6th Mexican Mathematical Olympiad Problems 1992

6th Mexican Mathematical Olympiad Problems 1992

A1.  The tetrahedron OPQR has the ∠POQ = ∠POR = ∠QOR = 90^o. X, Y, Z are the midpoints of PQ, QR and RP. Show that the four faces of the tetrahedron OXYZ have equal area.

A2.  Given a prime number p, how many 4-tuples (a, b, c, d) of positive integers with 0 < a, b, c, d < p-1 satisfy ad = bc mod p?

A3.  Given 7 points inside or on a regular hexagon, show that three of them form a triangle with area ≤ 1/6 the area of the hexagon.

B1.  Show that 1 + 11¹¹ + 111¹¹¹ + 1111¹¹¹¹ + ... + 1111111111^1111111111 is divisible by 100.

B2.  x, y, z are positive reals with sum 3. Show that 6 < √(2x+3) + √(2y+3) + √(2z+3) < 3√5.

B3.  ABCD is a rectangle. I is the midpoint of CD. BI meets AC at M. Show that the line DM passes through the midpoint of BC. E is a point outside the rectangle such that AE = BE and ∠AEB = 90^o. If BE = BC = x, show that EM bisects ∠AMB. Find the area of AEBM in terms of x.

Solutions

Problem A1

The tetrahedron OPQR has the ∠POQ = ∠POR = ∠QOR = 90^o. X, Y, Z are the midpoints of PQ, QR and RP. Show that the four faces of the tetrahedron OXYZ have equal area.

Solution

Use vectors.

Take origin O, put p = OP, q = OQ, r = OR. Then OX = (p + q)/2, OY = (q + r)/2, OZ = (r + p)/2. Now the area of OXY is the magnitude of the vector (p + q) x (q + r)/4 = (p x q + p x r + q x r)/4. But the vectors p, q, r are orthogonal, so this is just (p²q² + p²r² + q²r²)^1/2/4 (*), where p = |p| etc. Similarly, the area of OYZ is the magnitude of (q + r) x (r + p)/4 = (-p x q - p x r + q x r)/4. This has the same value (*) - because of the orthogonality the minus signs make no difference. Similarly for area OZX.

Finally, area XYZ is the magnitude of the vector (OY - OX) x (OZ - OX) = (r - p) x (r - q)/4 = (p x q - p x r + q x r)/4, which again has the value (*).

Problem A2

Given a prime number p, how many 4-tuples (a, b, c, d) of positive integers with 0 < a, b, c, d < p-1 satisfy ad = bc mod p?

Solution

Given a, b, c ≠ 0 mod p, we can always find d ≠ 0 mod p such that ad = bc mod p. However, it is possible that d = p-1 mod p. If bc = 1 mod p, then a(p-1) ≠ 1 mod p, so d is not p-1 for any choice of a. On the other hand if bc ≠ 1 mod p, then if a = -bc mod p we will get d = p-1 mod p. Thus we may choose b arbitrarily (p-2 choices). Then for c = 1/b (one choice), we have p-2 choices for a, and d is then determined. For c ≠ 1/b (p-3 choices), we have p-3 choices for a, and d is then determined. Thus in total there are (p-2)(p-2 + (p-3)²) = (p-2)(p²-5p+7) possibilities.

Thanks to Suat Namli

Problem B1

Show that 1 + 11¹¹ + 111¹¹¹ + 1111¹¹¹¹ + ... + 1111111111^1111111111 is divisible by 100.

Solution

We have 1111 = 11 mod 100, and 1110 = 1 mod 100, so sum given = 1 + 11 + 11 + ... + 11 = 1 + 99 = 0 mod 100.

Thanks to Suat Namli

Problem B3

ABCD is a rectangle. I is the midpoint of CD. BI meets AC at M. Show that the line DM passes through the midpoint of BC. E is a point outside the rectangle such that AE = BE and ∠AEB = 90^o. If BE = BC = x, show that EM bisects ∠AMB. Find the area of AEBM in terms of x.

Answer

x²(1/2 + (√2)/3)

Solution

The diagonals of the rectangle bisect each other, so AC is a median of the triangle BCD, BI is another median, so M is the centroid. Hence DM is the third median and so passes through the midpoint of BC.

AB = x√2, so CI = x/√2. Hence BI = x√(3/2) and BM = (2/3)BI = x√(2/3). Now AC = x√3, so CM = (2/3)AC/2 = x/√3. Hence BC² = BM² + CM², so ∠AMB = 90^o. So M and E lie on the circle diameter AB. Hence ∠EMA = ∠EBA = 45^o, so EM bisects ∠AMB.

area AEB = x²/2, area AMB = BM·AM/2 = x√(2/3)x/√3 = x²(√2)/3. Hence area AEBM = x²(1/2 + (√2)/3)

Thanks to Suat Namli