2nd USA Mathematical Olympiad 1973 Problems

2nd USA Mathematical Olympiad 1973 Problems

1.  Show that if two points lie inside a regular tetrahedron the angle they subtend at a vertex is less than π/3.  

2.  The sequence a_n is defined by a₁ = a₂ = 1, a_n+2 = a_n+1 + 2a_n. The sequence b_n is defined by b₁ = 1, b₂ = 7, b_n+2 = 2b_n+1 + 3b_n. Show that the only integer in both sequences is 1.

3.  Three vertices of a regular 2n+1 sided polygon are chosen at random. Find the probability that the center of the polygon lies inside the resulting triangle.

4.  Find all complex numbers x, y, z which satisfy x + y + z = x² + y² + z² = x³ + y³ + z³ = 3. 

5.  Show that the cube roots of three distinct primes cannot be terms in an arithmetic progression (whether consecutive or not).  

Solution

2nd USA Mathematical Olympiad 1973

Problem 1
Show that if two points lie inside a regular tetrahedron the angle they subtend at a vertex is less than π/3.
Solution

Let the tetrahedron be ABCD and the points be P and Q. Note that we are asked to prove the result for any vertex, not just some. So consider angle PAQ. Let the rays AP, AQ meet the plane BCD at P', Q' respectively. So we have to show that angle P'AQ' < 60^o for P' and Q' interior points of the triangle BCD. Extend P'Q' to meet the sides of the triangle at X and Y. Without loss of generality, X lies BC and Y lies on CD. Obviously if is sufficient to show that angle XAY < 60^o.

X and Y cannot both be vertices (or P', Q' would not have been interior points of the triangle and hence P and Q would not have been strictly inside the tetrahedron). So suppose X is not a vertex. We show that XY <= XD. Consider triangle XYD. ∠XDY < ∠BDC = 60^o, but ∠XYD = ∠XCD + ∠CXY ≥ 60^o, so ∠XDY < ∠XYD. Hence XY ≤ XD. But XD = AX (consider, for example, the congruent triangles AXB and DXB). Hence XY < AX. Similarly, XY ≤ AY. Hence angle XAY < 60^o. Problem 2

The sequence a_n is defined by a₁ = a₂ = 1, a_n+2 = a_n+1 + 2a_n. The sequence b_n is defined by b₁ = 1, b₂ = 7, b_n+2 = 2b_n+1 + 3b_n. Show that the only integer in both sequences is 1.
Solution

We can solve the first recurrence relation to give a_n = A (-1)ⁿ + B 2ⁿ. Using a₁ and a₂, we get a_n = (2ⁿ + (-1)ⁿ⁺¹)/3. Similarly, for the second recurrence relation we get b_n = 2.3^n-1 + (-1)ⁿ. So if a_m = b_n then 2.3ⁿ + 3 (-1)ⁿ = 2^m + (-1)^m+1 or 2^m = 2.3ⁿ + 3 (-1)ⁿ + (-1)^m.
If m = 1 or 2, then we find n = 1 is the only solution, corresponding to the fact that the term 1 is in both sequences. If m > 2, then 2^m = 0 mod 8. But 3ⁿ = (-1)ⁿ mod 4, so 2.3ⁿ + 3 (-1)ⁿ + (-1)^m = 5 (-1)ⁿ + (-1)^m mod 8 which cannot be 0 mod 8. So there are no solutions for m > 2.

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Problem 3
Three vertices of a regular 2n+1 sided polygon are chosen at random. Find the probability that the center of the polygon lies inside the resulting triangle.
Solution

Answer: (n+1)/(4n-2).

Label the first vertex picked as 1 and the others as 2, 3, ... , 2n+1 (in order). There are 2n(2n-1)/2 ways of choosing the next two vertices. If the second vertex is 2 (or 2n+1), then there is just one way of picking the third vertex so that the center lies in the triangle (vertex n+2). If the second vertex is 3 (or 2n), then there are two (n+2, n+3) and so on. So the total number of favourable triangles is 2(1 + 2 + ... + n) = n(n+1). Thus the required probability is (n+1)/(4n-2).

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Problem 4
Find all complex numbers x, y, z which satisfy x + y + z = x² + y² + z² = x³ + y³ + z³ = 3.
Solution

Answer: 1, 1, 1.
We have (x + y + z)² = (x² + y² + z²) + 2(xy + yz + zx), so xy + yz + zx = 3.
(x + y + z)³ = (x³ + y³ + z³) + 3(x²y + xy² + y²z + yz² + z²x + zx²) + 6xyz, so 8 = (x²y + xy² + y²z + yz² + z²x + zx²) + 2xyz. But (x + y + z)(xy + yz + zx) = (x²y + xy² + y²z + yz² + z²x + zx²) + 3xyz, so xyz = -1. Hence x, y, z are the roots of the cubic w³ - 3w² + 3w - 1 = (w - 1)³. Hence x = y = z = 1.

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Problem 5
Show that the cube roots of three distinct primes cannot be terms in an arithmetic progression (whether consecutive or not).
Solution

Suppose the primes are p, q, r so that q^1/3 = p^1/3 + md, r^1/3 = p^1/3 + nd, where m and n (but not necessarily d) are integers. Then nq^1/3 - mr^1/3 = (n - m)p^1/3. Cubing: n³q - 3n²mq^2/3r^1/3 + 3nm²q^1/3r^2/3 - m³r = (n-m)³p, or q^1/3r^1/3(mr^1/3 - nq^1/3) = ( (n-m)³p + m³r - n³q)/(3mn). But mr^1/3 - nq^1/3 = (m - n)p^1/3, so we have (pqr)^1/3 = ( (n-m)³p + m³r - n³q)/(3mn(m-n) ) (*).

It is now clear that we do not need p, q, r prime, just that pqr is not a cube, for then by the usual argument it must be irrational so that (*) is impossible.