2nd Mexican Mathematical Olympiad Problems 1988

2nd Mexican Mathematical Olympiad Problems 1988

A1.  In how many ways can we arrange 7 white balls and 5 black balls in a line so that there is at least one white ball between any two black balls?

A2.  If m and n are positive integers, show that 19 divides 11m + 2n iff it divides 18m + 5n.

A3.  Two circles of different radius R and r touch externally. The three common tangents form a triangle. Find the area of the triangle in terms of R and r.

A4.  How many ways can we find 8 integers a₁, a₂, ... , a₈ such that 1 ≤ a₁ ≤ a₂ ≤ ... ≤ a₈ ≤ 8?

B1.  a and b are relatively prime positive integers, and n is an integer. Show that the greatest common divisor of a²+b²-nab and a+b must divide n+2.

B2.  B and C are fixed points on a circle. A is a variable point on the circle. Find the locus of the incenter of ABC as A varies.

B3.  [unclear]

B4.  Calculate the volume of an octahedron which has an inscribed sphere of radius 1. 

Solutions

Problem A1

In how many ways can we arrange 7 white balls and 5 black balls in a line so that there is at least one white ball between any two black balls?

Answer

56

Solution

We have BWBWBWBWB and 3 spare W which can go anywhere. There are 6 different places to put each W (at either end or with one of the existing W). If we put each W in a different place then there are 6·5·4/3! = 20 possibilities. If we put two in one place and one in another, there are 6·5 = 30 possibilities. If we put them all in the same place, there are 6 possibilities. Total 56. As a check, put down the 7 W first. Then there are 8 places to put the five B, and each must go in a different place. Hence 8·7·6·5·4/5! = 56. 

Problem A2

If m and n are positive integers, show that 19 divides 11m + 2n iff it divides 18m + 5n.

Solution

19 divides 18m + 5n iff it divides 8(18m + 5n) = 19(7m + 2n) + 11m + 2n and hence iff it divides 11m + 2n. 

Problem A3

Two circles of different radius R and r touch externally. The three common tangents form a triangle. Find the area of the triangle in terms of R and r.

Solution

Let the triangles have centers O, O'. Let the common tangents meet at A, B, C as shown. Take the radii to be r, R as shown. Let the circle touch at Z. Note that BZA, OXA and O'YA are all similar. So area ABC = 2 area BZA = 2 AZ²/AX² area OXA = AZ²r/AX.

By the similarity of AXO, AYO', r/AO = R/(AO+R+r), so AO = r(R+r)/(R-r). Hence AZ = AO + r = 2Rr/(R-r). By Pythagoras, AX² = AO² - r² = (r²/(R-r)²)((R+r)² - (R-r)²) = 4Rr³/(R-r)². Hence area ABC = 2(Rr)^3/2/(R-r). 

Problem A4

How many ways can we find 8 integers a₁, a₂, ... , a₈ such that 1 ≤ a₁ ≤ a₂ ≤ ... ≤ a₈ ≤ 8?

Answer

15C7 = 15!/(8!7!) = 6435

Solution

Arrange 8 black balls and 7 white balls in a row. Then the 7 white balls divide the line into 8 parts corresponding to a_i = 1, 2, ... , 8. The number of black balls in each part gives the number of ai with that value. For example WBBWWBWBWBBWWBB corresponds to a₁ = a₂ = 2, a₃ = 4, a₄ = 5, a₅ = a₆ = 6, a₇ = a₈ = 8. 

Problem B1

a and b are relatively prime positive integers, and n is an integer. Show that the greatest common divisor of a²+b²-nab and a+b must divide n+2.

Solution

Let d = gcd(a²+b²-nab, a+b). Then d divides (a+b)² - (a²+b²-nab) = (n+2)ab. If any integer k divides a, then it divides a² - nab, but not b² (because a and b are relatively prime) and hence not (a²+b²-nab). So d must be relatively prime to a, and similarly to b. Hence d divides n+2. 

Problem B2

B and C are fixed points on a circle. A is a variable point on the circle. Find the locus of the incenter of ABC as A varies.

Solution

Let I be the incenter. ∠BIC = 180^o - ∠B/2 - ∠C/2 = 90^o + ∠A/2 = constant. So the locus is the arc of the circle through B and C which lies inside the circle and such that points on the arc give the angle 90^o + ∠A/2. 

Problem B4

Calculate the volume of an octahedron which has an inscribed sphere of radius 1.

Answer

4√3

Solution

Opposite faces of the octahedron are parallel and hence a distance 2 apart. Suppose the edge length is x. Then the distance from the center of a face to a vertex of the face is x/√3. A section through the centers of two opposite faces, perpendicular to the faces, and through a vertex of one, is shown above. Thus 4 = 3x²/4 - x²/12 = 2x²/3, so x = √6.

We can also view the octahedron as two square pyramids with bases coincident. The square base has area x² = 6. The height of each pyramid is x/√2 = √3, so volume = (2/3)6√3 = 4√3.