CEMC Banner

2015 Galois Contest
Solutions
(Grade 10)

Thursday, April 16, 2015
(in North America and South America)

Friday, April 17, 2015
(outside of North American and South America)

©2015 University of Waterloo


    1. The x-intercept of a line occurs at the point whose y-coordinate is 0.

      Substituting y=0 into the equation of line 1, we get 0=2x+6 or 2x=6 and so x=3.

      Line 1 has x-intercept 3. (Point P has coordinates (3,0).)

    2. Solution 1

      The equation of line 2 with slope 3 and y-intercept b is y=3x+b.

      Line 2 passes through Q(3,12) and so x=3 and y=12 satisfies the equation of line 2.

      Substituting, we get 12=3(3)+b or 12=9+b and so b=21.

      The equation of line 2 is y=3x+21.

      Solution 2

      A line with slope m and passing through the point (x1,y1) has equation yy1=m(xx1).

      Since line 2 has slope m=3 and passes through Q(3,12), the slope-point equation of line 2 is y12=3(x3).

    3. We find the coordinates of point R by substituting y=0 into the equation of line 2.

      This gives 0=3x+21 or 3x=21 and so x=7.

      That is, line 2 has x-intercept 7 (point R has coordinates (7,0)).

      If we let the base of PQR be side PR, then the height of the triangle is the vertical distance from Q to PR.

      Since Q has y-coordinate 12, then this height is 12.

      The x-coordinate of P is 3 and the x-coordinate of R is 7. (Both y-coordinates are 0.)

      Therefore, PR has length 7(3)=10.

      Finally, the area of PQR is 12×10×12=60.

    1. The total number of students at School A is the sum of the number of students who received a ride and the number of students who did not, or 330+420=750.

      Since 330 of 750 students received a ride, the percentage of students at School A who received a ride is 330750×100%=0.44×100%=44%.

    2. Solution 1

      At school B, 30% of 240 students or 30100×240=7200100=72 students received a ride.

      If 50% or 12 of the students in School B were to receive a ride, then 12×240=120 students would get a ride.

      Therefore, 12072=48 more students needed to receive a ride so that 50% of the students in School B got a ride.

      Solution 2

      As a percent, the difference between 50% of students receiving a ride and the 30% of students who did receive a ride is 20%.

      Therefore, 20% of 240 students or 20100×240=4800100=48 additional students would need to receive a ride so that 50% of students in School B got a ride.

    3. Solution 1

      At school C, 45% of 200 students or 45100×200=9000100=90 students received a ride.

      At school D, x% of 300 students or x100×300=300x100=3x students received a ride.

      The total number of students at School C and School D is 200+300=500.

      The total number of students receiving a ride at School C and School D is 90+3x.

      Since 57.6% of the combined group of students from the two schools received a ride, then 90+3x500=57.6100.

      Multiplying both sides of this equation by 500, we get 90+3x=57.6×5 or 90+3x=288 or 3x=198 and so x=66.

      Solution 2

      At school C, 45% of 200 students or 45100×200=9000100=90 students received a ride.

      The total number of students at School C and School D is 200+300=500.

      Since 57.6% of the combined group of students from the two schools received a ride, then 57.6100×500=28800100=288 students received a ride.

      Out of the 288 students who received a ride, 90 students were from School C and so the remaining 28890=198 students were from School D.

      Since there are 300 students at School D, the percentage of students receiving a ride is 198300×100%=0.66×100%=66%.

      Therefore, the value of x is 66.

    4. At school E, n% of 200 students or n100×200=200n100=2n students received a ride.

      At school F, 2n% of 250 students or 2n100×250=500n100=5n students received a ride.

      The total number of students at School E and School F is 200+250=450.

      The total number of students receiving a ride at School E and School F is 2n+5n=7n.

      Between 55% and 60% of the 450 students from the two schools received a ride.

      Since 55% of 450 is 247.5 and 60% of 450 is 270, then 7n>247.5 and 7n<270.

      Solving 7n>247.5 we get n>35.35, after rounding, and 7n<270 gives n<38.57, after rounding.

      Since n is a positive integer and n>35.35 and n<38.57, then the possible values of n are 36,37 and 38.

    1. Since 5 is an odd integer, then n must be an odd integer for the sum n+5 to be an even integer.

      (If n was an even integer, then n+5 would be the sum of an even integer and an odd integer, which is an odd integer.)

    2. We first note that the product of an even integer and any other integers, even or odd, is always an even integer.

      Let N=cd(c+d).

      If c or d is an even integer (or both c and d are even integers), then N is the product of an even integer and some other integers and thus is even.

      The only remaining possibility is that both c and d are odd integers.

      If c and d are odd integers, then the sum c+d is an even integer.

      In this case, N is again the product of an even integer and some other integers and so it is an even integer.

      Therefore, for any integers c and d, cd(c+d) is always an even integer.

    3. Since e and f are positive integers so that ef=300, then we may begin by determining the factor pairs of positive integers whose product is 300.

      Written as ordered pairs (x,y) with x<y, these are: (1,300),(2,150),(3,100),(4,75),(5,60),(6,50),(10,30),(12,25),(15,20). It is also required that the sum e+f be odd and so exactly one of e or f must be odd.

      Therefore, the factor pairs whose sum is odd are: (1,300),(3,100),(4,75),(5,60),(12,25),(15,20). There are 6 ordered pairs (e,f) satisfying the given conditions.

    4. Since both m and n are positive integers, then 2n>1 and so 2n+m>m+1.

      Let a=m+1 and b=2n+m or a=2n+m and b=m+1 so that ab=9000.

      We must first determine all factor pairs (a,b) of positive integers whose product is 9000.

      We begin by considering the parity (whether each is even or odd) of the factors a and b.

      Since 2 is even, then 2n is even for all positive integers n.

      If m is even then 2n+m is even since the sum of two even integers is even.

      However if m is even, then m+1 is odd since the sum of an even integer and an odd integer is odd.

      That is, if m is even, then a is odd and b is even or a is even and b is odd.

      We say that the factors a and b have different parity since one is even and one is odd.

      If m is odd then 2n+m is odd. If m is odd then m+1 is even.

      That is, if m is odd, then a is even and b is odd or a is odd and b is even and so the factors a and b have different parity for all possible values of m.

      Now we are searching for all factor pairs (a,b) of positive integers whose product is 9000 with a and b having different parity.

      Written as a product of its prime factors, 9000=23×32×53 and so ab=23×32×53.

      Since exactly one of a or b is odd, then one of them does not have a factor of 2 and so the other must have all factors of 2.

      That is, either a=23r=8r and b=s, or a=r and b=8s for positive integers r and s.

      In both cases, ab=8rs=9000 and so rs=90008=1125=3253.

      We now determine all factor pairs (r,s) of positive integers whose product is 1125.

      These are (r,s)=(1,1125),(3,375),(5,225),(9,125),(15,75),(25,45).

      Therefore (a,b)=(8r,s)=(8,1125),(24,375),(40,225),(72,125),(120,75),(200,45), or (a,b)=(r,8s)=(1,9000),(3,3000),(5,1800),(9,1000),(15,600),(25,360).

      Since 2n+m>m+1>1, then the pair (1,9000) is not possible.

      This leaves 11 factor pairs (a,b) such that ab=9000 with a and b having different parity.

      Each of these 11 factor pairs (a,b) gives an ordered pair (m,n).

      To see this, let m+1 equal the smaller of a and b, and let 2n+m equal the larger (since 2n+m>m+1).

      For example when (a,b)=(8,1125), then m+1=8 or m=7 and so 2n+m=2n+7=1125 or 2n=1118 or n=559.

      That is, the factor pair (a,b)=(8,1125) corresponds to the ordered pair (m,n)=(7,559) so that (m+1)(2n+m)=9000.

      Each of the 11 pairs (a,b) gives an ordered pair (m,n) such that (m+1)(2n+m)=9000.

      We determine the corresponding ordered pair (m,n) for each (a,b) in the table below (although this work is not necessary since we were only asked for the number of ordered pairs).

      (a,b) m+1 2n+m (m,n)
      (8,1125) 8 1125 (7,559)
      (24,375) 24 375 (23,176)
      (40,225) 40 225 (39,93)
      (72,125) 72 125 (71,27)
      (120,75) 75 120 (74,23)
      (200,45) 45 200 (44,78)
      (3,3000) 3 3000 (2,1499)
      (5,1800) 5 1800 (4,898)
      (9,1000) 9 1000 (8,496)
      (15,600) 15 600 (14,293)
      (25,360) 25 360 (24,168)

      There are 11 ordered pairs (m,n) of positive integers satisfying (m+1)(2n+m)=9000.

    1. Since EXD is a straight line, then YXE+YXZ=180.

      Since XYZ is an equilateral triangle, then YXZ=60.

      Thus, YXE=18060=120.

    2. We make a table to track the position of the square as it rotates about the triangle:

      After
      Move #
      Coinciding
      Vertices
      Second Vertex of XYZ
      on Side of Square
      Centre of Rotation
      of Next Move
      Angle of Rotation
      of Next Move
      Initial D and Z X on DE X 120
      1 E and Y X on DE Y 30
      2 E and Y Z on EB Z 120
      3 B and X Z on EB X 30
      4 B and X Y on BC Y 120
      5 C and Z Y on BC Z 30
      6 C and Z X on CD X 120
      7 D and Y X on CD Y 30
      8 D and Y Z on DE Z 120

      The information in this table comes from the diagrams here:

      Square BCDE goes through eight moves, and the diagram
shows its state after each move. A description of the diagram is given in the following list.

      Each angle of rotation is either 18060=120 or 9060=30.

      Therefore, D next coincides with a vertex of the triangle after 7 moves.

      (We continued the table through the 8th move to be more useful in part (c).)

    3. After 0 moves, the square has D at Z and X on DE and the next rotation is a rotation of 120 about X.

      After 8 moves, the square has D at Y and Z on DE and the next rotation is a rotation of 120 about Z.

      Therefore, through 8 moves, the net change of position of the square was moving clockwise around 2 sides of the triangle. After these 8 moves, the square is in a similar position, relative to the triangle, to its initial position: D is at a vertex of the triangle and DE lies along a side of the triangle.

      Starting from after the 8th move, the square starts in this similar position, and so 8 more moves will take the square again to a similar position. The net change will again be that the square has moved clockwise around 2 sides of the triangle. (We note from part (b) that after 8 moves is the first time (after the original position) that vertex D coincides with a vertex of the triangle at the same relative position.)

      Thus, after 16 moves, the square will have D at X and Y on DE and the next rotation is a rotation of 120 about Y. (This is the second time that vertex D coincides with a vertex of the triangle at the same relative position.)

      Using similar reasoning, after 24 moves, the square will have D at Z and X on DE and the next rotation is a rotation of 120 about X, which is the original position.

      Therefore, we need to determine the total distance travelled by E through these 24 moves.

      This total distance equals 3 times the distance travelled by E through the first 8 moves.

      This is because the relative position of the square with respect to the triangle (D at a vertex of the triangle, DE along a side of the triangle) is the same after 8 moves as it was after 0 moves, so the relative sequence of rotations undergone by E from after move 8 to after move 16, and from after move 16 to after move 24, will be the same as they were for the first 8 moves.

      Since EBCD has side length 2, then ED=EB=2.

      Also, the distance from E to the midpoint of ED is 1, as is the distance from E to the midpoint of EB.

      Since EBCD has side length 2, then EC=22.

      Also, the distance from E to the midpoint of each of DC and BC is 22+12=5, by the Pythagorean Theorem.

      We make a chart of the rotation undergone by E through each of the first 8 moves:

      Move
      #
      Centre of
      Rotation
      Distance of
      Centre from E
      Angle of Rotation
      of Square
      1 X 1 120
      2 Y 0 30
      3 Z 1 120
      4 X 2 30
      5 Y 5 120
      6 Z 22 30
      7 X 5 120
      8 Y 2 30

      During each rotation, each point on the square (except for the point of the square in contact with the centre of rotation) is rotated about the centre of rotation through the angle of rotation of the square.

      Thus, during each rotation, the distance travelled by E is the fraction of the circumference of the whole circle given by the angle of rotation as compared to 360.

      Therefore, the distance travelled by E through the first 8 moves is 1203602π(1)+303602π(0)+1203602π(1)+303602π(2)+ 1203602π(5)+303602π(22)+1203602π(5)+303602π(2) which simplifies to 23π+0+23π+13π+235π+132π+235π+13π or 2π+435π+132π Therefore, when the square returns to its initial position after 24 moves, E has travelled 3 times this far, or a total distance of 6π+45π+2π.