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.
+
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)
+
The quantity of milk with <math>1\%</math> fat, that was added is (in kg)
  
A. <math>1000</math>
+
<math>\mathrm{(A)}\ 1000\qquad\mathrm{(B)}\ 600\qquad\mathrm{(C)}\ 800\qquad\mathrm{(D)}\ 120\qquad\mathrm{(E)}\ 480</math>
 
 
B. <math>600</math>
 
 
 
C.  <math>800</math>
 
 
 
D. <math>120</math>
 
 
 
E. <math>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</math>  <math>\forall \alpha , \beta \in R</math>.
+
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
+
The value of the expression <math>K=\left[\left(1+\sqrt{3}\right)*2\right]*\sqrt{2}</math> is
  
A. <math>3</math>
+
<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ \sqrt{3}\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 1</math>
 
 
B. <math>0</math>
 
 
 
C.  <math>\sqrt{3}</math>
 
 
 
D. <math>9</math>
 
 
 
E. <math>1</math>
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 2|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 2|Solution]]
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The domain of the function <math>f(x)=\sqrt{4+2x}</math> is
 
The domain of the function <math>f(x)=\sqrt{4+2x}</math> is
  
A. <math>(-2,+\infty)</math>
+
<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>
 
 
B. <math>[0,+\infty)</math>
 
 
 
C.  <math>[-2,+\infty)</math>
 
 
 
D. <math>[-2,0]</math>
 
 
 
E. <math>R</math>
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 3|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 3|Solution]]
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Which of the following is correct, about the graph of <math>f</math>?
 
Which of the following is correct, about the graph of <math>f</math>?
  
A. intersects x-axis
+
<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>
 
 
B. touches y-axis
 
 
 
C. touches x-axis
 
 
 
D. has minimum point
 
 
 
E. has maximum point
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 4|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 4|Solution]]
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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
 
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
  
A. <math>384</math>
+
<math>\mathrm{(A)}\ 384\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 56\qquad\mathrm{(E)}\ 512</math>
 
 
B. <math>2</math>
 
 
 
C.  <math>15</math>
 
 
 
D. <math>56</math>
 
 
 
E. <math>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+7\sqrt{3}}-\sqrt{7+4\sqrt{3}}</math> is
+
The value of the expression <math>K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}</math> is
  
A. <math>4</math>
+
<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>
 
 
B. <math>4\sqrt{3}</math>
 
 
 
C.  <math>12+4\sqrt{3}</math>
 
 
 
D. <math>-2</math>
 
 
 
E. <math>2</math>
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 6|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 6|Solution]]
  
 
== Problem 7 ==
 
== Problem 7 ==
<div style="float:right">
+
[[Image:2006 CyMO-7.PNG|250px|right]]
[[Image:2006 CyMO-7.PNG|250px]]
 
</div>
 
  
In the figure, <math>ABC</math> is an equilateral triangle and <math>AD\perp BC</math>, <math>DE\perp AC</math>, <math>EZ\perp BC</math>. If <math>EZ=\sqrt{3}</math>, then the length of the side os the triangle ABC is
+
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
  
A. <math>\frac{3\sqrt{3}}{2}</math>
+
<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>
 
 
B. <math>8</math>
 
 
 
C.  <math>4</math>
 
 
 
D. <math>3</math>
 
 
 
E. <math>9</math>
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 7|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 7|Solution]]
  
 
== Problem 8 ==
 
== Problem 8 ==
<div style="float:right">
+
[[Image:2006 CyMO-8.PNG|250px|right]]
[[Image:2006 CyMO-8.PNG|250px]]
 
</div>
 
 
 
In the figure <math>ABCDE</math> is a regular 5-sided polygon and <math>Z</math>, <math>H</math>, <math>L</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>AHCLCIDKEZA</math> is 1, then the area of the shaded quadrilateral <math>ACIZ</math> is
 
  
A. <math>\frac{2}{3}</math>
+
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
  
B. <math>\frac{1}{2}</math>
+
<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>
 
 
C. <math>\frac{3}{7}</math>
 
 
 
D. <math>\frac{3}{10}</math>
 
 
 
E. None of these
 
  
 
[[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
+
If <math>x=\sqrt[3]{4}</math> and <math>y=\sqrt[3]{6}-\sqrt[3]{3}</math>, then which of the following is correct?
  
A. <math>x=y</math>
+
<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>
 
 
B. <math>x<y</math>
 
 
 
C. <math>x=2y</math>
 
 
 
D. <math>x>2y</math>
 
 
 
E. None of these
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 9|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 9|Solution]]
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If <math>2^x=15</math> and <math>15^y=256</math>, then the product <math>xy</math> equals
 
If <math>2^x=15</math> and <math>15^y=256</math>, then the product <math>xy</math> equals
  
A. <math>7</math>
+
<math>\mathrm{(A)}\ 7\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 6</math>
 
 
B. <math>3</math>
 
 
 
C. <math>1</math>
 
 
 
D. <math>8</math>
 
 
 
E. <math>6</math>
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 10|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 10|Solution]]
  
 
== Problem 11 ==
 
== Problem 11 ==
<div style="float:right">
+
[[Image:2006 CyMO-11.PNG|250px|right]]
[[Image:2006 CyMO-11.PNG|250px]]
 
</div>
 
 
 
The lines <math>(\epsilon):x-2y=0</math> and <math>(\delta):x+y=4</math> intersect at the point <math>C</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>OAC</math> to the area of the triangle <math>OBC</math> equals
 
 
 
A. <math>\frac{1}{3}</math>
 
 
 
B. <math>\frac{2}{3}</math>
 
  
C. <math>\frac{3}{5}</math>
+
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
  
D. <math>\frac{1}{2}</math>
+
<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>
 
 
E. <math>\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} \displaystyle \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>  
+
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
 
then <math>f(28,17)</math> equals
  
A. <math>8</math>
+
<math>\mathrm{(A)}\ 8\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 1</math>
 
 
B. <math>0</math>
 
 
 
C. <math>11</math>
 
 
 
D. <math>5</math>
 
 
 
E. <math>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
 
The sum of the digits of the number <math>10^{2006}-2006</math> is
  
A. <math>18006</math>
+
<math>\mathrm{(A)}\ 18006\qquad\mathrm{(B)}\ 20060\qquad\mathrm{(C)}\ 2006\qquad\mathrm{(D)}\ 18047\qquad\mathrm{(E)}\ \text{None of these}</math>
 
 
B. <math>20060</math>
 
 
 
C. <math>2006</math>
 
 
 
D. <math>18047</math>
 
 
 
E. None of these
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 13|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 13|Solution]]
  
 
== Problem 14 ==
 
== Problem 14 ==
<div style="float:right">
+
[[Image:2006 CyMO-14.PNG|250px|right]]
[[Image:2006 CyMO-17.PNG|250px]]
 
</div>
 
 
 
The rectangle <math>ABCD</math> is a small garden divided to the rectangle <math>AXED</math> and to the square <math>ZBCE</math>, so that <math>AE=2\sqrt{5}m</math> and the shaded area of the triangle <math>DBE</math> is <math>4m^2</math>. The area of the whole garden is
 
  
A. <math>24m^2</math>
+
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
  
B. <math>20m^2</math>
+
<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>
 
 
C. <math>16m^2</math>
 
 
 
D. <math>32m^2</math>
 
 
 
E. <math>10\sqrt5m^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
  
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>
 
 
A. <math>3/4</math>
 
 
 
B. <math>\frac{3}{2}</math>
 
 
 
C. <math>\frac{2}{5}</math>
 
 
 
D. <math>\frac{1}{2}</math>
 
 
 
E. <math>\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
 
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
  
A. <math>x^2-2k^2x+2m^2=0</math>
+
<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>
 
 
B. <math>x^2-\frac{k}{m}x+\frac{1}{2m}=0</math>
 
 
 
C. <math>x^2-\frac{m}{k}x+\frac{1}{2m}=0</math>
 
 
 
D. <math>2mx^2-kx+1=0</math>
 
 
 
E. <math>2kx^2-2mx+1=0</math>
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 16|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 16|Solution]]
  
 
== Problem 17 ==
 
== Problem 17 ==
<div style="float:right">
+
[[Image:2006 CyMO-17.PNG|250px|right]]
[[Image:2006 CyMO-17.PNG|250px]]
 
</div>
 
 
 
<math>ABC</math> is equilateral triangle of side <math>\alpha</math> and <math>AD=BE=\frac{\alpha}{3}</math>. The measure of the angle <math>\ang CPE</math> is
 
 
 
A. <math>60^{\circ}</math>
 
 
 
B. <math>50^{\circ}</math>
 
 
 
C. <math>40^{\circ}</math>
 
  
D. <math>45^{\circ}</math>
+
<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
  
E. <math>70^{\circ}</math>
+
<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 ==
<div style="float:right">
+
[[Image:2006 CyMO-18.PNG|250px|right]]
[[Image:2006 CyMO-18.PNG|250px]]
 
</div>
 
 
 
<math>K(k,0)</math> is the minimum point of the parabola and the parabola intersects the y-axis at the point <math>C(0,k)</math>.
 
If the area if the rectangle <math>OABC</math> is <math>8</math>, then the equation of the parabola is
 
 
 
A. <math>y=\frac{1}{2}(x+2)^2</math>
 
 
 
B. <math>y=\frac{1}{2}(x-2)^2</math>
 
  
C. <math>y=x^2+2</math>
+
<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
  
D. <math>y=x^2-2x+1</math>
+
<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>
 
 
E. <math>y=x^2-4x+4</math>
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 18|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 18|Solution]]
  
 
== Problem 19 ==
 
== Problem 19 ==
<div style="float:right">
+
[[Image:2006 CyMO-19.PNG|250px|right]]
[[Image:2006 CyMO-19.PNG|250px]]
+
 
</div>
+
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>.
 
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
 
Given that <math>f(1)=f(2)=1</math>, then <math>f(3n)</math> equals
  
A. <math>3</math>
+
<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ -3\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 1\qquad\mathrm{(E)}\ 0</math>
 
 
B. <math>-3</math>
 
 
 
C. <math>2</math>
 
 
 
D. <math>1</math>
 
 
 
E. <math>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
 
A convex polygon has <math>n</math> sides and <math>740</math> diagonals. Then <math>n</math> equals
  
A. <math>30</math>
+
<math>\mathrm{(A)}\ 30\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 60\qquad\mathrm{(E)}\ \text{None of these}</math>
 
 
B. <math>40</math>
 
 
 
C. <math>50</math>
 
 
 
D. <math>60</math>
 
 
 
E. None of these
 
  
 
[[2006 Cyprus MO/Lyceum/Problem 21|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 21|Solution]]
  
 
== Problem 22 ==
 
== Problem 22 ==
<div style="float:right">
+
[[Image:2006 CyMO-22.PNG|250px|right]]
[[Image:2006 CyMO-22.PNG|250px]]
+
 
</div>
+
<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
 
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
  
A. <math>0</math>
+
<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 1</math>
 
 
B. <math>5</math>
 
 
 
C. <math>2</math>
 
 
 
D. <math>4</math>
 
 
 
E. <math>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
 
The number of divisors of the number <math>2006</math> is
  
A. <math>3</math>
+
<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6</math>
 
 
B. <math>4</math>
 
 
 
C. <math>8</math>
 
 
 
D. <math>5</math>
 
 
 
E. <math>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 ==
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[[2006 Cyprus MO/Lyceum/Problem 26|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 26|Solution]]
  
 
== Problem 27 ==
 
== Problem 27 ==
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[[2006 Cyprus MO/Lyceum/Problem 27|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 27|Solution]]
  
 
== Problem 28 ==
 
== Problem 28 ==
 
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[[2006 Cyprus MO/Lyceum/Problem 28|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 28|Solution]]
  
 
== Problem 29 ==
 
== Problem 29 ==
 
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[[2006 Cyprus MO/Lyceum/Problem 29|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 29|Solution]]
  
 
== Problem 30 ==
 
== Problem 30 ==
 
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[[2006 Cyprus MO/Lyceum/Problem 30|Solution]]
 
[[2006 Cyprus MO/Lyceum/Problem 30|Solution]]

Latest revision as of 10:56, 27 April 2008

Problem 1

A diary industry, in a quantity of milk with $4\%$ fat adds a quantity of milk with $1\%$ fat and produces $1200$kg of milk with $2\%$ fat. The quantity of milk with $1\%$ fat, that was added is (in kg)

$\mathrm{(A)}\ 1000\qquad\mathrm{(B)}\ 600\qquad\mathrm{(C)}\ 800\qquad\mathrm{(D)}\ 120\qquad\mathrm{(E)}\ 480$

Solution

Problem 2

The operation $\alpha*\beta$ is defined by $\alpha*\beta=\alpha^2-\beta^2\ \forall\alpha,\beta\in\mathbb{R}$. The value of the expression $K=\left[\left(1+\sqrt{3}\right)*2\right]*\sqrt{2}$ is

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ \sqrt{3}\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 1$

Solution

Problem 3

The domain of the function $f(x)=\sqrt{4+2x}$ is

$\mathrm{(A)}\ (-2,+\infty)\qquad\mathrm{(B)}\ [0,+\infty)\qquad\mathrm{(C)}\ [-2,+\infty)\qquad\mathrm{(D)}\ [-2,0]\qquad\mathrm{(E)}\ \mathbb{R}$

Solution

Problem 4

Given the function $f(x)=\alpha x^2 +9x+ \frac{81}{4\alpha}$ , $\alpha \neq 0$ Which of the following is correct, about the graph of $f$?

$\mathrm{(A)}\ \text{intersects x-axis}\qquad\mathrm{(B)}\ \text{touches y-axis}\qquad\mathrm{(C)}\ \text{touches x-axis}\qquad\mathrm{(D)}\ \text{has minimum point}\qquad\mathrm{(E)}\ \text{has maximum point}$

Solution

Problem 5

If both integers $\alpha,\beta$ are bigger than 1 and satisfy $a^7=b^8$, then the minimum value of $\alpha+\beta$ is

$\mathrm{(A)}\ 384\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 56\qquad\mathrm{(E)}\ 512$

Solution

Problem 6

The value of the expression $K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}$ is

$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4\sqrt{3}\qquad\mathrm{(C)}\ 12+4\sqrt{3}\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 2$

Solution

Problem 7

2006 CyMO-7.PNG

In the figure, $AB\Gamma$ is an equilateral triangle and $A\Delta \perp B\Gamma$, $\Delta E\perp A\Gamma$, $EZ\perp B\Gamma$. If $EZ=\sqrt{3}$, then the length of the side of the triangle $AB\Gamma$ is

$\mathrm{(A)}\ \frac{3\sqrt{3}}{2}\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 9$

Solution

Problem 8

2006 CyMO-8.PNG

In the figure $AB\Gamma \Delta E$ is a regular 5-sided polygon and $Z$, $H$, $\Theta$, $I$, $K$ are the points of intersections of the extensions of the sides. If the area of the "star" $AHB\Theta \Gamma I\Delta KEZA$ is 1, then the area of the shaded quadrilateral $A\Gamma IZ$ is

$\mathrm{(A)}\ \frac{2}{3}\qquad\mathrm{(B)}\ \frac{1}{2}\qquad\mathrm{(C)}\ \frac{3}{7}\qquad\mathrm{(D)}\ \frac{3}{10}\qquad\mathrm{(E)}\ \text{None of these}$

Solution

Problem 9

If $x=\sqrt[3]{4}$ and $y=\sqrt[3]{6}-\sqrt[3]{3}$, then which of the following is correct?

$\mathrm{(A)}\ x=y\qquad\mathrm{(B)}\ x<y\qquad\mathrm{(C)}\ x=2y\qquad\mathrm{(D)}\ x>2y\qquad\mathrm{(E)}\ \text{None of these}$

Solution

Problem 10

If $2^x=15$ and $15^y=256$, then the product $xy$ equals

$\mathrm{(A)}\ 7\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 6$

Solution

Problem 11

2006 CyMO-11.PNG

The lines $(\epsilon):x-2y=0$ and $(\delta):x+y=4$ intersect at the point $\Gamma$. If the line $(\delta)$ intersects the axes $Ox$ and $Oy$ to the points $A$ and $B$ respectively, then the ratio of the area of the triangle $OA\Gamma$ to the area of the triangle $OB\Gamma$ equals

$\mathrm{(A)}\ \frac{1}{3}\qquad\mathrm{(B)}\ \frac{2}{3}\qquad\mathrm{(C)}\ \frac{3}{5}\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{4}{9}$

Solution

Problem 12

If $f(\alpha,\beta)= \begin{cases}\alpha & \textrm {if} \qquad \alpha=\beta \\ f(\alpha-\beta,\beta) & \textrm {if} \qquad \alpha>\beta \\ f(\beta-\alpha,\alpha) & \textrm {if} \qquad \alpha<\beta \end{cases}$

then $f(28,17)$ equals

$\mathrm{(A)}\ 8\qquad\mathrm{(B)}\ 0\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 1$

Solution

Problem 13

The sum of the digits of the number $10^{2006}-2006$ is

$\mathrm{(A)}\ 18006\qquad\mathrm{(B)}\ 20060\qquad\mathrm{(C)}\ 2006\qquad\mathrm{(D)}\ 18047\qquad\mathrm{(E)}\ \text{None of these}$

Solution

Problem 14

2006 CyMO-14.PNG

The rectangle $AB\Gamma \Delta$ is a small garden divided to the rectangle $AZE\Delta$ and to the square $ZB\Gamma E$, so that $AE=2\sqrt{5}\ \text{m}$ and the shaded area of the triangle $\Delta BE$ is $4\ \text{m}^2$. The area of the whole garden is

$\mathrm{(A)}\ 24\ \text{m}^2\qquad\mathrm{(B)}\ 20\ \text{m}^2\qquad\mathrm{(C)}\ 16\ \text{m}^2\qquad\mathrm{(D)}\ 32\ \text{m}^2\qquad\mathrm{(E)}\ 10\sqrt{5}\ \text{m}^2$

Solution

Problem 15

The expression $\frac{1}{2+\sqrt7} + \frac{1}{\sqrt7+\sqrt{10}}+ \frac{1}{\sqrt{10}+\sqrt{13}} + \frac{1}{\sqrt{13}+4}$ equals

$\mathrm{(A)}\ \frac{3}{4}\qquad\mathrm{(B)}\ \frac{3}{2}\qquad\mathrm{(C)}\ \frac{2}{5}\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{2}{3}$

Solution

Problem 16

If $x_1,x_2$ are the roots of the equation $x^2-2kx+2m=0$, then $\frac{1}{x_1},\frac{1}{x_2}$ are the roots of the equation

$\mathrm{(A)}\ x^2-2k^2x+2m^2=0\qquad\mathrm{(B)}\ x^2-\frac{k}{m}x+\frac{1}{2m}=0\qquad\mathrm{(C)}\ x^2-\frac{m}{k}x+\frac{1}{2m}=0\\ \mathrm{(D)}\ 2mx^2-kx+1=0\qquad\mathrm{(E)}\ 2kx^2-2mx+1=0$

Solution

Problem 17

2006 CyMO-17.PNG

$AB\Gamma$ is equilateral triangle of side $\alpha$ and $A\Delta=BE=\frac{\alpha}{3}$. The measure of the angle $\ang \Gamma PE$ (Error compiling LaTeX. Unknown error_msg) is

$\mathrm{(A)}\ 60^\circ\qquad\mathrm{(B)}\ 50^\circ\qquad\mathrm{(C)}\ 40^\circ\qquad\mathrm{(D)}\ 45^\circ\qquad\mathrm{(E)}\ 70^\circ$

Solution

Problem 18

2006 CyMO-18.PNG

$K(k,0)$ is the minimum point of the parabola and the parabola intersects the y-axis at the point $\Gamma (0,k)$. If the area if the rectangle $OAB\Gamma$ is $8$, then the equation of the parabola is

$\mathrm{(A)}\ y=\frac{1}{2}(x+2)^2\qquad\mathrm{(B)}\ y=\frac{1}{2}(x-2)^2\qquad\mathrm{(C)}\ y=x^2+2\qquad\mathrm{(D)}\ y=x^2-2x+1\qquad\mathrm{(E)}\ y=x^2-4x+4$

Solution

Problem 19

2006 CyMO-19.PNG

In the figure, $AB\Gamma$ is an isosceles triangle with$AB=A\Gamma=\sqrt2$ and $\ang A=45^\circ$ (Error compiling LaTeX. Unknown error_msg). If $B\Delta$ is an altitude of the triangle and the sector $B\Lambda \Delta KB$ belongs to the circle $(B,B\Delta )$, the area of the shaded region is

$\mathrm{(A)}\ \frac{4\sqrt3-\pi}{6}\qquad\mathrm{(B)}\ 4\left(\sqrt2-\frac{\pi}{3}\right)\qquad\mathrm{(C)}\ \frac{8\sqrt2-3\pi}{16}\qquad\mathrm{(D)}\ \frac{\pi}{8}\qquad\mathrm{(E)}\ \text{None of these}$

Solution

Problem 20

The sequence $f:N \to R$ satisfies $f(n)=f(n-1)-f(n-2),\forall n\geq 3$. Given that $f(1)=f(2)=1$, then $f(3n)$ equals

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ -3\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 1\qquad\mathrm{(E)}\ 0$

Solution

Problem 21

A convex polygon has $n$ sides and $740$ diagonals. Then $n$ equals

$\mathrm{(A)}\ 30\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 60\qquad\mathrm{(E)}\ \text{None of these}$

Solution

Problem 22

2006 CyMO-22.PNG

$AB\Gamma \Delta$ is rectangular and the points $K,\Lambda ,M,N$ lie on the sides $AB, B\Gamma , \Gamma \Delta, \Delta A$ respectively so that $\frac{AK}{KB}=\frac{BL}{L\Gamma}=\frac{\Gamma M}{M\Delta}=\frac{\Delta N}{NA}=2$. If $E_1$ is the area of $K\Lambda MN$ and $E_2$ is the area of the rectangle $AB\Gamma \Delta$, the ratio $\frac{E_1}{E_2}$ equals

$\mathrm{(A)}\ \frac{5}{9}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{9}{5}\qquad\mathrm{(D)}\ \frac{3}{5}\qquad\mathrm{(E)}\ \text{None of these}$

Solution

Problem 23

Of $21$ 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

$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 1$

Solution

Problem 24

The number of divisors of the number $2006$ is

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$

Solution

Problem 25

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Solution

Problem 26

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Solution

Problem 27

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Solution

Problem 28

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Solution

Problem 29

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Solution

Problem 30

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Solution

See also