Solutions.

304.

Prove that, for any complex numbers z and w,

( |z |+ |w |) ê ê  z |z | +  w |w | ê ê £ 2 |z + w | .

Solution 2. Let z = ae^{i a} and w = be^{i b}, with a and b real and positive. Then the left side is equal to

Observe that

Solution 3. Let z = ae^{i a} and w = be^{i b}, where a and b are positive reals. Then the inequality is equivalent to

Solution 4. [G. Goldstein] Observe that, for each m Î C,

Let t = r(cosq+ i sinq). Then the inequality becomes

Solution 5. [R. Mong] Consider complex numbers as vectors in the plane. q = (|z |/|w |)w is a vector of magnitude z in the direction w and p = (|w |/|z |)z is a vector of magnitude w in the direction z. A reflection about the angle bisector of vectors z and w interchanges p and w, q and z. Hence |p + q | = |w + z |. Therefore

305.

Suppose that u and v are positive integer divisors of the positive integer n and that uv < n. Is it necessarily so that the greatest common divisor of n/u and n/v exceeds 1?

Solution 2. Let g = gcd (u, v), u = gs and v = gt. Then gst £ g²st < n so that st < n/g. Now s and t are a coprime pair of integers, each of which divides n/g. Therefore, n/g = dst for some d > 1. Therefore n/u = n/(gs) = dt and n/v = n/(gt) = ds, so that n/u and n/v are divisible by d, and so their greatest common divisor exceeds 1.

Solution 3. uv < n Þnuv < n² Þ n < (n/u)(n/v). Suppose, if possible, that n/u and n/v have greatest common divisor 1. Then the least common multiple of n/u and n/v must equal (n/u)(n/v). But n is a common multiple of n/u and n/v, so that (n/u)(n/v) £ n, a contradiction. Hence the greatest common divisor of n/u and n/v exceeds 1.

Solution 4. Let P be the set of prime divisors of n, and for each p Î P. Let a(p) be the largest integer k for which p^k divides n. Since u and v are divisors of n, the only prime divisors of either u or v must belong to P. Suppose that b(p) is the largest value of the integer k for which p^k divides uv.

If b(p) ³ a(p) for each p Î P, then n would divide uv, contradicting uv < n. (Note that b(p) > a(p) may occur for some p.) Hence there is a prime q Î P for which b(q) < a(q). Then q^a(q) is not a divisor of either u or v, so that q divides both n/u and n/v. Thus, the greatest common divisor of n/u and n/v exceeds 1.

Solution 5. [D. Shirokoff] If n/u and n/v be coprime, then there are integers x and y for which (n/u)x + (n/v) = 1, whence n(xv + yu) = uv. Since n and uv are positive, then so is the integer xv + yu. But uv < nÞ 0 < xv + yu < 1, an impossibility. Hence the greatest common divisor of n/u and n/v exceeds 1.

306.

The circumferences of three circles of radius r meet in a common point O. They meet also, pairwise, in the points P, Q and R. Determine the maximum and minimum values of the circumradius of triangle PQR.

Solution 1. [M. Lipnowski] ÐQPO = ÐQRO, since OQ is a common chord of two congruent circles, and so subtends equal angles at the respective circumferences. (Why are angle QPO and QRO not supplementary?) Similarly, ÐOPR = ÐOQR. Let P¢ be the reflected image of P in the line QR so that triangle P¢QR and PQR are congruent. Then

Solution 2. [P. Shi; A. Wice] Let U, V, W be the centres of the circle. Then OVPW is a rhombus, so that OP and VW intersect at right angles. Let H, J, K be the respective intersections of the pairs (OP, VW), (OQ, UW), (OP, UV). Then H (respectively J, K) is the midpoint of OP and VW (respectively OQ and UW, OP and UV). Triangle PQR is carried by a dilation with centre O and factor ¹/₂ onto HJK. Also, HJK is similar with factor ¹/₂ to triangle UVW (determined by the midlines of the latter triangle). Hence triangles PQR and UVW are congruent. But the circumcircle of triangle UVW has centre O and radius r, so the circumradius of triangle PQR is also r.

Solution 3. [G. Zheng] Let U, V, W be the respective centres of the circumcircles of OQR, ORP, OPQ. Place O at the centre of coordinates so that

Solution 4. Let U, V, W be the respective centres of the circles QOR, ROP, POQ. Suppose that ÐOVR = 2b; then ÐOPR = b. Suppose that ÐOWQ = 2g; then ÐOPQ = g. Hence ÐQPR = b+ g. Let r be the circumradius of triangle PQR. Then |QR | = 2rsin(b+ g).

Consider triangle QUR. The reflection in the axis OQ takes W to U so that ÐQUO = ÐQWO = 2g. Similarly, ÐRUO = 2g, whence ÐQUR = 2(b+ g). Thus triangle QUR is isosceles with |QU | = |QR | = r and apex angle QUR equal to 2 (b+ g). Hence |QR | = 2rsin(b+ g). It follows that r = r.

Comment. This problem was the basis of the logo for the 40th International Mathematical Olympiad held in 1999 in Romania.

307.

Let p be a prime and m a positive integer for which m < p and the greatest common divisor of m and p is equal to 1. Suppose that the decimal expansion of m/p has period 2k for some positive integer k, so that

m p = .ABABABAB ... = (10k A + B)(10-2k +10-4k + ¼)

where A and B are two distinct blocks of k digits. Prove that

A + B = 10k - 1 .

(For example, 3/7 = 0.428571 ... and 428 + 571 = 999.)

Comment. This problem appeared in the College Mathematics Journal 35 (2004), 26-30. In writing up the solution, it is clearer to set up the equation and clear fractions, so that you can argue in terms of factors of products.

308.

Let a be a parameter. Define the sequence { f_n (x) : n = 0, 1, 2, ¼} of polynomials by

f0 (x) º 1

fn+1 (x) = x fn (x) + fn (ax)

for n ³ 0.

(a) Prove that, for all n, x,

fn (x) = xn fn (1/x) .

(b) Determine a formula for the coefficient of x^k (0 £ k £ n) in f_n (x).

When a = 1, it can be established by induction that f_n (x) = (x + 1)ⁿ = å_k=0ⁿ (n || k) xⁿ. Also, when a = 0, f_n (x) = xⁿ + x^n-1 + ¼+ x + 1 = (xⁿ⁺¹ - 1)(x - 1)^-1. Thus, (a) holds in these cases and b(n, k) is respectively equal to (n || k) and 1.

Suppose, henceforth, that a ¹ 1. For n ³ 0,

We conjecture what the coefficients b(n, k) are from an examination of the first few terms of the sequence:

We establish this conjecture. Let c(n, k) be the right side of (3) for 1 £ k £ n-1 and c(n, n) = 1. Then c(n, 0) = b(n, 0) = c(n, n) = b(n, n) = 1 for each n. In particular, c(n, k) = b(n, k) when n = 1.

We show that

Finally, (a) can be established in a straightforward way, either from the formula (3) or using the pair of recursions (1) and (2).

Solution 2. (a) Observe that f₀ (x) = 1, f₁ (x) = x + 1 and f₁ (x) - f₀ (x) = x = a⁰ x f₀ (x/a). Assume as an induction hypothesis that f_k (x) = x^k f(1/x) and

Then

Comment. Because of the appearance of the factor a - 1 in denominators, you should dispose of the case a = 1 separately. Failure to do so on a competition would likely cost a mark.

309.

Let ABCD be a convex quadrilateral for which all sides and diagonals have rational length and AC and BD intersect at P. Prove that AP, BP, CP, DP all have rational length.

Note that |AB |² = p² + q², |AC |² = c², |BC |² = (p² - 2pc + c²)+ q², |CD |² = (c² - 2cr + r²) + s² and |AD |² = r² + s², we have that

Similarly

Solution 2. By the cosine law, the cosines of all of the angles of the triangle ACD, BCD, ABC and ABD are rational. Now

310.

(a) Suppose that n is a positive integer. Prove that

(x + y)n = n å k=0  æ è n k ö ø x (x + k)k-1 (y - k)n-k .

(b) Prove that

(x + y)n = n å k=0  æ è n k ö ø x (x - kz)k-1 (y + kz)n-k .

The establishment of the identities involves the recognition of a certain sum which arise in the theory of finite differences. Let f(x) be a function of x and define the following operators that take functions to functions:

Now let f(x) be a polynomial of degree d ³ 0. If f(x) is constant (d = 0), then Df(x) = 0. If d ³ 1, then Df(x) is a polynomial of degree d - 1. It follows that D^d f(x) is constant, and Dⁿ f(x) = 0 whenever n > d. This yields the identity

Solution 1. [G. Zheng]

Let m = n - j so that 1 £ m £ n. Then

Solution 2. [M. Lipnowski] We prove that

That f¢(y) = g¢(y) is a consequence of the induction hypothesis and the identity (n || k)(n - k) = n ((n-1) || k). Also