26th British Mathematical Olympiad 1990 Problems

26th British Mathematical Olympiad 1990 Problems

1.  Show that if a polynomial with integer coefficients takes the value 1990 at four different integers, then it cannot take the value 1997 at any integer.

2.  The fractional part { x } of a real number is defined as x - [x]. Find a positive real x such that { x } + { 1/x } = 1 (*). Is there a rational x satisfying (*)?

3.  Show that √(x² + y² - xy) + √(y² + z² - yz) ≥ √(z² + x² + zx) for any positive real numbers x, y, z.

4.  A rectangle is inscribed in a triangle if its vertices all lie on the boundary of the triangle. Given a triangle T, let d be the shortest diagonal for any rectangle inscribed in T. Find the maximum value of d²/area T for all triangles T.

5.  ABC is a triangle with incenter I. X is the center of the excircle opposite A. Show that AI·AX = AB·AC and AI·BX·CX = AX·BI·CI. 

Solutions

Problem 1

Show that if a polynomial with integer coefficients takes the value 1990 at four different integers, then it cannot take the value 1997 at any integer.

Solution

Let the polynomial be p(x) and the integers at which it is 1990 be a, b, c, d. Let q(x) = (x - a)(x - b)(x - c)(x - d). Then p(x) - 1990 = q(x) r(x), where r(x) has integer coefficients. [We can successively subtract off integral multiples of q(x) to zero the terms of p(x) - 1990 starting at the highest until we get a polynomial of degree less than 4, but then that has 4 distinct roots and so must be zero].

But now if p(k) = 1997, then 7 = q(k) r(k) = (k - a)(k - b)(k - c) r(k). At most two of the linear factors can be ±1, so this is not possible. 

Problem 2

The fractional part { x } of a real number is defined as x - [x]. Find a positive real x such that { x } + { 1/x } = 1 (*). Is there a rational x satisfying (*)?

Solution

There is no loss of generality in taking x > 1. So {1/x} = 1/x and {x} = x - n for some n. So we have x - n + 1/x = 1 or x² - (n+1)x + 1 = 0, so x = (n+1)/2 + 1/2 √(n²+2n-3). Evidently n = 0 and n = 1 do not work. But n = 2 gives x = (√5 + 3)/2, which works. Similarly, for example n = 3 gives x = 2 + √3, which works. But √(n²+2n-3) cannot be integral (because n is too small and n+1 is too large) and hence, by the usual argument, not rational either. So there are no rational solutions. 

Problem 3

Show that √(x² + y² - xy) + √(y² + z² - yz) ≥ √(z² + x² + zx) for any positive real numbers x, y, z.

Solution

Note that there is a + on the rhs not a - .

A neat geometric approach is take OA = x, ∠AOB = 60^o, OB = y, ∠BOC = 60^o, OC = z. Then, by the cosine rule, AB = √(x² + y² - xy), BC = √(y² + z² - yz) and AC = √(z² + x² + zx). 

Problem 4

A rectangle is inscribed in a triangle if its vertices all lie on the boundary of the triangle. Given a triangle T, let d be the shortest diagonal for any rectangle inscribed in T. Find the maximum value of d²/area T for all triangles T.

Solution

Answer: (4√3)/7.

Suppose two vertices are on the side length a and that the corresponding altitude is length h. Let the side parallel to the altitude have length x. Then the other side has length (h - x)a/h. So the square of the diagonal is x² + a²(1 - x/h)² = (1 + a²/h²)(x - a²h/(a²+h²) )² + a²h²/(a²+h²). Hence d² = a²h²/(a²+h²), or rather that is true provided we have picked the right side.

Suppose b is another side length with corresponding altitude k and. Let A = area T, then ah = bk = 2A. So we require a² + h² ≥ b² + k² or a² + 4A²/a² ≥ b² + 4A²/b² or (a² - b²)(1 - 4A²/(a²b²) ) ≥ 0. But ab ≥ 2A (with equality iff the sides a and b are perpendicular, but the longest side is never perpendicular to another side). So we get the shortest diagonal by taking the longest side.

The required ratio d²/A is thus 2ah/(a² + h²), where a is the longest side. We may assume A = 1/2 (since the ratio is unaffected by the scale of the triangles). So we have to maximise 2/(a² + h²) or equivalently to minimise a² + 1/a². The area of an equilateral triangle with side 1 is (√3)/4 < 1/2, so a > 1, but a² + 1/a² is an increasing function of a for a > 1, so we want to minimise a subject to the constraint that A is fixed. That obviously occurs for an equilateral triangle. An equilateral triangle with side 1 has altitude (√3)/2, so d²/A has maximum value √3/(1 + 3/4) = (4√3)/7. 

Problem 5

ABC is a triangle with incenter I. X is the center of the excircle opposite A. Show that AI·AX = AB·AC and AI·BX·CX = AX·BI·CI.

Solution

We show that triangles AIB and ACX are similar. Obviously ∠BAI = ∠XAC. ∠XIC = A/2 + C/2, ∠ICX = 90^o, so B/2 = ∠IXC = ∠AXC (same angle). Hence ∠ABI = ∠AXC, and the triangles are similar. So AI/AB = AC/AX, which gives the first equation.

The same argument shows that AIC and ABX are similar. So BI/AI = XC/AC and CI/AC = XB/AX. Multiplying gives the second equation.