17th British Mathematical Olympiad 1981 Problems

17th British Mathematical Olympiad 1981 Problems

1.  ABC is a triangle. Three lines divide the triangle into four triangles and three pentagons. One of the triangle has its three sides along the new lines, the others each have just two sides along the new lines. If all four triangles are congruent, find the area of each in terms of the area of ABC.

2.  An axis of a solid is a straight line joining two points on its boundary such that a rotation about the line through an angle greater than 0 deg and less than 360 deg brings the solid into coincidence with itself. How many such axes does a cube have? For each axis indicate the minimum angle of rotation and how the vertices are permuted.

3.  Find all real solutions to x²y² + x²z² = axyz, y²z² + y²x² = bxyz, z²x² + z²y² = cxyz, where a, b, c are fixed reals.

4.  Find the remainder on dividing x⁸¹ + x⁴⁹ + x²⁵ + x⁹ + x by x³ - x.

5.  The sequence u₀, u₁, u₂, ... is defined by u₀ = 2, u₁ = 5, u_n+1u_n-1 - u_n² = 6^n-1. Show that all terms of the sequence are integral.

6.  Show that for rational c, the equation x³ - 3cx² - 3x + c = 0 has at most one rational root.

7.  If x and y are non-negative integers, show that there are non-negative integers a, b, c, d such that x = a + 2b + 3c + 7d, y = b + 2c + 5d iff 5x ≥ 7y. 

17th BMO 1981

Problem 1

ABC is a triangle. Three lines divide the triangle into four triangles and three pentagons. One of the triangle has its three sides along the new lines, the others each have just two sides along the new lines. If all four triangles are congruent, find the area of each in terms of the area of ABC.

Solution

 

The obvious configuration has ∠EDF = ∠EJI, so KH is parallel to AC and similarly LI parallel to BC, and GJ parallel to AB. Then to get equal area we need E,F to trisect LI and similarly for the other two lines. Suppose GH = kBC. Then GJ = 3kAB, so CG = 3kBC. Hence CH = 2kBC and similarly BG = 2kBC, so k = 1/5 and the area of each small triangle is 1/25 area ABC. 

However, there are other configurations. For example, suppose we take GJ parallel to AB and KH parallel to AC, but LI not parallel to BC. Take GH = a, DH = b, GD = c. Then GDH is evidently similar to BAC because of the parallel lines. We find DE = b, EJ = b, so GJ = 2b+c. GJC is similar to GDH, so GC = a(2b+c)/c. Similarly, DF = KF = c, so KH = b+2c. Hence BH = a(b+2c)/b. Hence BC = a(bc+2b²+2c²)/bc. So BC/GH = (bc+2b²+2c²)/bc and hence the area of a small triangle is 1/(1 + 2b/c + 2c/b)² times the area of ABC. Here the ratio can take any value less than 1/25 depending on the ratio b/c. 

17th BMO 1981

Problem 2

An axis of a solid is a straight line joining two points on its boundary such that a rotation about the line through an angle greater than 0 deg and less than 360 deg brings the solid into coincidence with itself. How many such axes does a cube have? For each axis indicate the minimum angle of rotation and how the vertices are permuted.

Answer

3 through the centers of two opposite faces, minimum angle 90^o
6 through the midpoints of two opposite edges, minimum angle 180^o
4 through opposite vertices, minimum angle 120^o

17th BMO 1981

Problem 3

Find all real solutions to x²y² + x²z² = axyz, y²z² + y²x² = bxyz, z²x² + z²y² = cxyz, where a, b, c are fixed reals.

Answer

(x,y,z) = (t,0,0), (0,t,0), (0,0,t) are solutions for any t and any a,b,c.
If a, b, c are the sides of a triangle (so a,b,c > 0 and a+b > c etc), or -a, -b, -c are the sides of a triangle, then x = ±√((s-b)(s-c)), y = ±√((s-a)(s-c)), z = ±√((s-a)(s-b)), where we have either 0 or 2 minus signs, are solutions.

Solution

If x = 0, then y²z² = 0, so y or z = 0. If x = y = 0, then z can be arbitrary. Similarly for the other pairs. Thus we get solutions (x,y,z) = (0,0,t), (0,t,0), (t,0,0) for any t. So suppose xyz ≠ 0.

Subtracting the second equation from the first, x²z²-y²z² = (a-b)xyz. Adding to the last equation, x²z² = (a-b+c)xyz/2. Similarly y²z² = (-a+b+c)xyz/2. Hence x²y²z⁴ = (a-b+c)(-a+b+c)x²y²z²/4, so z² = (a-b+c)(-a+b+c)/4 = (s-a)(s-b), where s = (a+b+c)/2. Similarly, x² = (s-b)(s-c), y² = (s-a)(s-c). However, this solution is only possible if a, b, c satisfy certain conditions.

If s-a > 0, then z² > 0 implies s-b > 0 and x² > 0 implies s-c > 0. Adding s-a > 0 and s-b > 0 gives c > 0. Similarly, a > 0 and b > 0. So a, b, c must be the sides of a triangle. Similarly, if s-a < 0, then s-b < 0 and s-c < 0, and hence a, b, c < 0. In this case -a, -b, -c are the sides of a triangle. 

17th BMO 1981

Problem 4

Find the remainder on dividing x⁸¹ + x⁴⁹ + x²⁵ + x⁹ + x by x³ - x.

Solution

Put p(x) = x⁸¹ + x⁴⁹ + x²⁵ + x⁹ + x = q(x)(x³ - x) + ax² + bx + c. Putting x = 0 gives c = 0. Putting x = 1 gives 5 = a + b, putting x = -1 gives -5 = a - b, so a = 0, b = 5. Hence the remainder is 5x. 

17th BMO 1981

Problem 5

The sequence u₀, u₁, u₂, ... is defined by u₀ = 2, u₁ = 5, u_n+1u_n-1 - u_n² = 6^n-1. Show that all terms of the sequence are integral.

Solution

We show by induction that u_n = 2ⁿ + 3ⁿ. True for n = 0 and 1. Suppose it is true for n and n-1, then u_n+1(2^n-1+3^n-1) - (2ⁿ+3ⁿ)² = 6^n-1. Hence u_n+1(2^n-1+3^n-1) = 2²ⁿ + 3²ⁿ + 13·6^n-1 = (2ⁿ⁺¹+3ⁿ⁺¹)(2^n-1+3^n-1). Hence u_n+1 = 2ⁿ⁺¹+3ⁿ⁺¹, so the result is true for n+1 and hence for all n. 

17th BMO 1981

Problem 6

Show that for rational c, the equation x³ - 3cx² - 3x + c = 0 has at most one rational root.

Solution

Suppose it has a rational root k. Then c = (k³-3k)/(3k²-1), so the equation is (3k²-1)x³ - 3(k³-3k)x² - 3(3k²-1)x + k³-3k = 0, which factorises as (x-k)( (3k²-1)x² + 8kx - (k²-3) ) = 0. The roots of the quadratic are -4k/(3k²-1) ± (√3)(k²+1)/(3k²-1), which are irrational. 

17th BMO 1981

Problem 7

If x and y are non-negative integers, show that there are non-negative integers a, b, c, d such that x = a + 2b + 3c + 7d, y = b + 2c + 5d iff 5x ≥ 7y.

Solution

If 5x = 7y, then y = 5d for some non-negative integer d, and we can take a = b = c = 0. If 5x = 7y+1, then x = 7d+3 for some non-negative integer d, so y = 5d+2, c = 1, a = b = 0. If 5x = 7y+2, then x = 7d+6, so y = 5d+4, c = 2, a = b = 0. If 5x = 7y+3, then x = 7d+2, y = 5d+1, b = 1, a = c = 0. If 5x = 7y+4, then x = 7d+5, y = 5d+3, a = 0, b = c = 1. If 5x = 7y + n, for n > 4, then we can take m as the residue of n mod 5, a = n-m and b,c,d as above.

The implication the other way is trivial.