4th Swedish Mathematical Society Problems 1964

1.  Find the side lengths of the triangle ABC with area S and ∠BAC = x such that the side BC is as short as possible.

2.  Find all positive integers m, n such that n + (n+1) + (n+2) + ... + (n+m) = 1000.

3.  Find a polynomial with integer coefficients which has √2 + √3 and √2 + 3^1/3 as roots.

4.  Points H₁, H₂, ... , H_n are arranged in the plane so that each distance H_iH_j ≤ 1. The point P is chosen to minimise max(PH_i). Find the largest possible value of max(PH_i) for n = 3. Find the best upper bound you can for n = 4.

5.  a₁, a₂, ... , a_n are constants such that f(x) = 1 + a₁ cos x + a₂ cos 2x + ... + a_n cos nx ≥ 0 for all x. We seek estimates of a₁. If n = 2, find the smallest and largest possible values of a₁. Find corresponding estimates for other values of n. 

Solutions

Problem 1

Find the side lengths of the triangle ABC with area S and ∠BAC = x such that the side BC is as short as possible.

Answer

BC = 2√(S tan(x/2)), AB = AC = √(S/(sin(x/2)cos(x/2)) )

Solution

For given BC, the locus of A with ∠BAC = x is the arc of a circle. The area is BC times the height, so we maximise the area by taking AB = AC. Since we want BC as small as possible, we must have AB = AC. So suppose BC = 2k, height = h. Then S = hk, and h = k cot(x/2). So S = k² cot(x/2). Hence k = √(S tan(x/2)). Then the other sides have length k/sin(x/2). 

Problem 2

Find all positive integers m, n such that n + (n+1) + (n+2) + ... + (n+m) = 1000.

Answer

(m,n) = (15,55), (24,28), (4,198).

Solution

There are m+1 terms, average size (2n+m)/2. So (m+1)(2n+m) = 2000 = 16·125. Note that m+1 and 2n+m have opposite parity. Also 1 < m+1 < 2n+m. So we have m+1 = 16, 25 or 5. 

Problem 3

Find a polynomial with integer coefficients which has √2 + √3 and √2 + 3^1/3 as roots.

Answer

(x⁴ - 10x² + 1)(x⁶ - 6x⁴ - 6x³ + 12x² - 36x + 1) = 0

Solution

(x + √2 + √3)(x - √2 - √3) = x² - 5 - 2√6. Similarly, (x + √2 - √3)(x - √2 + √3) = x² - 5 + 2√6, so √2 + √3 is a root of (x² - 5 - 2√6)(x² - 5 + 2√6) = x⁴ - 10x² + 1.

√2 + 3^1/3 is a root of (x - √2)³ - 3 = 0, or (x³ + 6x - 3) - (3x² + 2)√2 = 0. Multiplying by (x³ + 6x - 3) + (3x² + 2)√2 we get (x³ + 6x - 3)² - 2(3x² + 2)2 = 0 or x⁶ - 6x⁴ - 6x³ + 12x² - 36x + 1 = 0.

So, (x⁴ - 10x² + 1)(x⁶ - 6x⁴ - 6x³ + 12x² - 36x + 1) = 0 has both required roots.