2nd Swedish Mathematical Society Problems 1962

1.  Find all polynomials f(x) such that f(2x) = f '(x) f ''(x).

2.  ABCD is a square side 1. P and Q lie on the side AB and R lies on the side CD. What are the possible values for the circumradius of PQR?

3.  Find all pairs (m, n) of integers such that n² - 3mn + m - n = 0.

4.  Which of the following statements are true?

(A) X implies Y, or Y implies X, where X is the statement, the lines L₁, L₂, L₃ lie in a plane, and Y is the statement, each pair of the lines L₁, L₂, L₃ intersect.

(B) Every sufficiently large integer n satisfies n = a⁴ + b⁴ for some integers a, b.

(C) There are real numbers a₁, a₂, ... , a_n such that a₁ cos x + a₂ cos 2x + ... + a_n cos nx > 0 for all real x.

5.  Find the largest cube which can be placed inside a regular tetrahedron with side 1 so that one of its faces lies on the base of the tetrahedron. 

Solutions

Problem 1

Find all polynomials f(x) such that f(2x) = f '(x) f ''(x).

Answer

4x³/9

Solution

Suppose deg f = n. Then deg f ' = n-1 and deg f " = n-2, so n = n-1+n-2. Hence n = 3. So put f(x) = ax³ + bx² + cx + d. Then we have 8ax³ + 4bx² + 2cx + d = (3ax² + 2bx + c)(6ax + 2b) = 18a²x³ + 18abx² + (6ac+4b²)x + 2bc. Comparing x³, a = 4/9. Then comparing x², b = 0. Comparing x, c = 0. Comparing constant term, d = 0. 

Problem 2

ABCD is a square side 1. P and Q lie on the side AB and R lies on the side CD. What are the possible values for the circumradius of PQR?

Answer

any value in [1/2, 1/√2]

Solution

Let O be the circumcenter. Then OP + OR ≥ PR ≥ AD = 1, so the radius is at least 1/2. P,Q,R always lie inside or on the circle through A,B,C,D which has radius 1/√2, so the radius is at most 1/√2.

Now take ∠RPQ = 90^o, then QR is a diameter of the circumcircle and so the circumradius is ½QR. By adjusting the positions of Q, R we can obviously get QR to have any value in [1,√2]. 

Problem 3

Find all pairs (m, n) of integers such that n² - 3mn + m - n = 0.

Answer

(m,n) = (0,0) or (0,1)

Solution

(3m+1)(3n-1) = 3n²-1, so if p divides 3n-1, then it also divides 3n²-1, and hence also (3n²-1) - n(3n-1) = n-1, and hence also (3n-1) - 3(n-1) = 2. So 3n-1 = 0, ±1 or ±2. But n must be an integer, so 0, 1, -2 do not work. -1 gives n = 0 and hence m = 0, which is a solution. 2 gives n = 1 and hence m = 0, which is a solution.