Difference between revisions of "2007 Cyprus MO/Lyceum/Problems"

Latest revision as of 07:29, 19 December 2009

Problem 1

If $x-y=1$ , then the value of the expression $K=x^2+x-2xy+y^2-y$ is

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

Problem 2

Given the formula $f(x) = 4^x$ , then $f(x+1)-f(x)$ equals to

$\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4^x\qquad\mathrm{(C)}\ 2\cdot4^x\qquad\mathrm{(D)}\ 4^{x+1}\qquad\mathrm{(E)}\ 3\cdot4^x$

Solution

Problem 3

A cyclist drives form town A to town B with velocity $40 {}^{km}/{}_h$ and comes back with velocity $60 {}^{km}/{}_h$ . The mean velocity in ${}^{km}/{}_h$ for the total distance is

$\mathrm{(A)}\ 45\qquad\mathrm{(B)}\ 48\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 55\qquad\mathrm{(E)}\ 100$

Solution

Problem 4

We define the operation $a*b = \frac{1+a}{1+b^2}$ , $\forall a,b \in \real$ .

The value of $(2*0)*1$ is

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

Solution

Problem 5

If the remainder of the division of $a$ with $35$ is $23$ , then the remainder of the division of $a$ with $7$ is

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

Solution

Problem 6

$ABCD$ is a square of side length 2 and $FG$ is an arc of the circle with centre the midpoint $K$ of the side $AB$ and radius 2. The length of the segments $FD=GC=x$ is

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

Solution

Problem 7

If a diagonal $d$ of a rectangle forms a $60^\circ$ angle with one of its sides, then the area of the rectangle is

$\mathrm{(A)}\ \frac{d^2 \sqrt{3}}{4}\qquad\mathrm{(B)}\ \frac{d^2}{2}\qquad\mathrm{(C)}\ 2d^2\qquad\mathrm{(D)}\ d^2 \sqrt{2}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}$

Solution

Problem 8

If we subtract from 2 the inverse number of $x-1$ , we get the inverse of $x-1$ . Then the number $x+1$ equals to

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

Solution

Problem 9

We consider the sequence of real numbers $a_1,a_2,a_3,...$ such that $a_1=0$ , $a_2=1$ and $a_n=a_{n-1}-a_{n-2}$ , $\forall n \in \{3,4,5,6,...\}$ . The value of the term $a_{138}$ is

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

Solution

Problem 10

The volume of an orthogonal parallelepiped is $132\;\mathrm{cm}^3$ and its dimensions are integers. The minimum sum of the dimensions is

$\mathrm{(A)}\ 27\ \mathrm{cm}\qquad\mathrm{(B)}\ 19\ \mathrm{cm}\qquad\mathrm{(C)}\ 20\ \mathrm{cm}\qquad\mathrm{(D)}\ 18\ \mathrm{cm}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}$

Solution

Problem 11

If $X=\frac{1}{2007 \sqrt{2006}+2006 \sqrt{2007}}$ and $Y=\frac{1}{\sqrt{2006}}-\frac{1}{\sqrt{2007}}$ , which of the following is correct?

$\mathrm{(A)}\ X=2Y\qquad\mathrm{(B)}\ Y=2X\qquad\mathrm{(C)}\ X=Y\qquad\mathrm{(D)}\ X=Y^2\qquad\mathrm{(E)}\ Y=X^2$

Solution

Problem 12

The function $f : \Re \rightarrow \Re$ has the properties $f(0) = -1$ and $f(xy)+f(x)+f(y)=x+y+xy+k$ $\forall x,y \in \Re$ , where $k \in \Re$ is a constant. The value of $f(-1)$ is

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

Solution

Problem 13

If $x_1,x_2$ are the roots of the equation $x^2+ax+1=0$ and $x_3,x_4$ are the roots of the equation $x^2+bx+1=0$ , then the expression $\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4}+ \frac{x_3}{x_1x_2x_4}+\frac{x_4}{x_1x_2x_3}$ equals to

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

Solution

Problem 14

In the square $ABCD$ the segment $KB$ equals a side of the square. The ratio of areas $\frac{S_1}{S_2}$ is

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

Solution

Problem 15

The reflex angles of the concave octagon $ABCDEFGH$ measure $240^\circ$ each. Diagonals $AE$ and $GC$ are perpendicular, bisect each other, and are both equal to $2$ .

The area of the octagon is

$\mathrm{(A)}\ \frac{6-2\sqrt{3}}{3}\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ \frac{6+2\sqrt{3}}{3}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}$

Solution

Problem 16

The full time score of a football match was $3$ - $2$ . How many possible half time results could there have been in this match?

$\mathrm{(A)}\ 5\qquad\mathrm{(B)}\ 6\qquad\mathrm{(C)}\ 10\qquad\mathrm{(D)}\ 11\qquad\mathrm{(E)}\ 12$

Solution

Problem 17

The last digit of the number $a=2^{2007}+3^{2007}+5^{2007}+7^{2007}$ is

$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$

Solution

Problem 18

How many subsets are there for the set $A=\{1,2,3,4,5,6,7\}$ ?

$\mathrm{(A)}\ 7\qquad\mathrm{(B)}\ 14\qquad\mathrm{(C)}\ 49\qquad\mathrm{(D)}\ 64\qquad\mathrm{(E)}\ 128$

Solution

Problem 19

120 five-digit numbers can be written with the digits $1,2,3,4,5$ . If we place these numbers in increasing order, then the position of the number $41253$ is

$\mathrm{(A)}\ 71^{\mathrm{st}}\qquad\mathrm{(B)}\ 72^{\mathrm{nd}}\qquad\mathrm{(C)}\ 73^{\mathrm{rd}}\qquad\mathrm{(D)}\ 74^{\mathrm{th}}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}$

Solution

Problem 20

The mean value for 9 Math-tests that a student succeded was $10$ (in scale $0$ - $20$ ). If we put the grades of these tests in incresing order, then the maximum grade of the $5^{th}$ test is

$\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 19$

Solution

Problem 21

In the following figure, three equal cycles of diameter $20\,\mathrm{ cm}$ represent pulleys, that are connected with a strap. If the distances between any two pulley center points are $AB=3\,\mathrm{m}$ , $AC=4\,\mathrm{m}$ and $BC=5\,\mathrm{m}$ , then the length of the strap is

$\mathrm{(A)}\ 12+20\pi)\ \mathrm{m}\qquad\mathrm{(B)}\ (12+\pi)\ \mathrm{m}\qquad\mathrm{(C)}\ (12+4\pi)\ \mathrm{m}\qquad\mathrm{(D)}\ \left(12+\frac{\pi}{5}\right)\ \mathrm{m}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}$

Solution

Problem 22

In the following figure $ABCD$ is an orthogonal trapezium with $\ang A= \ang D=90^\circ$ (Error compiling LaTeX. Unknown error_msg) and bases $AB = a$ , $DC = 2a$ . If $AD = 3a$ and $M$ is the midpoint of the side $BC$ , then $AM$ equals to

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

Solution

Problem 23

In the figure above the right section of a parabolic tunnel is presented. Its maximum height is $OC=8\,\mathrm{m}$ and its maximum width is $AB=20\,\mathrm{m}$ . If M is the midpoint of $OB$ , then the height $MK$ of the tunnel at the point $M$ is

$\mathrm{(A)}\ 5\ \mathrm{m}\qquad\mathrm{(B)}\ 5.2\ \mathrm{m}\qquad\mathrm{(C)}\ 5.5\ \mathrm{m}\qquad\mathrm{(D)}\ 5.8\ \mathrm{m}\qquad\mathrm{(E)}\ 6\ \mathrm{m}$

Solution

Problem 24

Costas sold two televisions for €198 each. From the sale of the first one he made a profit of 10% on its value and from the sale of the second one, he had a loss of 10% on its value. After the sale of the two televisions Costas had in total

$\mathrm{(A)}$ profit €4

$\mathrm{(B)}$ neither profit nor loss

$\qquad\mathrm{(C)}$ loss €8

$\qquad\mathrm{(D)}$ profit €8

$\qquad\mathrm{(E)}$ loss €4

Solution

Problem 25

A jeweler makes crosses, according to the pattern shown above. The crosses are made from golden cyclical discs, with diameter of 1cm each. The height of a cross, which is made from 402 such discs is

$\mathrm{(A)}\ 198\ \mathrm{cm}\qquad\mathrm{(B)}\ 2\ \mathrm{m}\qquad\mathrm{(C)}\ 201\ \mathrm{cm}\qquad\mathrm{(D)}\ 202\ \mathrm{cm}\qquad\mathrm{(E)}\ 204\ \mathrm{cm}$

Solution

Problem 26

The number of boys in a school is 3 times the number of girls and the number of girls is 9 times the number of teachers. Let us denote with $b$ , $g$ and $t$ , the number of boys, girls and teachers respectively. Then the total number of boys, girls and teachers equals to

$\mathrm{(A)}\ 31b\qquad\mathrm{(B)}\ \frac{37b}{27}\qquad\mathrm{(C)}\ 13g\qquad\mathrm{(D)}\ \frac{37g}{27}\qquad\mathrm{(E)}\ \frac{37t}{27}$

Solution

Problem 27

In the following diagram, the light beam $\epsilon$ is reflected on the $x$ -axis and the beam $d$ , being reflected on a mirror parallel to the $y$ -axis at distance 6, intersects the $y$ -axis at point $B$ .
The equation of line $f$ is given by

$\mathrm{(A)}\ x+y-11=0\qquad\mathrm{(B)}\ x+y+11=0\qquad\mathrm{(C)}\ x-y+11=0\qquad\mathrm{(D)}\ x-y-11=0\qquad\mathrm{(E)}\ y=-x+10$

Solution

Problem 28

The product of $15^8\cdot28^6\cdot5^{11}$ is an integer number whose last digits are zeros. How many zeros are there?

$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 12\qquad\mathrm{(E)}\ 19$

Solution

Problem 29

The minimum value of a positive integer $k$ , for which the sum $S=k+(k+1)+(k+2)+\ldots+(k+10)$ is a perfect square, is

$\mathrm{(A)}\ 5\qquad\mathrm{(B)}\ 6\qquad\mathrm{(C)}\ 10\qquad\mathrm{(D)}\ 11\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}$

Solution

Problem 30

A coin with a shape of a regular hexagon of side 1 is tangent to a square of side 6, as shown in the figure.

The coin rotates on the perimeter of the square, until it reaches its original position.

The length of the line which is being inscribed by the center of the hexagon is

$\mathrm{(A)}\ \frac{34\pi}{3}\qquad\mathrm{(B)}\ 24\qquad\mathrm{(C)}\ \frac{28\pi}{3}\qquad\mathrm{(D)}\ 6 \pi\sqrt{2}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}$

Solution

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Difference between revisions of "2007 Cyprus MO/Lyceum/Problems"

Latest revision as of 07:29, 19 December 2009

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

See also

@@ Line 2: / Line 2: @@
 If <math>x-y=1</math>, then the value of the expression <math>K=x^2+x-2xy+y^2-y</math> is
-A. <math>2</math>
+<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ -2\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ -1\qquad\mathrm{(E)}\ 0</math>
-B. <math>-2</math>
-C.  <math>1</math>
-D. <math>-1</math>
-E. <math>0</math>
 [[2007 Cyprus MO/Lyceum/Problem 1|Solution]]
@@ Line 17: / Line 9: @@
 Given the formula <math>f(x) = 4^x</math>, then <math>f(x+1)-f(x)</math> equals to
-A. <math>4</math>
+<math>\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4^x\qquad\mathrm{(C)}\ 2\cdot4^x\qquad\mathrm{(D)}\ 4^{x+1}\qquad\mathrm{(E)}\ 3\cdot4^x</math>
-B. <math>4^x</math>
-C. <math>2\cdot4^x</math>
-D. <math>4^{x+1}</math>
-E. <math>3\cdot4^x</math>
 [[2007 Cyprus MO/Lyceum/Problem 2|Solution]]
 == Problem 3 ==
-A cyclist drives form town A to town B with velocity <math>40 \frac{\mathrm{km}}{\mathrm{h}}</math> and comes back with velocity <math> 60 \frac{\mathrm{km}}{\mathrm{h}}</math>. The mean velocity in <math>\frac{\mathrm{km}}{\mathrm{h}}</math> for the total distance is
+A cyclist drives form town A to town B with velocity <math>40  {}^{km}/{}_h</math> and comes back with velocity <math> 60 {}^{km}/{}_h</math>. The mean velocity in <math>{}^{km}/{}_h</math> for the total distance is
-A. <math>45</math>
-B. <math>48</math>
-C. <math>50</math>
+<math>\mathrm{(A)}\ 45\qquad\mathrm{(B)}\ 48\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 55\qquad\mathrm{(E)}\ 100</math>
-D. <math>55</math>
-E. <math>100</math>
 [[2007 Cyprus MO/Lyceum/Problem 3|Solution]]
@@ Line 49: / Line 25: @@
 The value of <math>(2*0)*1</math> is
-A. <math>2</math>
+<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 0\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{5}{2}</math>
-B. <math>1</math>
-C. <math>0</math>
-D. <math>\frac{1}{2}</math>
-E. <math>\frac{5}{2}</math>
 [[2007 Cyprus MO/Lyceum/Problem 4|Solution]]
@@ Line 65: / Line 32: @@
 If the remainder of the division of <math>a</math> with <math>35</math> is <math>23</math>, then the remainder of the division of <math>a</math> with <math>7</math> is
-A. <math>1</math>
+<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 5</math>
-B. <math>2</math>
-C. <math>3</math>
-D. <math>4</math>
-E. <math>5</math>
 [[2007 Cyprus MO/Lyceum/Problem 5|Solution]]
 == Problem 6 ==
-<div style="float:right">
+[[Image:2007 CyMO-6.PNG|250px|right]]
-[[Image:2007 CyMO-6.PNG|250px]]
-</div>
 <math>ABCD</math> is a square of side length 2 and <math>FG</math> is an arc of the circle with centre the midpoint <math>K</math> of the side <math>AB</math> and radius 2. The length of the segments <math>FD=GC=x</math> is
-A. <math>\frac{1}{4}</math>
+<math>\mathrm{(A)}\ \frac{1}{4}\qquad\mathrm{(B)}\ \frac{\sqrt{2}}{2}\qquad\mathrm{(C)}\ 2-\sqrt{3}\qquad\mathrm{(D)}\ \sqrt{3}-1\qquad\mathrm{(E)}\ \sqrt{2}-1</math>
-B. <math>\frac{\sqrt{2}}{2}</math>
-C. <math>2-\sqrt{3}</math>
-D. <math>\sqrt{3}-1</math>
-E. <math>\sqrt{2}</math><math>-1</math>
 [[2007 Cyprus MO/Lyceum/Problem 6|Solution]]
@@ Line 99: / Line 48: @@
 If a diagonal <math>d</math> of a rectangle forms a <math>60^\circ</math> angle with one of its sides, then the area of the rectangle is
-A. <math>\frac{d^2 \sqrt{3}}{4}</math>
+<math>\mathrm{(A)}\ \frac{d^2 \sqrt{3}}{4}\qquad\mathrm{(B)}\ \frac{d^2}{2}\qquad\mathrm{(C)}\ 2d^2\qquad\mathrm{(D)}\ d^2 \sqrt{2}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}</math>
-B. <math>\frac{d^2}{2}</math>
-C. <math>2d^2</math>
-D. <math>d^2 \sqrt{2}</math>
-E. None of these
 [[2007 Cyprus MO/Lyceum/Problem 7|Solution]]
@@ Line 114: / Line 55: @@
 If we subtract from 2 the inverse number of <math>x-1</math>, we get the inverse of <math>x-1</math>. Then the number <math>x+1</math> equals to
-A. <math>0</math>
+<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ -1\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ \frac{1}{2}</math>
-B. <math>1</math>
-C. <math>-1</math>
-D. <math>3</math>
-E. <math>\frac{1}{2}</math>
 [[2007 Cyprus MO/Lyceum/Problem 8|Solution]]
@@ Line 129: / Line 62: @@
 We consider the sequence of real numbers <math>a_1,a_2,a_3,...</math> such that <math>a_1=0</math>, <math>a_2=1</math> and <math>a_n=a_{n-1}-a_{n-2}</math>, <math>\forall n \in \{3,4,5,6,...\}</math>. The value of the term <math>a_{138}</math> is
-A. <math>0</math>
+<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ -1\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ 2\qquad\mathrm{(E)}\ -2</math>
-B. <math>-1</math>
-C. <math>1</math>
-D. <math>2</math>
-E. <math>-2</math>
 [[2007 Cyprus MO/Lyceum/Problem 9|Solution]]
@@ Line 144: / Line 69: @@
 The volume of an orthogonal parallelepiped is <math>132\;\mathrm{cm}^3 </math> and its dimensions are integers. The minimum sum of the dimensions is
-A. <math>27\;\mathrm{ cm}</math>
+<math>\mathrm{(A)}\ 27\ \mathrm{cm}\qquad\mathrm{(B)}\ 19\ \mathrm{cm}\qquad\mathrm{(C)}\ 20\ \mathrm{cm}\qquad\mathrm{(D)}\ 18\ \mathrm{cm}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}</math>
-B. <math>19\;\mathrm{ cm}</math>
-C. <math>20\;\mathrm{ cm}</math>
-D. <math>18\;\mathrm{ cm}</math>
-E. None of these
 [[2007 Cyprus MO/Lyceum/Problem 10|Solution]]
@@ Line 159: / Line 76: @@
 If <math>X=\frac{1}{2007 \sqrt{2006}+2006 \sqrt{2007}}</math> and <math>Y=\frac{1}{\sqrt{2006}}-\frac{1}{\sqrt{2007}}</math>, which of the following is correct?
-A. <math>X=2Y</math>
+<math>\mathrm{(A)}\ X=2Y\qquad\mathrm{(B)}\ Y=2X\qquad\mathrm{(C)}\ X=Y\qquad\mathrm{(D)}\ X=Y^2\qquad\mathrm{(E)}\ Y=X^2</math>
-B. <math>Y=2X</math>
-C. <math>X=Y</math>
-D. <math>X=Y^2</math>
-E. <math>Y=X^2</math>
 [[2007 Cyprus MO/Lyceum/Problem 11|Solution]]
@@ Line 174: / Line 83: @@
 The function <math>f : \Re \rightarrow \Re</math> has the properties <math>f(0) = -1</math> and <math>f(xy)+f(x)+f(y)=x+y+xy+k</math> <math>\forall x,y \in \Re</math>, where <math>k \in \Re</math> is a constant. The value of <math>f(-1)</math> is
-A. <math>1</math>
+<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ -1\qquad\mathrm{(C)}\ 0\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 3</math>
-B. <math>-1</math>
-C. <math>0</math>
-D. <math>-2</math>
-E. <math>3</math>
 [[2007 Cyprus MO/Lyceum/Problem 12|Solution]]
@@ Line 189: / Line 90: @@
 If <math>x_1,x_2</math> are the roots of the equation <math>x^2+ax+1=0</math> and <math>x_3,x_4</math> are the roots of the equation <math>x^2+bx+1=0</math>, then the expression <math>	\frac{x_1}{x_2x_3x_4}+\frac{x_2}{x_1x_3x_4}+ \frac{x_3}{x_1x_2x_4}+\frac{x_4}{x_1x_2x_3}</math>equals to
-A. <math>a^2+b^2-2</math>
+<math>\mathrm{(A)}\ a^2+b^2-2\qquad\mathrm{(B)}\ a^2+b^2\qquad\mathrm{(C)}\ \frac{a^2+b^2}{2}\qquad\mathrm{(D)}\ a^2+b^2+1\qquad\mathrm{(E)}\ a^2+b^2-4</math>
-B. <math>a^2+b^2</math>
-C. <math>\frac{a^2+b^2}{2}</math>
-D. <math>a^2+b^2+1</math>
-E. <math>a^2+b^2-4</math>
 [[2007 Cyprus MO/Lyceum/Problem 13|Solution]]
 == Problem 14 ==
-<div style="float:right">
+[[Image:2007 CyMO-14.PNG|250px|right]]
-[[Image:2007 CyMO-14.PNG|250px]]
-</div>
 In the square <math>ABCD</math> the segment <math>KB</math> equals a side of the square. The ratio of areas <math>\frac{S_1}{S_2}</math> is
-A. <math>\frac{1}{3}</math>
+<math>\mathrm{(A)}\ \frac{1}{3}\qquad\mathrm{(B)}\ \frac{2}{3}\qquad\mathrm{(C)}\ \frac{1}{\sqrt{2}}\qquad\mathrm{(D)}\ \sqrt{2}-1\qquad\mathrm{(E)}\ \frac{\sqrt{2}}{4}</math>
-B. <math>\frac{2}{3}</math>
-C. <math>\frac{1}{\sqrt{2}}</math>
-D. <math>\sqrt2 -1</math>
-E. <math>\frac{\sqrt{2}}{4}</math>
 [[2007 Cyprus MO/Lyceum/Problem 14|Solution]]
 == Problem 15 ==
-<div style="float:right">
+[[Image:2007 CyMO-15.PNG|250px|right]]
-[[Image:2007 CyMO-15.PNG|250px]]
-</div>
 The reflex angles of the concave octagon <math>ABCDEFGH</math> measure <math>240^\circ</math> each. Diagonals <math>AE</math> and <math>GC</math> are perpendicular, bisect each other, and are both equal to <math>2</math>.
 The area of the octagon is
-A. <math>\frac{6-2\sqrt{3}}{3}</math>
+<math>\mathrm{(A)}\ \frac{6-2\sqrt{3}}{3}\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ \frac{6+2\sqrt{3}}{3}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}</math>
-B. <math>8</math>
-C. <math>1</math>
-D. <math>\frac{6+2\sqrt{3}}{3}</math>
-E. None of these
 [[2007 Cyprus MO/Lyceum/Problem 15|Solution]]
 == Problem 16 ==
-The full time score of a football match was <math>3</math>-<math>2</math>. how many possible half time results can we have for this match?
+The full time score of a football match was <math>3</math>-<math>2</math>. How many possible half time results could there have been in this match?
-A. <math>5</math>
+<math>\mathrm{(A)}\ 5\qquad\mathrm{(B)}\ 6\qquad\mathrm{(C)}\ 10\qquad\mathrm{(D)}\ 11\qquad\mathrm{(E)}\ 12</math>
-B. <math>6</math>
-C. <math>10</math>
-D. <math>11</math>
-E. <math>12</math>
 [[2007 Cyprus MO/Lyceum/Problem 16|Solution]]
@@ Line 257: / Line 124: @@
 The last digit of the number <math>a=2^{2007}+3^{2007}+5^{2007}+7^{2007}</math> is
-A. <math>0</math>
+<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8</math>
-B. <math>2</math>
-C. <math>4</math>
-D. <math>6</math>
-E. <math>8</math>
 [[2007 Cyprus MO/Lyceum/Problem 17|Solution]]
@@ Line 272: / Line 131: @@
 How many subsets are there for the set <math>A=\{1,2,3,4,5,6,7\}</math>?
-A. <math>7</math>
+<math>\mathrm{(A)}\ 7\qquad\mathrm{(B)}\ 14\qquad\mathrm{(C)}\ 49\qquad\mathrm{(D)}\ 64\qquad\mathrm{(E)}\ 128</math>
-B. <math>14</math>
-C. <math>49</math>
-D. <math>64</math>
-E. <math>128</math>
 [[2007 Cyprus MO/Lyceum/Problem 18|Solution]]
@@ Line 287: / Line 138: @@
 five-digit numbers can be written with the digits <math>1,2,3,4,5</math>. If we place these numbers in increasing order, then the position of the number <math>41253</math> is
-A. <math>71^{st}</math>
+<math>\mathrm{(A)}\ 71^{\mathrm{st}}\qquad\mathrm{(B)}\ 72^{\mathrm{nd}}\qquad\mathrm{(C)}\ 73^{\mathrm{rd}}\qquad\mathrm{(D)}\ 74^{\mathrm{th}}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}</math>
-B. <math>72^{nd}</math>
-C. <math>73^{rd}</math>
-D. <math>74^{th}</math>
-E. None of these
 [[2007 Cyprus MO/Lyceum/Problem 19|Solution]]
@@ Line 302: / Line 145: @@
 The mean value for 9 Math-tests that a student succeded was <math>10</math> (in scale <math>0</math>-<math>20</math>). If we put the grades of these tests in incresing order, then the maximum grade of the <math>5^{th}</math> test is
-A. <math>15</math>
+<math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 19</math>
-B. <math>16</math>
+[[2007 Cyprus MO/Lyceum/Problem 20|Solution]]
-C. <math>17</math>
+== Problem 21 ==
+[[Image:2007 CyMO-21.PNG|250px|right]]
-D. <math>18</math>
+In the following figure, three equal cycles of diameter <math>20\,\mathrm{ cm}</math> represent pulleys, that are connected with a strap. If the distances between any two pulley center points are <math>AB=3\,\mathrm{m}</math>, <math>AC=4\,\mathrm{m}</math> and <math>BC=5\,\mathrm{m}</math>, then the length of the strap is
-E. <math>19</math>
+<math>\mathrm{(A)}\ 12+20\pi)\ \mathrm{m}\qquad\mathrm{(B)}\ (12+\pi)\ \mathrm{m}\qquad\mathrm{(C)}\ (12+4\pi)\ \mathrm{m}\qquad\mathrm{(D)}\ \left(12+\frac{\pi}{5}\right)\ \mathrm{m}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}</math>
-[[2007 Cyprus MO/Lyceum/Problem 20|Solution]]
+[[2007 Cyprus MO/Lyceum/Problem 21|Solution]]
-== Problem 21 ==
+== Problem 22 ==
-<div style="float:right">
+[[Image:2007 CyMO-22.PNG|200px|right]]
-[[Image:2007 CyMO-21.PNG|250px]]
-</div>
-[[2007 Cyprus MO/Lyceum/Problem 21|Solution]]
+In the following figure <math>ABCD</math> is an orthogonal trapezium with <math>\ang A= \ang D=90^\circ</math> and bases <math>AB = a</math> , <math>DC = 2a</math> . If <math>AD = 3a</math> and <math>M</math> is the midpoint of the side <math>BC</math>, then <math>AM</math> equals to
-== Problem 22 ==
+<math>\mathrm{(A)}\ \frac{3a}{2}\qquad\mathrm{(B)}\ \frac{3a}{\sqrt{2}}\qquad\mathrm{(C)}\ \frac{5a}{2}\qquad\mathrm{(D)}\ \frac{3a}{\sqrt{3}}\qquad\mathrm{(E)}\ 2a</math>
-<div style="float:right">
-[[Image:2007 CyMO-22.PNG|250px]]
-</div>
 [[2007 Cyprus MO/Lyceum/Problem 22|Solution]]
 == Problem 23 ==
-<div style="float:right">
+[[Image:2007 CyMO-23.PNG|500px|center]]
-[[Image:2007 CyMO-23.PNG|250px]]
-</div>
+In the figure above the right section of a parabolic tunnel is presented. Its maximum height is <math>OC=8\,\mathrm{m}</math> and its maximum width is <math>AB=20\,\mathrm{m}</math>. If M is the midpoint of <math>OB</math>, then the height <math>MK</math> of the tunnel at the point <math>M</math> is
+<math>\mathrm{(A)}\ 5\ \mathrm{m}\qquad\mathrm{(B)}\ 5.2\ \mathrm{m}\qquad\mathrm{(C)}\ 5.5\ \mathrm{m}\qquad\mathrm{(D)}\ 5.8\ \mathrm{m}\qquad\mathrm{(E)}\ 6\ \mathrm{m}</math>
 [[2007 Cyprus MO/Lyceum/Problem 23|Solution]]
@@ Line 338: / Line 179: @@
 Costas sold two televisions for €198 each. From the sale of the first one he made a profit of 10% on its value and from the sale of the second one, he had a loss of 10% on its value. After the sale of the two televisions Costas had in total
-A. profit €4
+<math>\mathrm{(A)}</math> profit €4
-B. neither profit nor loss
+<math>\mathrm{(B)}\</math> neither profit nor loss
-C. loss €8
+<math>\qquad\mathrm{(C)}</math> loss €8
-D. profit €8
+<math>\qquad\mathrm{(D)}</math> profit €8
-E. loss €4
+<math>\qquad\mathrm{(E)}</math> loss €4
 [[2007 Cyprus MO/Lyceum/Problem 24|Solution]]
 == Problem 25 ==
-<div style="float:right">
+[[Image:2007 CyMO-25.PNG|500px|center]]
-[[Image:2007 CyMO-25.PNG|250px]]
-</div>
+A jeweler makes crosses, according to the pattern shown above. The crosses are made from golden cyclical discs, with diameter of 1cm each. The height of a cross, which is made from 402 such discs is
+<math>\mathrm{(A)}\ 198\ \mathrm{cm}\qquad\mathrm{(B)}\ 2\ \mathrm{m}\qquad\mathrm{(C)}\ 201\ \mathrm{cm}\qquad\mathrm{(D)}\ 202\ \mathrm{cm}\qquad\mathrm{(E)}\ 204\ \mathrm{cm}</math>
 [[2007 Cyprus MO/Lyceum/Problem 25|Solution]]
@@ Line 360: / Line 203: @@
 The number of boys in a school is 3 times the number of girls and the number of girls is 9 times the number of teachers. Let us denote with <math>b</math>, <math>g</math> and <math>t</math>, the number of boys, girls and teachers respectively. Then the total number of boys, girls and teachers equals to
-A. <math>31b</math>
+<math>\mathrm{(A)}\ 31b\qquad\mathrm{(B)}\ \frac{37b}{27}\qquad\mathrm{(C)}\ 13g\qquad\mathrm{(D)}\ \frac{37g}{27}\qquad\mathrm{(E)}\ \frac{37t}{27}</math>
-B. <math>\frac{37b}{27}</math>
+[[2007 Cyprus MO/Lyceum/Problem 26|Solution]]
-C. <math>13g</math>
+== Problem 27 ==
+[[Image:2007 CyMO-27.PNG|380px|right]]
-D. <math>\frac{37g}{27}</math>
-E. <math>\frac{37t}{27}</math>
+In the following diagram, the light beam <math>\epsilon</math> is reflected on the <math>x</math>-axis and the beam <math>d</math>, being reflected on a mirror parallel to the <math>y</math>-axis at distance 6, intersects the <math>y</math>-axis at point <math>B</math>. <br>
+The equation of line <math>f</math> is given by
-[[2007 Cyprus MO/Lyceum/Problem 26|Solution]]
+<math>\mathrm{(A)}\ x+y-11=0\qquad\mathrm{(B)}\ x+y+11=0\qquad\mathrm{(C)}\ x-y+11=0\qquad\mathrm{(D)}\ x-y-11=0\qquad\mathrm{(E)}\ y=-x+10</math>
-== Problem 27 ==
-<div style="float:right">
-[[Image:2007 CyMO-27.PNG|250px]]
-</div>
 [[2007 Cyprus MO/Lyceum/Problem 27|Solution]]
@@ Line 382: / Line 220: @@
 The product of <math>15^8\cdot28^6\cdot5^{11}</math> is an integer number whose last digits are zeros. How many zeros are there?
-A. <math>6</math>
+<math>\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 12\qquad\mathrm{(E)}\ 19</math>
-B. <math>8</math>
-C. <math>11</math>
-D. <math>12</math>
-E. <math>19</math>
 [[2007 Cyprus MO/Lyceum/Problem 28|Solution]]
@@ Line 397: / Line 227: @@
 The minimum value of a positive integer <math>k</math>, for which the sum <math>S=k+(k+1)+(k+2)+\ldots+(k+10)</math> is a perfect square, is
-A. <math>5</math>
+<math>\mathrm{(A)}\ 5\qquad\mathrm{(B)}\ 6\qquad\mathrm{(C)}\ 10\qquad\mathrm{(D)}\ 11\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}</math>
-B. <math>6</math>
+[[2007 Cyprus MO/Lyceum/Problem 29|Solution]]
-C. <math>10</math>
+== Problem 30 ==
+[[Image:2007 CyMO-30.PNG|200px|right]]
-D. <math>11</math>
+A coin with a shape of a regular hexagon of side 1 is tangent to a square of side 6, as shown in the figure.
-E. None of these
+The coin rotates on the perimeter of the square, until it reaches its original position.
-[[2007 Cyprus MO/Lyceum/Problem 29|Solution]]
+The length of the line which is being inscribed by the center of the hexagon is
-== Problem 30 ==
+<math>\mathrm{(A)}\ \frac{34\pi}{3}\qquad\mathrm{(B)}\ 24\qquad\mathrm{(C)}\ \frac{28\pi}{3}\qquad\mathrm{(D)}\ 6 \pi\sqrt{2}\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}</math>
-<div style="float:right">
-[[Image:2007 CyMO-30.PNG|250px]]
-</div>
 [[2007 Cyprus MO/Lyceum/Problem 30|Solution]]
 == See also ==
-* [[ 2007 Cyprus MO/Lyceum]]
+* [[2007 Cyprus MO/Lyceum]]
 * [[Cyprus Mathematical Olympiad]]
 * [[Cyprus MO Problems and Solutions]]
 * [[Mathematics competition resources]]