1st Mexican Mathematical Olympiad Problems 1987

1st Mexican Mathematical Olympiad Problems 1987

A1.  a/b and c/d are positive fractions in their lowest terms such that a/b + c/d = 1. Show that b = d.

A2.  How many positive integers divide 20! ?

A3.  L and L' are parallel lines and P is a point midway between them. The variable point A lies L, and A' lies on L' so that ∠APA' = 90^o. X is the foot of the perpendicular from P to the line AA'. Find the locus of X as A varies.

A4.  Let N be the product of all positive integers ≤ 100 which have exactly three positive divisors. Find N and show that it is a square.

B1.  ABC is a triangle with ∠A = 90^o. M is a variable point on the side BC. P, Q are the feet of the perpendiculars from M to AB, AC. Show that the areas of BPM, MQC, AQMP cannot all be equal.

B2.  Prove that (n³ - n)(5⁸ⁿ⁺⁴ + 3⁴ⁿ⁺²) is a multiple of 3804 for all positive integers n.

B3.  Show that n² + n - 1 and n² + 2n have no common factor.

B4.  ABCD is a tetrahedron. The plane ABC is perpendicular to the line BD. ∠ADB = ∠CDB = 45^o and ∠ABC = 90^o. Find ∠ADC. A plane through A perpendicular to DA meets the line BD at Q and the line CD at R. If AD = 1, find AQ, AR, and QR. 

Solutions

Problem A1
a/b and c/d are positive fractions in their lowest terms such that a/b + c/d = 1. Show that b = d.
Solution
ad + bc = bd, so b(d-c) = ad. So b divides ad, but is prime to a, hence it divides d. Similarly, d divides b. Hence b = d.

Problem A2
How many positive integers divide 20! ?
Answer
2⁴3³5·19 = 41040
Solution
20! is divisible by [20/2] + [20/4] + [20/8] + [20/16] = 18 powers of 2, and [20/3] + [20/9] = 8 powers of 3. It is divisible by 54, 72, 11, 13, 17 and 19. In other words, 20! = 2¹⁸3⁸5⁴7²11·13·17·19. So it has (18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1) = 2⁴3³5·19.

Problem A3
L and L' are parallel lines and P is a point midway between them. The variable point A lies L, and A' lies on L' so that ∠APA' = 90^o. X is the foot of the perpendicular from P to the line AA'. Find the locus of X as A varies.
Answer
the circle center P touching L and L' (except for the points of contact with L and L')
Solution

Note that ∠A'PB' = 180^o - ∠APA' - ∠APB = 90^o - ∠APB = ∠PAB. So AP = PB/sin PAB, and A'P = PB'/cos A'PB' = PB/cos PAB. Hence A'P/AP = tan PAB. But A'P/AP = tan PAA', so ∠PAA' = ∠PAB. But ∠XPA' = ∠PAA', so ∠XPA' = ∠B'PA'. Hence triangles PB'A' and PXA' are congruent. So PX = PB', which is constant. So X lies on the circle diameter BB'.

Now suppose X is any point on this circle except B and B'. Let the tangent at X meet L and L' at A and A'. Then ∠PBA = ∠PXA = 90^o, so ABPX is cyclic, so ∠PAX = ∠PBX. But P is the center of the circle through B', X, B, so ∠PBX = ½ ∠B'PX = ∠A'PX = 90^o - ∠AA'P. Hence ∠PAX + ∠AA'P = 90^o, so ∠APA' = 90^o. Thus every point X of the circle except B and B' belong to the locus.

Problem A4
Let N be the product of all positive integers ≤ 100 which have exactly three positive divisors. Find N and show that it is a square.
Answer
210²
Solution
If n = p^aq^br^c..., where p, q, r are primes, then it has (a+1)(b+1)(c+1)... divisors. Hence n must be p² for some prime p. Thus N = (2·3·5·7)².



Problem B1
ABC is a triangle with ∠A = 90^o. M is a variable point on the side BC. P, Q are the feet of the perpendiculars from M to AB, AC. Show that the areas of BPM, MQC, AQMP cannot all be equal.
Solution

Triangles BPM, MQC are similar, so if they have equal areas, then M is the midpoint of BC. But then P, Q are the midpoints of AB, AC. So in this case area AQMP = ½ area ABC, area BPM = ¼ area ABC.



Problem B2
Prove that (n³ - n)(5⁸ⁿ⁺⁴ + 3⁴ⁿ⁺²) is a multiple of 3804 for all positive integers n.
Solution
We have A²ⁿ⁺¹ + B²ⁿ⁺¹ = (A + B)(A²ⁿ - A^2n-1B + A²ⁿB² - ... + B²ⁿ). Hence (5⁴ + 3²) divides 5⁸ⁿ⁺⁴ + 3⁴ⁿ⁺². Also one of n-1, n, n+1 must be a multiple of 3, and at least one must be even. Hence 6·(625+9) =3804 always divides (n³ - n)(5⁸ⁿ⁺⁴ + 3⁴ⁿ⁺²).

Problem B3
Show that n² + n - 1 and n² + 2n have no common factor.
Solution
If d divides both then it also divides (n² + 2n) - (n² + n - 1) = n+1 and hence also (n+1)² - (n² + 2n) = 1.

Problem B4
ABCD is a tetrahedron. The plane ABC is perpendicular to the line BD. ∠ADB = ∠CDB = 45^o and ∠ABC = 90^o. Find ∠ADC. A plane through A perpendicular to DA meets the line BD at Q and the line CD at R. If AD = 1, find AQ, AR, and QR.
Answer
60^o, 1, √3, √2
Solution
The lines BA, BC, BD are mutually perpendicular. The triangles BAD and BDC are isosceles, so BA = BD = BC. Hence ACD is equilateral and ∠ADC = 60^o.
It is convenient to take Cartesian axes with origin B. Take A as (1/√2, 0, 0), C as (0, 1/√2, 0, 0), D as (0, 0, 1/√2). Then Q is (0, 0, -1/√2) and R is (0, √2, -1/√2). So we find (by Pythagoras) AQ = 1, AR = √3, QR = √2.