December Exam 2

This is a 4 hour paper.

Various questions from Ukraine

1. Does there exist a function f : ℝ → ℝ such that for all x, y ∈ ℝ the following equality holds:

f(xy) = max{f(x), y} + min{f(y), x}?

Solution Put x = y = 1 to obtain that

f(1) = max{f(1), 1} + min{f(1), 1} = f(1) + 1

which is absurd.

2. Positive integers a and n are such that n divides a^{2} + 1. Prove that there is a positive integer b such that n(n^{2} + 1) divides b^{2} + 1. Solution We have a^{2} ≡ −1 mod n and n^{2} ≡ −1 mod n^{2} +1. Now n, n^{2} + 1 are coprime (a common divisor would divide 1) so we may apply the Chinese Remainder Theorem to find an integer b such that b ≡ a mod n and b ≡ n mod n^{2}+1 so b^{2} ≡ a^{2} ≡ −1 mod n and b^{2} ≡ n^{2} ≡ −1 mod n^{2}+ 1. Thus b^{2} ≡ −1 mod n(n^{2} +1) and therefore n(n^{2} +1) | b^{2} +1. Of course, if you are Tom Coker or Paul Jefferys, then you are at liberty to sidestep the Chinese Remainder Theorem and simply hurl a bolt of lightning by letting b = an^{2} + a + n which visibly has the appropriate divisibility properties!

3. An acute angled triangle ABC is such that AC and BC are of different lengths, and is inscribed in a circle ω. Let N be the midpoint of the arc AC (and B does not lie on this arc). Let M be the midpoint of the arc BC (and A does not lie on this arc). Let D be the point on the arc MN such that DC ∥ NM. Let K be an arbitrary point on the arc AB (and C does not lie on this arc). Let O, O_{1} and O_{2} be the incentres of the triangles ABC, ACK, CBK respectively. Suppose that L is the intersection of the line DO and the circle ω such that L ≠ D. Prove that the points K, O_{1}, O_{2}, L are concyclic. So far no solution has ben offered.

4. Let a, b, c and α, β, γ be positive real numbers such that α + β + γ = 1. Prove the inequality

αa + βb + γc + 2\sqrt{(αβ + βγ + γα)(ab + bc + ca)} ≤ a + b + c.

Solution (Thanks to Tim Austin and Paul Jefferys) Replacing a by a/(a+ b + c), b by b/(a + b + c) and c by c/(a + b + c) we see that it suffices to assume that a + b + c = 1 and to show that

αa + βb + γc + 2\sqrt{(αβ + βγ + γα)(ab + bc + ca)} ≤ 1.

Now by the Cauchy-Schwarz inequality,

αa + βb + γc ≤ \sqrt{α^{2} + β^{2} + γ^{2}}\sqrt{a^{2} + b^{2} + c^{2}} (†)

Let p = αβ + βγ + γα and q = ab + bc + ca. Now α^{2} + β^{2} + γ^{2} = 1 − 2p and a^{2} + b^{2} + c^{2} = 1 − 2q. Now (†) implies that

αa + βb + γc + 2\sqrt{pq} ≤ \sqrt{1 − 2p}\sqrt{1 − 2q} + 2\sqrt{pq}
≤ (1 − 2p + 1 − 2q)/2 + p + q = 1

using the GM-AM inequality.