3rd Mexican Mathematical Olympiad Problems 1989

3rd Mexican Mathematical Olympiad Problems 1989

A1.  The triangle ABC has AB = 5, the medians from A and B are perpendicular and the area is 18. Find the lengths of the other two sides.

A2.  Find integers m and n such that n² is a multiple of m, m³ is a multiple of n², n⁴ is a multiple of m³, m⁵ is a multiple of n⁴, but n⁶ is not a multiple of m⁵.

A3.  Show that there is no positive integer of 1989 digits, at least three of them 5, such that the sum of the digits is the same as the product of the digits.

B1.  Find a positive integer n with decimal expansion a_ma_m-1...a₀ such that a₁a₀a_ma_m-1...a₂0 = 2n.

B2.  C₁ and C₂ are two circles of radius 1 which touch at the center of a circle C of radius 2. C₃ is a circle inside C which touches C, C₁ and C₂. C₄ is a circle inside C which touches C, C₁ and C₃. Show that the centers of C, C₁, C₃ and C₄ form a rectangle.

B3.  How many paths are there from A to B which do not pass through any vertex twice and which move only downwards or sideways, never up?

Solutions

Problem A1

The triangle ABC has AB = 5, the medians from A and B are perpendicular and the area is 18. Find the lengths of the other two sides.

Answer

2√13, √73

Solution

Let the medians be AD and BE, meeting at F. Let G be the foot of the perpendicular from F to AB. F is the centroid, so it is one third of the way along towards C on the median from C, and hence FG is one third of the height of the triangle. Hence area AFB = 6, so FG = 2.4. Let AG = x. Then since AFG, FGB are similar, we have FG/AG = FG/FB or x(5-x) - 5.76. Solving the quadratic, x = 1.8 or 3.2. wlog we can take AG = 3.2, so FGA is a 3,4,5 triangle and hence AF = 4. Similarly BF = 3. So EF = BF/2 = 1.5.

Now by Pythagoras AE² = 4² + (3/2)² = 73/4. So AC = 2AE = √73. Similarly, BD2 = 3² + 2² = 13, so BC = 2√13

Problem 2

Find integers m and n such that n² is a multiple of m, m³ is a multiple of n², n⁴ is a multiple of m³, m⁵ is a multiple of n⁴, but n⁶ is not a multiple of m⁵.

Solution

For example, take n = 2⁴, m = 2⁵.

Problem 3

Show that there is no positive integer of 1989 digits, at least three of them 5, such that the sum of the digits is the same as the product of the digits.

Solution

Suppose there is such a number. None of the digits can be zero, or the product would be zero. Hence the sum of the digits is at least 1986·1 + 3·5 = 2001. But the product of the digits is a multiple of 5³, so the sum must be at least 2125. That is 136 larger than 1989·1. Any one digit can contribute at most 9-1 = 8 to the excess, so at least [136/8] = 17 digits must be larger than 1. So the product of the digits must be at least 2¹⁷ > 2¹⁵ = 32768 > 9·1989, which is the largest possible sum of 1989 digits. Contradiction.

Problem B1

Find a positive integer n with decimal expansion a_ma_m-1...a₀ such that a₁a₀a_ma_m-1...a₂0 = 2n.

Answer

10 526 315 789 473 684 210

or 789 473 684 210 526 315

Solution

Clearly a₀ = 0 or 5 and a₁ = 1. Take a₀ = 0. Now we can calculate the digits successively from the right. Doubling ...10 gives ...20, so a₂ = 2. Doubling ...210 gives ...420, so a₃ = 4. Doubling ...4210 gives ...8420 so a₄ = 8. Doubling ...84210 gives ...68420 so a₅ = 6, and so on. We stop when the result of the doubling starts with 10. That gives the first answer above. Similarly, starting with ...15 gives the second answer above.

Problem B2

C₁ and C₂ are two circles of radius 1 which touch at the center of a circle C of radius 2. C₃ is a circle inside C which touches C, C₁ and C₂. C₄ is a circle inside C which touches C, C₁ and C₃. Show that the centers of C, C₁, C₃ and C₄ form a rectangle.

Solution

Let O be the center of C, A the center of C₁, B the center of C₂ and X the center of C₃. Suppose C₃ has radius x. Then the height of AXB must be 2-x if C₃ touches C. But if it touches C₁ and C₂, then it has sides 2, 1+x, 1+x, so by Pythagoras √((1+x)²-1) = 2-x. Hence x = 2/3.

Let Y be the center of C₄. If AOXY is a rectangle, then C₄ must have radius 1/3. It is easy to see that with this center and radius it will touch C₁ and C₃. We have to show that it also touches C.

But AYO is a 3,4,5 triangle, so YO = 5/3. Hence YO + radius C₄ = radius C, as required, so C₄ does touch C.