Canadian Senior Mathematics Contest 2017 Answers

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• The CENTRE for EDUCATION

  in MATHEMATICS and COMPUTING cemc.uwaterloo.ca

  2018 Canadian Intermediate

  Mathematics Contest

  Wednesday, November 21, 2018 (in North America and South America)

  Thursday, November 22, 2018 (outside of North America and South America)

  Solutions

  ©2018 University of Waterloo

• 2018 Canadian Intermediate Mathematics Contest Solutions Page 2

  Part A

  1. Since the angles in any triangle add to 180◦, then

  60◦ + (5x)◦ + (3x)◦ = 180◦

  60 + 5x + 3x = 180

  8x = 120

  and so x = 15.

  Answer: x = 15

  2. Since 10% of the 500 animals are chickens, then 1 10 × 500 = 50 of the animals are chickens.

  The remaining 500− 50 = 450 animals are goats and cows. Since there are twice as many goats as cows, then the total number of goats and cows is three times the number of cows. Therefore, there are 1

  3 × 450 = 150 cows.

  (Since there are 150 cows, then there are 2×150 = 300 goats. Combining with the 50 chickens, we have 150 + 300 + 50 = 500 animals, as expected.)

  Answer: 150 cows

  3. Since 4ADB and 4CBD are right-angled, we can apply the Pythagorean Theorem in each triangle.

  A

  B C

  D

  52

  48

  21

  Therefore, AB2 = AD2+BD2 or 522 = 482+BD2 and so BD2 = 522−482 = 2704−2304 = 400. Since BD > 0, then BD =

  √ 400 = 20.

  Furthermore, DC2 = BD2 + BC2 = 202 + 212 = 400 + 441 = 841. Since DC > 0, then DC =

  √ 841 = 29.

  Answer: DC = 29

  4. Let x be the number in the bottom right corner of the 8× 8 square. Since there are 8 numbers across the bottom row of this square, then the number in the bottom left corner is x − 7. (The 8 numbers in the bottom row are x − 7, x − 6, x − 5, x − 4, x − 3, x− 2, x− 1, and x.) In the larger grid, each number is 24 greater than the number directly above it. The number in the bottom right corner of the 8× 8 square is seven rows beneath the number in the upper right corner. Since there are 24 numbers in each row, each number in the grid above the bottom row is 24 less than the number directly below it. This means the number in the top right corner is 7 × 24 = 168 less than the number in the

• 2018 Canadian Intermediate Mathematics Contest Solutions Page 3

  bottom right corner, as it is 7 rows above it. Since the number in the bottom right corner is x, then the number in the top right corner is x− 168. Moving to the left across the first row of the 8× 8 square, the number in the top left corner is (x− 168)− 7 or x− 175. Since the sum of the numbers in the four corners of the square is to be 1646, then

  x + (x− 7) + (x− 168) + (x− 175) = 1646 4x− 350 = 1646

  4x = 1996

  x = 499

  Therefore, the number in the bottom right corner is 499. We should also verify that, starting with 499 in the bottom right corner, we can construct an 8× 8 square that doesn't extend past the top or left border of the grid. Since the bottom right number in the grid is 576, then the numbers up the right side of the grid are 576, 552, 528, 504, which means that, moving from right to left, 499 is sixth number of the 4th row from the bottom. Since the grid is 24× 24, then the square does not meet the left or top borders.

  Answer: 499

  5. Consider the point R(4, 1). Here, PR is horizontal and RQ is vertical and so 4PQR is right-angled at R. Because PR = 4− 1 = 3 and RQ = 5− 1 = 4, then the area of 4PQR is 1

  2 ×PR×RQ which

  equals 1 2 × 3× 4 or 6.

  Therefore, the point R(4, 1) has the property that 4PQR has area 6. If a point X lies on the line through R parallel to PQ, then the perpendicular distance from X to the line through PQ is the same as the perpendicular distance from R to PQ. This means that the height of 4PQX equals the height of 4PQR and so the areas of these triangles are equal.

  y

  xO R (4, 1)P (1, 1)

  Q (4, 5)

  In other words, any point X on the line through R parallel to PQ has the property that the area of 4PQX is 6. To get from P to Q, we move 3 units to the right and 4 units up. Therefore, starting at R(4, 1) and moving to the points (4+3, 1+4) = (7, 5) will get to another point on the line through R(4, 1) parallel to PQ (since moving right 3 and up 4 preserves the

• 2018 Canadian Intermediate Mathematics Contest Solutions Page 4

  slope of 4

  3 .).

  Similarly, the point (7 + 3, 5 + 4) = (10, 9) is on this line. In other words, the triangle with vertices at (1, 1), (4, 5) and (7, 5) and the triangle with vertices at (1, 1), (4, 5) and (10, 9) both have areas of 6. (To do this more formally, we could determine that the equation of the line through R parallel

  to PQ is y = 4

  3 x− 13

  3 and then check the possible integer x-coordinates of points on this line

  (between 0 and 10, inclusive) to get the points (1,−3), (4, 1), (7, 5), (10, 9), omitting (1,−3) since −3 is not in the correct range). Next, consider S(1, 5). Again, the area of 4PQS is 6 because PS is vertical (with length 4) and QS is horizontal (with length 3). Moving right 3 and up 4 from (1, 5) gives the point (4, 9).

  Since we are told that there are five such points, then we have found all of the points and they are (4, 1), (7, 5), (10, 9), (1, 5), and (4, 9). (We could note that any point T for which the area of 4PQT is 6 must lie on one of these two lines. This is because T must have a particular perpendicular distance to the line through PQ and must be on one side of this line or the other. Because these two lines are fixed, there are indeed only five points with these properties.)

  Answer: (4, 1), (7, 5), (10, 9), (1, 5), and (4, 9)

  6. We know that 1 ≤ n ≤ 20 and 1 ≤ k ≤ 20. We focus on the possible values of k. Consider k = 20. Since there are 20 chairs, then moving 20 chairs around the circle moves each person back to their original seat. Therefore, any configuration of people sitting in chairs is preserved by this movement. In other words, n can take any value from 1 to 20, inclusive, when k = 20. Thus, there are 20 pairs (n, k) that work when k = 20.

  Consider k = 10. Here, any people in chairs 1 to 10 move to chairs 11 to 20 and any people in chairs 11 to 20 move to chairs 1 to 10. This means that the two halves (1 to 10 and 11 to 20) must contain the same number of people in the same configuration. (That is, if there are people in chairs 1, 3, 4, 8, then there are people in chairs 11, 13, 14, 18, and vice versa.) This means that the total number of people in chairs is even. There are 10 possibilities for the number of chairs occupied among the first 10 chairs (1 to 10), and so 10 possibilities for the total number of chairs occupied (the even numbers from 2 to 20, inclusive). Thus, there are 10 pairs (n, k) that work when k = 10.

  Consider k = 5. Here, any people in chairs 1 to 5 move to chairs 6 to 10, any people in chairs 6 to 10 move to chairs 11 to 15, any people in chairs 11 to 15 move to chairs 16 to 20, and any people in chairs 16 to 20 move to chairs 1 to 5. This means that the four sections containing 5 chairs must contain the same number of people in the same configuration.

• 2018 Canadian Intermediate Mathematics Contest Solutions Page 5

  This means that the total number of people in chairs is a multiple of 4 and can be 4, 8, 12, 16, or 20, which gives 5 possible values of n. Thus, there are 5 pairs (n, k) that work when k = 5.

  If we consider k = 4 and k = 2, in a similar way, we can determine that there are 4 and 2 possible values for n, respectively. Thus, there are 4 pairs (n, k) that work when k = 4 (namely, (5, 4), (10, 4), (15, 4), (20, 4)) and 2 pairs (n, k) that work when k = 2 (namely (10, 2) and (20, 2)).

  Consider k = 1. Here, each person in a chair moves 1 chair along the circle. If there is a person in chair 1, then they move to chair 2, which means that there must have been a person in chair 2. This person has moved to chair 3, which means that there must have been a person in chair 3, and so on. Continuing, we see that all 20 chairs must be full, and so n = 20 is the only possibility. Thus, there is 1 pair (n, k) that works when k = 1.

  Consider k = 3, k = 7 and k = 9. In each of these cases, n = 20 is the only possibility. If k = 3, following a similar argument to that for k = 1, the person in chair 1 moves to chair 4, the person in chair 4 moves to chair 7, and so on:

  1→ 4→ 7→ 10→ 13→ 16→ 19→ 2→ 5→ 8→ 11→ 14

  → 17→ 20→ 3→ 6→ 9→ 12→ 15→ 18→ 1 The cycle only concludes when all 20 chairs are included, and so if 1 chair is occupied, then all chairs are occupied. A similar cycle can be constructed for k = 7 and k = 9 and so in each case only n = 20 works.

  Consider k = 6. Looking at the cycles as in the previous case,

  1→ 7→ 13→ 19→ 5→ 11→ 17→ 3→ 9→ 15→ 1

  2→ 8→ 14→ 20→ 6→ 12→ 18→ 4→ 10→ 16→ 2 Note that starting at any chair other than 1 or 2 gives one of these two cycles. That is, all 20 chairs appear here. This means that each set of 10 chairs are either all occupied or are all not occupied. This means that there are two possible values of n, namely 10 or 20. Thus, there are 2 pairs (n, k) that work when k = 6.

  Consider k = 8. Here, there are 4 cycles:

  1→ 9→ 17→ 5→ 13→ 1 2→ 10→ 18→ 6→ 14→ 2

  3→ 11→ 19→ 7→ 15→ 3 4→ 12→ 20→ 8→ 16→ 4 and so there are 4 values of n. Note that starting at any chair other than 1, 2, 3 or 4 gives one of these four cycles. That is, all 20 chairs appear here. Thus, there a

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