Difference between revisions of "2006 Cyprus MO/Lyceum/Problems"
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== Problem 1 == | == Problem 1 == | ||
+ | A diary industry, in a quantity of milk with <math>4\%</math> fat adds a quantity of milk with <math>1\%</math> fat and produces <math>1200</math>kg of milk with <math>2\%</math> fat. | ||
+ | The quantity of milk with <math>1\%</math> fat, that was added is (in kg) | ||
+ | |||
+ | <math>\mathrm{(A)}\ 1000\qquad\mathrm{(B)}\ 600\qquad\mathrm{(C)}\ 800\qquad\mathrm{(D)}\ 120\qquad\mathrm{(E)}\ 480</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 1|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 1|Solution]] | ||
== Problem 2 == | == Problem 2 == | ||
+ | The operation <math>\alpha*\beta</math> is defined by <math>\alpha*\beta=\alpha^2-\beta^2\ \forall\alpha,\beta\in\mathbb{R}</math>. | ||
+ | The value of the expression <math>K=\left[\left(1+\sqrt{3}\right)*2\right]*\sqrt{2}</math> is | ||
+ | <math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ \sqrt{3}\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 1</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 2|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 2|Solution]] | ||
== Problem 3 == | == Problem 3 == | ||
+ | The domain of the function <math>f(x)=\sqrt{4+2x}</math> is | ||
+ | <math>\mathrm{(A)}\ (-2,+\infty)\qquad\mathrm{(B)}\ [0,+\infty)\qquad\mathrm{(C)}\ [-2,+\infty)\qquad\mathrm{(D)}\ [-2,0]\qquad\mathrm{(E)}\ \mathbb{R}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 3|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 3|Solution]] | ||
== Problem 4 == | == Problem 4 == | ||
+ | Given the function <math>f(x)=\alpha x^2 +9x+ \frac{81}{4\alpha}</math> , <math>\alpha \neq 0</math> | ||
+ | Which of the following is correct, about the graph of <math>f</math>? | ||
+ | <math>\mathrm{(A)}\ \text{intersects x-axis}\qquad\mathrm{(B)}\ \text{touches y-axis}\qquad\mathrm{(C)}\ \text{touches x-axis}\qquad\mathrm{(D)}\ \text{has minimum point}\qquad\mathrm{(E)}\ \text{has maximum point}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 4|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 4|Solution]] | ||
== Problem 5 == | == Problem 5 == | ||
+ | If both integers <math>\alpha,\beta</math> are bigger than 1 and satisfy <math>a^7=b^8</math>, then the minimum value of <math>\alpha+\beta</math> is | ||
+ | <math>\mathrm{(A)}\ 384\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 56\qquad\mathrm{(E)}\ 512</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 5|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 5|Solution]] | ||
== Problem 6 == | == Problem 6 == | ||
+ | The value of the expression <math>K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}</math> is | ||
+ | <math>\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4\sqrt{3}\qquad\mathrm{(C)}\ 12+4\sqrt{3}\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 2</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 6|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 6|Solution]] | ||
== Problem 7 == | == Problem 7 == | ||
− | + | [[Image:2006 CyMO-7.PNG|250px|right]] | |
− | [[Image:2006 CyMO-7.PNG|250px]] | + | |
− | </ | + | In the figure, <math>AB\Gamma</math> is an equilateral triangle and <math>A\Delta \perp B\Gamma</math>, <math>\Delta E\perp A\Gamma</math>, <math>EZ\perp B\Gamma</math>. If <math>EZ=\sqrt{3}</math>, then the length of the side of the triangle <math>AB\Gamma</math> is |
+ | |||
+ | <math>\mathrm{(A)}\ \frac{3\sqrt{3}}{2}\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 9</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 7|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 7|Solution]] | ||
== Problem 8 == | == Problem 8 == | ||
− | + | [[Image:2006 CyMO-8.PNG|250px|right]] | |
− | [[Image:2006 CyMO-8.PNG|250px]] | + | |
− | </ | + | In the figure <math>AB\Gamma \Delta E</math> is a regular 5-sided polygon and <math>Z</math>, <math>H</math>, <math>\Theta</math>, <math>I</math>, <math>K</math> are the points of intersections of the extensions of the sides. |
+ | If the area of the "star" <math>AHB\Theta \Gamma I\Delta KEZA</math> is 1, then the area of the shaded quadrilateral <math>A\Gamma IZ</math> is | ||
+ | |||
+ | <math>\mathrm{(A)}\ \frac{2}{3}\qquad\mathrm{(B)}\ \frac{1}{2}\qquad\mathrm{(C)}\ \frac{3}{7}\qquad\mathrm{(D)}\ \frac{3}{10}\qquad\mathrm{(E)}\ \text{None of these}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 8|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 8|Solution]] | ||
== Problem 9 == | == Problem 9 == | ||
+ | If <math>x=\sqrt[3]{4}</math> and <math>y=\sqrt[3]{6}-\sqrt[3]{3}</math>, then which of the following is correct? | ||
+ | |||
+ | <math>\mathrm{(A)}\ x=y\qquad\mathrm{(B)}\ x<y\qquad\mathrm{(C)}\ x=2y\qquad\mathrm{(D)}\ x>2y\qquad\mathrm{(E)}\ \text{None of these}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 9|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 9|Solution]] | ||
== Problem 10 == | == Problem 10 == | ||
+ | If <math>2^x=15</math> and <math>15^y=256</math>, then the product <math>xy</math> equals | ||
+ | |||
+ | <math>\mathrm{(A)}\ 7\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 6</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 10|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 10|Solution]] | ||
== Problem 11 == | == Problem 11 == | ||
− | + | [[Image:2006 CyMO-11.PNG|250px|right]] | |
− | [[Image:2006 CyMO- | ||
− | |||
+ | The lines <math>(\epsilon):x-2y=0</math> and <math>(\delta):x+y=4</math> intersect at the point <math>\Gamma</math>. If the line <math>(\delta)</math> intersects the axes <math>Ox</math> and <math>Oy</math> to the points <math>A</math> and <math>B</math> respectively, then the ratio of the area of the triangle <math>OA\Gamma</math> to the area of the triangle <math>OB\Gamma</math> equals | ||
+ | |||
+ | <math>\mathrm{(A)}\ \frac{1}{3}\qquad\mathrm{(B)}\ \frac{2}{3}\qquad\mathrm{(C)}\ \frac{3}{5}\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{4}{9}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 11|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 11|Solution]] | ||
== Problem 12 == | == Problem 12 == | ||
+ | If <math>f(\alpha,\beta)= \begin{cases}\alpha & \textrm {if} \qquad \alpha=\beta \\ f(\alpha-\beta,\beta) & \textrm {if} \qquad \alpha>\beta \\ f(\beta-\alpha,\alpha) & \textrm {if} \qquad \alpha<\beta \end{cases} </math> | ||
+ | then <math>f(28,17)</math> equals | ||
+ | |||
+ | <math>\mathrm{(A)}\ 8\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 1</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 12|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 12|Solution]] | ||
== Problem 13 == | == Problem 13 == | ||
+ | The sum of the digits of the number <math>10^{2006}-2006</math> is | ||
+ | <math>\mathrm{(A)}\ 18006\qquad\mathrm{(B)}\ 20060\qquad\mathrm{(C)}\ 2006\qquad\mathrm{(D)}\ 18047\qquad\mathrm{(E)}\ \text{None of these}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 13|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 13|Solution]] | ||
== Problem 14 == | == Problem 14 == | ||
− | + | [[Image:2006 CyMO-14.PNG|250px|right]] | |
− | [[Image:2006 CyMO- | ||
− | |||
+ | The rectangle <math>AB\Gamma \Delta</math> is a small garden divided to the rectangle <math>AZE\Delta</math> and to the square <math>ZB\Gamma E</math>, so that <math>AE=2\sqrt{5}\ \text{m}</math> and the shaded area of the triangle <math>\Delta BE</math> is <math>4\ \text{m}^2</math>. The area of the whole garden is | ||
+ | |||
+ | <math>\mathrm{(A)}\ 24\ \text{m}^2\qquad\mathrm{(B)}\ 20\ \text{m}^2\qquad\mathrm{(C)}\ 16\ \text{m}^2\qquad\mathrm{(D)}\ 32\ \text{m}^2\qquad\mathrm{(E)}\ 10\sqrt{5}\ \text{m}^2</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 14|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 14|Solution]] | ||
== Problem 15 == | == Problem 15 == | ||
+ | The expression <math>\frac{1}{2+\sqrt7} + \frac{1}{\sqrt7+\sqrt{10}}+ \frac{1}{\sqrt{10}+\sqrt{13}} + \frac{1}{\sqrt{13}+4}</math> equals | ||
+ | <math>\mathrm{(A)}\ \frac{3}{4}\qquad\mathrm{(B)}\ \frac{3}{2}\qquad\mathrm{(C)}\ \frac{2}{5}\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{2}{3}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 15|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 15|Solution]] | ||
== Problem 16 == | == Problem 16 == | ||
+ | If <math>x_1,x_2</math> are the roots of the equation <math>x^2-2kx+2m=0</math>, then <math>\frac{1}{x_1},\frac{1}{x_2}</math> are the roots of the equation | ||
+ | <math>\mathrm{(A)}\ x^2-2k^2x+2m^2=0\qquad\mathrm{(B)}\ x^2-\frac{k}{m}x+\frac{1}{2m}=0\qquad\mathrm{(C)}\ x^2-\frac{m}{k}x+\frac{1}{2m}=0\\ \mathrm{(D)}\ 2mx^2-kx+1=0\qquad\mathrm{(E)}\ 2kx^2-2mx+1=0</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 16|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 16|Solution]] | ||
== Problem 17 == | == Problem 17 == | ||
+ | [[Image:2006 CyMO-17.PNG|250px|right]] | ||
+ | <math>AB\Gamma</math> is equilateral triangle of side <math>\alpha</math> and <math>A\Delta=BE=\frac{\alpha}{3}</math>. The measure of the angle <math>\ang \Gamma PE</math> is | ||
+ | |||
+ | <math>\mathrm{(A)}\ 60^\circ\qquad\mathrm{(B)}\ 50^\circ\qquad\mathrm{(C)}\ 40^\circ\qquad\mathrm{(D)}\ 45^\circ\qquad\mathrm{(E)}\ 70^\circ</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 17|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 17|Solution]] | ||
== Problem 18 == | == Problem 18 == | ||
+ | [[Image:2006 CyMO-18.PNG|250px|right]] | ||
+ | <math>K(k,0)</math> is the minimum point of the parabola and the parabola intersects the y-axis at the point <math>\Gamma (0,k)</math>. | ||
+ | If the area if the rectangle <math>OAB\Gamma</math> is <math>8</math>, then the equation of the parabola is | ||
+ | |||
+ | <math>\mathrm{(A)}\ y=\frac{1}{2}(x+2)^2\qquad\mathrm{(B)}\ y=\frac{1}{2}(x-2)^2\qquad\mathrm{(C)}\ y=x^2+2\qquad\mathrm{(D)}\ y=x^2-2x+1\qquad\mathrm{(E)}\ y=x^2-4x+4</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 18|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 18|Solution]] | ||
== Problem 19 == | == Problem 19 == | ||
− | + | [[Image:2006 CyMO-19.PNG|250px|right]] | |
− | [[Image:2006 CyMO-19.PNG|250px]] | ||
− | |||
+ | In the figure, <math>AB\Gamma</math> is an isosceles triangle with<math> AB=A\Gamma=\sqrt2</math> and <math>\ang A=45^\circ</math>. If <math>B\Delta</math> is an altitude of the triangle and the sector <math>B\Lambda \Delta KB</math> belongs to the circle <math>(B,B\Delta )</math>, the area of the shaded region is | ||
+ | |||
+ | <math>\mathrm{(A)}\ \frac{4\sqrt3-\pi}{6}\qquad\mathrm{(B)}\ 4\left(\sqrt2-\frac{\pi}{3}\right)\qquad\mathrm{(C)}\ \frac{8\sqrt2-3\pi}{16}\qquad\mathrm{(D)}\ \frac{\pi}{8}\qquad\mathrm{(E)}\ \text{None of these}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 19|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 19|Solution]] | ||
== Problem 20 == | == Problem 20 == | ||
+ | The sequence <math>f:N \to R</math> satisfies <math>f(n)=f(n-1)-f(n-2),\forall n\geq 3</math>. | ||
+ | Given that <math>f(1)=f(2)=1</math>, then <math>f(3n)</math> equals | ||
+ | <math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ -3\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 1\qquad\mathrm{(E)}\ 0</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 20|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 20|Solution]] | ||
== Problem 21 == | == Problem 21 == | ||
+ | A convex polygon has <math>n</math> sides and <math>740</math> diagonals. Then <math>n</math> equals | ||
+ | |||
+ | <math>\mathrm{(A)}\ 30\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 60\qquad\mathrm{(E)}\ \text{None of these}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 21|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 21|Solution]] | ||
== Problem 22 == | == Problem 22 == | ||
+ | [[Image:2006 CyMO-22.PNG|250px|right]] | ||
+ | <math>AB\Gamma \Delta</math> is rectangular and the points <math>K,\Lambda ,M,N</math> lie on the sides <math>AB, B\Gamma , \Gamma \Delta, \Delta A</math> respectively so that <math>\frac{AK}{KB}=\frac{BL}{L\Gamma}=\frac{\Gamma M}{M\Delta}=\frac{\Delta N}{NA}=2</math>. If <math>E_1</math> is the area of <math>K\Lambda MN</math> and <math>E_2</math> is the area of the rectangle <math>AB\Gamma \Delta</math>, the ratio <math>\frac{E_1}{E_2}</math> equals | ||
+ | |||
+ | <math>\mathrm{(A)}\ \frac{5}{9}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{9}{5}\qquad\mathrm{(D)}\ \frac{3}{5}\qquad\mathrm{(E)}\ \text{None of these}</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 22|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 22|Solution]] | ||
== Problem 23 == | == Problem 23 == | ||
+ | Of <math>21</math> students taking Mathematics, Physics and Chemistry, no student takes one subject only. The number of students taking Mathematics and Chemistry only, equals to four times the number taking Mathematics and Physics only. If the number of students taking Physics and Chemistry only equals to three times the number of students taking all three subjects, then the number of students taking all three subjects is | ||
+ | <math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 1</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 23|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 23|Solution]] | ||
== Problem 24 == | == Problem 24 == | ||
+ | The number of divisors of the number <math>2006</math> is | ||
+ | <math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6</math> | ||
[[2006 Cyprus MO/Lyceum/Problem 24|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 24|Solution]] | ||
== Problem 25 == | == Problem 25 == | ||
− | + | {{problem}} | |
[[2006 Cyprus MO/Lyceum/Problem 25|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 25|Solution]] | ||
== Problem 26 == | == Problem 26 == | ||
+ | {{problem}} | ||
[[2006 Cyprus MO/Lyceum/Problem 26|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 26|Solution]] | ||
== Problem 27 == | == Problem 27 == | ||
+ | {{problem}} | ||
[[2006 Cyprus MO/Lyceum/Problem 27|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 27|Solution]] | ||
== Problem 28 == | == Problem 28 == | ||
− | + | {{problem}} | |
[[2006 Cyprus MO/Lyceum/Problem 28|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 28|Solution]] | ||
== Problem 29 == | == Problem 29 == | ||
− | + | {{problem}} | |
[[2006 Cyprus MO/Lyceum/Problem 29|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 29|Solution]] | ||
== Problem 30 == | == Problem 30 == | ||
− | + | {{problem}} | |
[[2006 Cyprus MO/Lyceum/Problem 30|Solution]] | [[2006 Cyprus MO/Lyceum/Problem 30|Solution]] |
Latest revision as of 09:56, 27 April 2008
Contents
- 1 Problem 1
- 2 Problem 2
- 3 Problem 3
- 4 Problem 4
- 5 Problem 5
- 6 Problem 6
- 7 Problem 7
- 8 Problem 8
- 9 Problem 9
- 10 Problem 10
- 11 Problem 11
- 12 Problem 12
- 13 Problem 13
- 14 Problem 14
- 15 Problem 15
- 16 Problem 16
- 17 Problem 17
- 18 Problem 18
- 19 Problem 19
- 20 Problem 20
- 21 Problem 21
- 22 Problem 22
- 23 Problem 23
- 24 Problem 24
- 25 Problem 25
- 26 Problem 26
- 27 Problem 27
- 28 Problem 28
- 29 Problem 29
- 30 Problem 30
- 31 See also
Problem 1
A diary industry, in a quantity of milk with fat adds a quantity of milk with fat and produces kg of milk with fat. The quantity of milk with fat, that was added is (in kg)
Problem 2
The operation is defined by . The value of the expression is
Problem 3
The domain of the function is
Problem 4
Given the function , Which of the following is correct, about the graph of ?
Problem 5
If both integers are bigger than 1 and satisfy , then the minimum value of is
Problem 6
The value of the expression is
Problem 7
In the figure, is an equilateral triangle and , , . If , then the length of the side of the triangle is
Problem 8
In the figure is a regular 5-sided polygon and , , , , are the points of intersections of the extensions of the sides. If the area of the "star" is 1, then the area of the shaded quadrilateral is
Problem 9
If and , then which of the following is correct?
Problem 10
If and , then the product equals
Problem 11
The lines and intersect at the point . If the line intersects the axes and to the points and respectively, then the ratio of the area of the triangle to the area of the triangle equals
Problem 12
If
then equals
Problem 13
The sum of the digits of the number is
Problem 14
The rectangle is a small garden divided to the rectangle and to the square , so that and the shaded area of the triangle is . The area of the whole garden is
Problem 15
The expression equals
Problem 16
If are the roots of the equation , then are the roots of the equation
Problem 17
is equilateral triangle of side and . The measure of the angle $\ang \Gamma PE$ (Error compiling LaTeX. Unknown error_msg) is
Problem 18
is the minimum point of the parabola and the parabola intersects the y-axis at the point . If the area if the rectangle is , then the equation of the parabola is
Problem 19
In the figure, is an isosceles triangle with and $\ang A=45^\circ$ (Error compiling LaTeX. Unknown error_msg). If is an altitude of the triangle and the sector belongs to the circle , the area of the shaded region is
Problem 20
The sequence satisfies . Given that , then equals
Problem 21
A convex polygon has sides and diagonals. Then equals
Problem 22
is rectangular and the points lie on the sides respectively so that . If is the area of and is the area of the rectangle , the ratio equals
Problem 23
Of students taking Mathematics, Physics and Chemistry, no student takes one subject only. The number of students taking Mathematics and Chemistry only, equals to four times the number taking Mathematics and Physics only. If the number of students taking Physics and Chemistry only equals to three times the number of students taking all three subjects, then the number of students taking all three subjects is
Problem 24
The number of divisors of the number is
Problem 25
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Problem 26
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Problem 27
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Problem 28
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Problem 29
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Problem 30
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