9th Eötvös Competition Problems 1902

1.  Let p(x) = ax² + bx + c be a quadratic with real coefficients. Show that we can find reals d, e, f so that p(x) = d/2 x(x - 1) + ex + f, and that p(n) is always integral for integral n iff d, e, f are integers.

2.  P is a variable point outside the fixed sphere S with center O. Show that the surface area of the sphere center P radius PO which lies inside S is independent of P.

3.  The triangle ABC has area k and angle A = θ, and is such that BC is as small as possible. Find AB and AC. 

Solutions

Problem 1

Let p(x) = ax² + bx + c be a quadratic with real coefficients. Show that we can find reals d, e, f so that p(x) = d/2 x(x - 1) + ex + f, and that p(n) is always integral for integral n iff d, e, f are integers.

Solution

We can take d = 2a, e = a+b, f = c. Note that n(n-1)/2 is an integer for all integers n, so if d, e, f are integers, then p(n) is an integer for all integers n.

Now consider the converse. p(0) is an integer, so f is an integer. p(1) is an integer, so e + f is an integer and hence e also. p(2) is an integer, so d + 2e + f is an integer and hence also d. 

Problem 2

P is a variable point outside the fixed sphere S with center O. Show that the surface area of the sphere center P radius PO which lies inside S is independent of P.

Solution

This is an unsatisfactory question because it requires the use of calculus. Given calculus, it is easy.

Area = ∫₀^φ 2πr sin θ r dθ, because the ring shown has radius r sin θ and thickness r dθ. Integrating we get 2πr²(1 - cos φ) = 4πr² sin²(φ/2). Now sin(φ/2) = a/(2r), where a is the radius of the fixed sphere. Hence area = πa², which is independent of r. 

Problem 3

The triangle ABC has area k and angle A = θ, and is such that BC is as small as possible. Find AB and AC.

Solution

Take AB = c, AC = b, then BC² = b² + c²- 2bc cos θ = (b-c)² + 2bc(1 - cos θ). But k = ½bc sin θ, or bc = 2k/sin θ. Hence BC² = (b-c)² + 4k(1 - cos θ)/sin θ. Since the second term is fixed, we minimise BC by taking b = c. So k = ½bc sin θ, so b = c = √(2k/sin θ).