10th Canadian Mathematical Olympiad Problems 1978

10th Canadian Mathematical Olympiad Problems 1978

1.  A square has tens digit 7. What is the units digit?
2.  Find all positive integers m, n such that 2m² = 3n³.

3.  Find the real solution x, y, z to x + y + z = 5, xy + yz + zx = 3 with the largest z.

4.  ABCD is a convex quadrilateral with area 1. The lines AD, BC meet at X. The midpoints of the diagonals AC and BD are Y and Z. Find the area of the triangle XYZ.

5.  Two players play a game on an initially empty 3 x 3 board. Each player in turn places a black or white piece on an unoccupied square of the board. Each player may play either color. When the board is full player A gets one point for every row, column or main diagonal with 0 or 2 black pieces on it. Player B gets one point for every row, column or main diagonal with 1 or 3 black pieces on it. Can the game end in a draw? Which player has a winning strategy if player A plays first? If player B plays first?

6.  Sketch the graph of x³ + xy + y³ = 3.

Solurtions

Problem

A square has tens digit 7. What is the units digit?

Solution

Suppose the unsquared number has tens digit a and units digit b. The carry from b² must be odd (since 7 is odd), so b = 4 or 6. In either case the units digit of the square is 6. An example is 26² = 576. 

Problem 2

Find all positive integers m, n such that 2m² = 3n³.

Solution

2 must divide n, so 8 must divide 2m², so 2 must divide M. Similarly, 3 must divide m, so 9 divides 3n³ so 3 divides n. Hence 81 divides 2m², so 9 divides m. Put m = 18M, n = 6N, then 648M² = 648N³, so M² = N³. Thus M is a cube, so finally we have for some integer k, m = 18k³ , n = 6k². 

Problem 3

Find the real solution x, y, z to x + y + z = 5, xy + yz + zx = 3 with the largest z.

Solution

We have 25 = (x + y + z)² = (x² + y² + z²) + 2(xy + yz + zx) = (x² + y² + z²) + 6, so (x² + y² + z²) = 19. Hence (x - 1/3)² + (y - 1/3)² + (z - 1/3)² = (x² + y² + z²) -2(x + y + z)/3 + 1/3 = 19 - 10/3 + 1/3 = 16. So the maximum possible value of z is 4 1/3 (with solution x = 1/3, y = 1/3, z = 4 1/3). 

Problem 4

ABCD is a convex quadrilateral with area 1. The lines AD, BC meet at X. The midpoints of the diagonals AC and BD are Y and Z. Find the area of the triangle XYZ.

Solution

Use vectors with origin X. Let XA = a, XB = b, XC = hb, XD = ka. Then area ABCD = area XCD - area XAB = (hk - 1) |a x b| = 1, and area XYZ = |(a + hb) x (b + ka)/4| = (hk - 1)/4 |a x b| = 1/4. ( b x a = - a x b, a x a = b x b = 0). 

Problem 5

Two players play a game on an initially empty 3 x 3 board. Each player in turn places a black or white piece on an unoccupied square of the board. Each player may play either color. When the board is full player A gets one point for every row, column or main diagonal with 0 or 2 black pieces on it. Player B gets one point for every row, column or main diagonal with 1 or 3 black pieces on it. Can the game end in a draw? Which player has a winning strategy if player A plays first? If player B plays first?

Solution

A draw is possible, for example:

W  B  W

B  W  B

W  B  W

If A plays first, A can play black in the center and thereafter the opposite color to B's last move on the opposite side of the center (so if B plays in white in the middle of the top row, then A returns black in the middle of the bottom row). This must give two black pieces on each main diagonal and two black pieces on the central row and on the central column (a score of 4 so far). Also two adjacent corners must be black and the other two white. If the piece between the two adjacent corners is white, then A scores one for that row/col, if not A scores one for the parallel line at the opposite side of the board (which must have no black pieces). So A scores at least 5 and wins.

If B plays first, then B can win by a similar strategy. B plays white in the center and thereafter the opposite color to A's last move on the opposite side of the center. A similar argument to the above shows that this is a win for B. 

Problem 6

Sketch the graph of x³ + xy + y³ = 3.

Solution

Close to the line y = -x far from the origin. A slight bulge away from the line (in the NE direction into the first quadrant) near the origin. Passes through (0, k), (1, 1), (k, 0), where k is 3^1/3 = 1.44.