+\frac{z x+1}{z}\right)(x+y+z)
$$
But
$$
\frac{x y+1}{x}+\frac{y z+1}{y}+\frac{z x+1}{z}=x+y+z+\frac{1}{x}+\fra... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 201 |
C1. Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{1... | ## Solution.
We will first calculate $S_{999}$, then $S_{1999}-S_{999}$, and then $S_{2016}-S_{1999}$.
Writing the integers from 1 to 999 as 001 to 999, adding eventually also 000 (since 0 digits actually do not matter), each digit appears exactly 100 times in each position(as unit, ten, or hundred). Hence
$$
S_{999... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 202 |
C2. The natural numbers from 1 to 50 are written down on the blackboard. At least how many of them should be deleted, in order that the sum of any two of the remaining numbers is not a prime?
|
Solution. Notice that if the odd, respectively even, numbers are all deleted, then the sum of any two remaining numbers is even and exceeds 2 , so it is certainly not a prime. We prove that 25 is the minimal number of deleted numbers. To this end, we group the positive integers from 1 to 50 in 25 pairs, such that the ... | 25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 203 |
C3. Consider any four pairwise distinct real numbers and write one of these numbers in each cell of a $5 \times 5$ array so that each number occurs exactly once in every $2 \times 2$ subarray. The sum over all entries of the array is called the total sum of that array. Determine the maximum number of distinct total su... |
Solution. We will prove that the maximum number of total sums is 60 .
The proof is based on the following claim.
Claim. Either each row contains exactly two of the numbers, or each column contains exactly two of the numbers.
Proof of the Claim. Indeed, let $R$ be a row containing at least three of the numbers. Then... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 204 |
C4. A splitting of a planar polygon is a finite set of triangles whose interiors are pairwise disjoint, and whose union is the polygon in question. Given an integer $n \geq 3$, determine the largest integer $m$ such that no planar $n$-gon splits into less than $m$ triangles.
|
Solution. The required maximum is $\lceil n / 3\rceil$, the least integer greater than or equal to $n / 3$. To describe a planar $n$-gon splitting into this many triangles, write $n=3 m-r$, where $m$ is a positive integer and $r=0,1,2$, and consider $m$ coplanar equilateral triangles $A_{3 i} A_{3 i+1} A_{3 i+2}$, $i=... | \lceiln/3\rceil | Geometry | math-word-problem | Yes | Yes | olympiads | false | 205 |
G1. Let $A B C$ be an acute angled triangle, let $O$ be its circumcentre, and let $D, E, F$ be points on the sides $B C, A C, A B$, respectively. The circle $\left(c_{1}\right)$ of radius $F A$, centred at $F$, crosses the segment $(O A)$ at $A^{\prime}$ and the circumcircle (c) of the triangle $A B C$ again at $K$. S... |
Solution. We will prove that the quadrilateral $B K F A^{\prime}$ is cyclic and its circumcircle passes through the center $O$ of the circle (c).
The triangle $A F K$ is isosceles, so $m(\widehat{K F B})=2 m(\widehat{K A B})=m(\widehat{K O B})$. It follows that the quadrilateral $B K F O$ is cyclic.
The triangles $O... | proof | Geometry | proof | Yes | Yes | olympiads | false | 206 |
G2. Let $A B C$ be a triangle with $m(\widehat{B A C})=60^{\circ}$. Let $D$ and $E$ be the feet of the perpendiculars from $A$ to the external angle bisectors of $\widehat{A B C}$ and $\widehat{A C B}$, respectively. Let $O$ be the circumcenter of the triangle $A B C$. Prove that the circumcircles of the triangles $\t... |
Solution. Let $X$ be the intersection of the lines $B D$ and $C E$.
We will prove that $X$ lies on the circumcircles of both triangles $\triangle A D E$ and $\triangle B O C$ and then we will prove that the centers of these circles and the point $X$ are collinear, which is enough for proving that the circles are tang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 207 |
G3. A trapezoid $A B C D(A B \| C D, A B>C D)$ is circumscribed. The incircle of triangle $A B C$ touches the lines $A B$ and $A C$ at $M$ and $N$, respectively. Prove that the incenter of the trapezoid lies on the line $M N$.
=90^{\circ}-\frac{1}{2} m(\widehat{M A N}) \quad \text { and } \quad m(\widehat{B I C})=90^{\circ}+\frac{1}{2} m(\widehat{M A N})
$$
the quadrilateral $I R N C$ is cyclic.
It fo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 208 |
G4. Let $A B C$ be an acute angled triangle whose shortest side is $[B C]$. Consider a variable point $P$ on the side $[B C]$, and let $D$ and $E$ be points on $(A B]$ and $(A C]$, respectively, such that $B D=B P$ and $C P=C E$. Prove that, as $P$ traces $[B C]$, the circumcircle of the triangle $A D E$ passes throug... |
Solution. We claim that the fixed point is the center of the incircle of $A B C$.
Let $I$ be the center of the incircle of $A B C$. Since $B D=B P$ and $[B I$ is the bisector of $\widehat{D B P}$, the line $B I$ is the perpendicular bisector of $[D P]$. This yields $D I=P I$. Analogously we get $E I=P I$. So, the poi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 209 |
G5. Let $A B C$ be an acute angled triangle with orthocenter $H$ and circumcenter $O$. Assume the circumcenter $X$ of $B H C$ lies on the circumcircle of $A B C$. Reflect $O$ across $X$ to obtain $O^{\prime}$, and let the lines $X H$ and $O^{\prime} A$ meet at $K$. Let $L, M$ and $N$ be the midpoints of $[X B],[X C]$ ... |
Solution. The circumcircles of $A B C$ and $B H C$ have the same radius. So, $X B=$ $X C=X H=X O=r$ (where $r$ is the radius of the circle $A B C$ ) and $O^{\prime}$ lies on $C(X, r)$. We conclude that $O X$ is the perpendicular bisector for $[B C]$. So, $B O X$ and $C O X$ are equilateral triangles.
It is known that... | proof | Geometry | proof | Yes | Yes | olympiads | false | 210 |
G6. Given an acute triangle $A B C$, erect triangles $A B D$ and $A C E$ externally, so that $m(\widehat{A D B})=m(\widehat{A E C})=90^{\circ}$ and $\widehat{B A D} \equiv \widehat{C A E}$. Let $A_{1} \in B C, B_{1} \in A C$ and $C_{1} \in A B$ be the feet of the altitudes of the triangle $A B C$, and let $K$ and $L$ ... |
Solution. Let $M, P$ and $Q$ be the midpoints of $[B C],[C A]$ and $[A B]$, respectively.
The circumcircle of the triangle $A_{1} B_{1} C_{1}$ is the Euler's circle. So, the point $M$ lies on this circle.
It is enough to prove now that $\left[A_{1} M\right]$ is a common chord of the three circles $\left(A_{1} B_{1} ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 211 |
G7. Let $[A B]$ be a chord of a circle $(c)$ centered at $O$, and let $K$ be a point on the segment $(A B)$ such that $A K<B K$. Two circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, meet again at $L$. Let $P$ be one of the points of intersection of the line $K L$ and the circle (c), and let... |
Solution. Let $\left(c_{1}\right)$ and $\left(c_{2}\right)$ be circles through $K$, internally tangent to (c) at $A$ and $B$, respectively, and meeting again at $L$, and let the common tangent to $\left(c_{1}\right)$ and $(c)$ meet the common tangent to $\left(c_{2}\right)$ and $(c)$ at $Q$. Then the point $Q$ is the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 212 |
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