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

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== Problem 1 ==
 
== Problem 1 ==
−
If <math>x-y=1</math>,then the value of the expression <math>K=x^2+x-2xy+y^2-y</math> is
+
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>
+
[[2007 Cyprus MO/Lyceum/Problem 1|Solution]]
 +
 
 +
== Problem 2 ==
 +
Given the formula <math>f(x) = 4^x</math>, then <math>f(x+1)-f(x)</math> equals to
 +
 
 +
<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>
 +
 
 +
[[2007 Cyprus MO/Lyceum/Problem 2|Solution]]
 +
 
 +
== Problem 3 ==
 +
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
 +
 
 +
<math>\mathrm{(A)}\ 45\qquad\mathrm{(B)}\ 48\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 55\qquad\mathrm{(E)}\ 100</math>
 +
 
 +
[[2007 Cyprus MO/Lyceum/Problem 3|Solution]]
 +
 
 +
== Problem 4 ==
 +
We define the operation <math>a*b = \frac{1+a}{1+b^2}</math>, <math>\forall a,b \in \real</math>.
 +
 
 +
The value of <math>(2*0)*1</math> is
 +
 
 +
<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>
 +
 
 +
[[2007 Cyprus MO/Lyceum/Problem 4|Solution]]
 +
 
 +
== Problem 5 ==
 +
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
 +
 
 +
<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 5</math>
 +
 
 +
[[2007 Cyprus MO/Lyceum/Problem 5|Solution]]
 +
 
 +
== Problem 6 ==
 +
[[Image:2007 CyMO-6.PNG|250px|right]]
 +
 
 +
<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
 +
 
 +
<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>
 +
 
 +
[[2007 Cyprus MO/Lyceum/Problem 6|Solution]]
 +
 
 +
== Problem 7 ==
 +
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
 +
 
 +
<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>
 +
 
 +
[[2007 Cyprus MO/Lyceum/Problem 7|Solution]]
 +
 
 +
== Problem 8 ==
 +
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
 +
 
 +
<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ -1\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ \frac{1}{2}</math>
 +
 
 +
[[2007 Cyprus MO/Lyceum/Problem 8|Solution]]
 +
 
 +
== Problem 9 ==
 +
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
  
−
C.  <math>1</math>
+
<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ -1\qquad\mathrm{(C)}\ 1\qquad\mathrm{(D)}\ 2\qquad\mathrm{(E)}\ -2</math>
  
−
D. <math>-1</math>
+
[[2007 Cyprus MO/Lyceum/Problem 9|Solution]]
  
−
E. <math>0</math>
+
== Problem 10 ==
 +
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
  
−
[[2007 Cyprus MO/Lyceum/Problem 1|Solution]]
+
<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>
 +
 
 +
[[2007 Cyprus MO/Lyceum/Problem 10|Solution]]
 +
 
 +
== Problem 11 ==
 +
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?
 +
 
 +
<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>
  
−
== Problem 2 ==
+
[[2007 Cyprus MO/Lyceum/Problem 11|Solution]]
−
Given the formula <math>f(x) = 4^x</math>, then <math>f(x+1)-f(x)</math> equals to
 
  
−
A. <math>4</math>
+
== Problem 12 ==
 +
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
  
−
B. <math>4^x</math>
+
<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ -1\qquad\mathrm{(C)}\ 0\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 3</math>
  
−
C. <math>2\cdot4^x</math>
+
[[2007 Cyprus MO/Lyceum/Problem 12|Solution]]
  
−
D. <math>4^{x+1}</math>
+
== Problem 13 ==
 +
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
  
−
E. <math>3\cdot4^x</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>
  
−
[[2007 Cyprus MO/Lyceum/Problem 2|Solution]]
+
[[2007 Cyprus MO/Lyceum/Problem 13|Solution]]
  
−
== Problem 3 ==
+
== Problem 14 ==
−
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
+
[[Image:2007 CyMO-14.PNG|250px|right]]
  
−
A. <math>45</math>
+
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
  
−
B. <math>48</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>
  
−
C. <math>50</math>
+
[[2007 Cyprus MO/Lyceum/Problem 14|Solution]]
  
−
D. <math>55</math>
+
== Problem 15 ==
 +
[[Image:2007 CyMO-15.PNG|250px|right]]
  
−
E. <math>100</math>
+
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>.
  
−
[[2007 Cyprus MO/Lyceum/Problem 3|Solution]]
+
The area of the octagon is
  
−
== Problem 4 ==
+
<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>
−
We define the operation <math>a*b = \frac{1+a}{1+b^2}</math>, <math>\forall a,b \in \real</math>.
 
  
−
The value of <math>(2*0)*1</math> is
+
[[2007 Cyprus MO/Lyceum/Problem 15|Solution]]
  
−
A. <math>2</math>
+
== Problem 16 ==
 +
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?
  
−
B. <math>1</math>
+
<math>\mathrm{(A)}\ 5\qquad\mathrm{(B)}\ 6\qquad\mathrm{(C)}\ 10\qquad\mathrm{(D)}\ 11\qquad\mathrm{(E)}\ 12</math>
  
−
C. <math>0</math>
+
[[2007 Cyprus MO/Lyceum/Problem 16|Solution]]
  
−
D. <math>\frac{1}{2}</math>
+
== Problem 17 ==
 +
The last digit of the number <math>a=2^{2007}+3^{2007}+5^{2007}+7^{2007}</math> is
  
−
E. <math>\frac{5}{2}</math>
+
<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8</math>
  
 +
[[2007 Cyprus MO/Lyceum/Problem 17|Solution]]
  
−
[[2007 Cyprus MO/Lyceum/Problem 4|Solution]]
+
== Problem 18 ==
 +
How many subsets are there for the set <math>A=\{1,2,3,4,5,6,7\}</math>?
  
−
== Problem 5 ==
+
<math>\mathrm{(A)}\ 7\qquad\mathrm{(B)}\ 14\qquad\mathrm{(C)}\ 49\qquad\mathrm{(D)}\ 64\qquad\mathrm{(E)}\ 128</math>
−
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>
+
[[2007 Cyprus MO/Lyceum/Problem 18|Solution]]
  
−
B. <math>2</math>
+
== Problem 19 ==
 +
120 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
  
−
C. <math>3</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>
  
−
D. <math>4</math>
+
[[2007 Cyprus MO/Lyceum/Problem 19|Solution]]
  
−
E. <math>5</math>
+
== Problem 20 ==
 +
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
  
−
[[2007 Cyprus MO/Lyceum/Problem 5|Solution]]
+
<math>\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 16\qquad\mathrm{(C)}\ 17\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 19</math>
  
−
== Problem 6 ==
+
[[2007 Cyprus MO/Lyceum/Problem 20|Solution]]
−
{{image}}
 
−
<math>ABCD</math> is a square of side 2 and <math>FG</math> is an arc of the circle with centre the midpoint <math>K</math> os 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>
+
== Problem 21 ==
 +
[[Image:2007 CyMO-21.PNG|250px|right]]
  
−
B. <math>\frac{\sqrt{2}}{2}</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
  
−
C. <math>2-\sqrt{3}</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>
  
−
D. <math>\sqrt{3}-1</math>
+
[[2007 Cyprus MO/Lyceum/Problem 21|Solution]]
  
−
E. <math>\sqrt{2}</math><math>-1</math>
+
== Problem 22 ==
 +
[[Image:2007 CyMO-22.PNG|200px|right]]
  
 +
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
  
−
[[2007 Cyprus MO/Lyceum/Problem 6|Solution]]
+
<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>
  
−
== Problem 7 ==
+
[[2007 Cyprus MO/Lyceum/Problem 22|Solution]]
−
If the angle of the diagonal d of a rectangle forms an angle <math>60^\circ</math> with one of its sides, then the area of the recangle is
 
  
−
A. <math>\frac{d^2 \sqrt{3}}{4}</math>
+
== Problem 23 ==
 +
[[Image:2007 CyMO-23.PNG|500px|center]]
  
−
B. <math>\frac{d^2}{2}</math>
+
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
  
−
C. <math>2d^2</math>
+
<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>
  
−
D. <math>d^2 \sqrt{2}</math>
+
[[2007 Cyprus MO/Lyceum/Problem 23|Solution]]
  
−
E. None of these
+
== 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
  
−
[[2007 Cyprus MO/Lyceum/Problem 7|Solution]]
+
<math>\mathrm{(A)}</math> profit €4
  
−
== Problem 8 ==
+
<math>\mathrm{(B)}\</math> neither profit nor loss
−
If we substract 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>\qquad\mathrm{(C)}</math> loss €8
  
−
B. <math>1</math>
+
<math>\qquad\mathrm{(D)}</math> profit €8
  
−
C. <math>-1</math>
+
<math>\qquad\mathrm{(E)}</math> loss €4
  
−
D. <math>3</math>
+
[[2007 Cyprus MO/Lyceum/Problem 24|Solution]]
  
−
E. <math>\frac{1}{2}</math>
+
== Problem 25 ==
 +
[[Image:2007 CyMO-25.PNG|500px|center]]
  
−
[[2007 Cyprus MO/Lyceum/Problem 8|Solution]]
+
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
  
−
== Problem 9 ==
+
<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>
−
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. 0
+
[[2007 Cyprus MO/Lyceum/Problem 25|Solution]]
  
−
B. -1
+
== 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 <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
  
−
C. 1
+
<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>
  
−
D. 2
+
[[2007 Cyprus MO/Lyceum/Problem 26|Solution]]
  
−
E. -2
+
== Problem 27 ==
 +
[[Image:2007 CyMO-27.PNG|380px|right]]
  
−
[[2007 Cyprus MO/Lyceum/Problem 9|Solution]]
+
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
  
−
== Problem 10 ==
+
<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>
  
−
[[2007 Cyprus MO/Lyceum/Problem 10|Solution]]
+
[[2007 Cyprus MO/Lyceum/Problem 27|Solution]]
  
−
== Problem 11 ==
+
== Problem 28 ==
 +
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?
  
 +
<math>\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 11\qquad\mathrm{(D)}\ 12\qquad\mathrm{(E)}\ 19</math>
  
−
[[2007 Cyprus MO/Lyceum/Problem 11|Solution]]
+
[[2007 Cyprus MO/Lyceum/Problem 28|Solution]]
  
−
== Problem 12 ==
+
== Problem 29 ==
 +
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
  
 +
<math>\mathrm{(A)}\ 5\qquad\mathrm{(B)}\ 6\qquad\mathrm{(C)}\ 10\qquad\mathrm{(D)}\ 11\qquad\mathrm{(E)}\ \mathrm{None\ of\ these}</math>
  
−
[[2007 Cyprus MO/Lyceum/Problem 12|Solution]]
+
[[2007 Cyprus MO/Lyceum/Problem 29|Solution]]
  
−
== Problem 13 ==
+
== Problem 30 ==
 +
[[Image:2007 CyMO-30.PNG|200px|right]]
  
−
[[2007 Cyprus MO/Lyceum/Problem 13|Solution]]
+
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.
  
−
== Problem 14 ==
+
The coin rotates on the perimeter of the square, until it reaches its original position.
  
−
[[2007 Cyprus MO/Lyceum/Problem 14|Solution]]
+
The length of the line which is being inscribed by the center of the hexagon is
  
−
== Problem 15 ==
+
<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>
  
−
[[2007 Cyprus MO/Lyceum/Problem 15|Solution]]
+
[[2007 Cyprus MO/Lyceum/Problem 30|Solution]]
  
 
== See also ==
 
== See also ==
−
* [[ 2007 Cyprus MO/Lyceum]]
+
* [[2007 Cyprus MO/Lyceum]]
 
* [[Cyprus Mathematical Olympiad]]
 
* [[Cyprus Mathematical Olympiad]]
 
* [[Cyprus MO Problems and Solutions]]
 
* [[Cyprus MO Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]

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

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

Solution

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

2007 CyMO-6.PNG

$ABCD$ 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

2007 CyMO-14.PNG

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

2007 CyMO-15.PNG

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

2007 CyMO-21.PNG

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

2007 CyMO-22.PNG

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

2007 CyMO-23.PNG

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

2007 CyMO-25.PNG

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

2007 CyMO-27.PNG

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

2007 CyMO-30.PNG

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

See also

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