19.

Let f(1) = 1 and f(2) = 3. Suppose that, for n ³ 3, f(n) = max{ f(r) + f(n-r) : 1 £ r £ n-1 }. Determine necessary and sufficient conditions on the pair (a, b) that f(a + b) = f(a) + f(b).

20.

Suppose that a and c are fixed real numbers with a £ 1 £ c. Determine the largest value of b which is compatible with a + bc £ b + ac £ c + ab.

21.

Solve the following system for real values of x and y:

22.

Let ABC be a triangle with BC = 2·AC - 2·AB and D be a point on the side BC. Prove that ÐABD = 2ÐADB if and only if BD = 3CD.

23.

Two unequal spheres in contact have a common tangent cone. The three surfaces divide space into various parts, only one of which is bounded by all three surfaces; it is ``ring-shaped''. Being given the radii r and R of the spheres with r < R, find the volume of the ``ring-shaped'' region in terms of r and R.

24.

Let ABC be a triangle. Select points D, E, F outside of DABC such that DDBC, DEAC, DFAB are all isosceles with the equal sides meeting at these outside points and with ÐD = ÐE = ÐF. Prove that the lines AD, BE and CF all intersect in a common point.

19.

First solution. From the first few values of f(n), we conjecture that f(2k) = 3k and f(2k + 1) = 3k + 1 for each positive integer k. We establish this by induction. It is easily checked for k = 1. Suppose that it holds up to k = m.

Suppose that 2m + 2 is the sum of two positive even numbers 2x and 2y. Then f(2x) + f(2y) = 3(x + y) = 3(m + 1). If 2m + 2 is the sum of two positive odd numbers 2u + 1 and 2v + 1, then

Suppose 2m + 3 is the sum of 2z and 2w + 1. Then z + w = m + 1 and

By checking cases on the parity of a and b, one verifies that f(a + b) = f(a) + f(b) if and only if at least one of a and b is even. (If a and b are both odd, the left side is divisible by 3 while the right side is not.)

19.

Second solution. [K. Yeats] By inspection, we conjecture that f(n+1) = f(n) + 2 when n is odd, and f(n+1) = f(n) + 1 when n is even. This is true for n = 1, 2. Suppose it holds up to n = 2k. If 2k + 1 = i + j with i even and j odd, then f(i-1) + f(j+1) = f(i) - 2 + f(j) + 2 = f(i) + f(j) and f(i+1) + f(j-1) = f(i) + 1 + f(j) - 1 = f(i) + f(j) (where defined), so in particular f(2k+1) = f(2k) + f(1) = f(2k) + 1. Note that this also tells us that f(2k+1) = f(i) + f(j) whenever i + j = 2k+1. Now consider 2k+2 = i + j. If i and j are both even, then

20.

First solution. Observe that (b + ac) - (a + bc) = (c - 1)(a - b) and (c + ab) - (b + ac) = (1 - a)(c - b), so that the inequalities are equivalent to

21.

Preliminary comments. With the surds in the second equation, we must restrict ourselves to nonnegative values of x. Because of the complexity of the expressions, it is probably impossible to eliminate one of the variables and solve for the other. Let us make a few preliminary observations:

(i) (x, y) = (1, 1) is an obvious solution;

(ii) Both equations are symmetric in x and y;

(iii) Taking f(x, y) = 2^{x2 + y} + 2^{x + y2} and g(x, y) = Öx + Öy, we have that f(0, y) = 2^y + 2^y2 and g(0, y) = Öy; thus, f(0, y) = 8 Þ 1 < y < 2 and g(0, y) = 2Û y = 4. The graphs of f(x, y) = 8 and g(x, y) = 2 may look like the figures in the diagram.

This suggests that f(x, y) = 8 Þ x + y £ 2 and g(x, y) = 2 Þ x + y ³ 2 with equality for both Û (x, y) = (1, 1). Hence we look for a relationship among f(x, y), g(x, y) and x + y.

First solution.

21.

Second solution. [A. Rodriguez] Wolog, we may assume that x ³ 1. Let Öx + Öy = 2; then y = (2 - Öx)². Define

21.

Third solution. [S. Yazdani] Set Öx = 1 + u and Öy = 1 - u. Then x² + y = (1 + u)⁴ + (1 - u)² and x + y² = (1 - u)⁴ + (1 + u)², so

21.

Fourth solution. [C. Hsia] With Öx = 1 + u and Öy = 1 - u, we can write the first equation as

21.

Fifth solution. [M. Boase] 2(x + y) ³ (x + y)+ 2Ö[xy] = (Öx + Öy)² = 4 so that x + y ³ 2. Let f(t) = t(t + 1). For positive values of t, f(t) is an increasing strictly convex function of t. Hence

Comment. Note that 2(x² + y²) £ (x + y)² with equality if and only if x = y. Hence

21.

Sixth solution. [J. Chui] Wolog, let x ³ y so that Öx ³ 1 ³ Öy. Suppose that Öx = 1 + u and Öy = 1 - u. Then x + y = 2 +2u² ³ 2 and xy = (1 - u²)² £ 1. Thus

22.

First solution. Case (i): Suppose that ÐB is acute. Let AH ^BC and E lie on CH such that AE = AB.

AC² - CH² = AB² - BH² implies that

Suppose that ÐABD = 2ÐADB. Then ÐAEB = 2ÐADB Þ DADE is isosceles. Hence

Case (ii): Suppose ÐB = 90^°. Then

Case (iii): Suppose ÐB exceeds 90^°. Let AH ^BC and E be on CH produced such that AE = AB. Then

Let ÐABD = 2ÐADB. Then

22.

Second solution. [R. Hoshino] Let ÐABD = 2q. By the Law of Cosines, with the usual conventions for a, b, c,

Suppose now that D is selected so that ÐADB = q. Then, by the Law of Sines,

On the other hand, suppose D is selected so that BD = 3CD. Then BD = ³/₂(b-c). Let ÐADB = f. Then

22.

Third solution. [J. Chui] First, recall Stewart's Theorem. Let XYZ be a triangle with sides x, y, z respectively opposite XYZ. Let W be a point on YZ so that |XW | = u, |YW | = v and |ZW | = w. Then x(u² + vw) = vy² + wz². This is an immediate consequence of the Law of Cosines. Let q = ÐYWX. Then z² = u² + v² - 2uv cosq and y² = u² + w² + 2uw cosq. Multiply these equations by u and v respectively, add and use x = v + w to obtain the result.

Now to the problem. Suppose BD = 3CD. Let |AC | = 2b, |AB | = 2c, so that |BC | = 4(b - c), |BD | = 3(b - c) and |CD | = b - c. If |AD | = d, then an application of Stewart's Theorem yields d² = 2c(3b - c). Applying the Law of Cosines to DABC and DABD respectively yields

On the other hand, let 2ÐADB = ÐABC. If D¢ is a point on BC with BD¢ = 3CD¢, the 2 ÐAD¢B = ÐABC = 2 ÐADB, so that D = D¢. The result follows.

22.

Fourth solution. Let |AB | = a, |AC | = a + 2, |BD | = 3, |CD | = 1, ÐABD = 2q, ÐADB = f. Then (a + 2)² = a² + 16 - 8acos2q, whence a = 3(1 + 2cos2q)^-1 (so 0 < q < 60^°). By the Law of Sines,

22.

Fifth solution. [A. Birka] First, note that, when BD = 3CD, we must have ÐADB < 90^°, since AC > AB and D is on the same side of the altitude from A as C. Also, when ÐABD = 2ÐADB, ÐADB < 90^°. Thus, we can assume that ÐADB is acute throughout.

We can select positive numbers u, v and w so that |BC | = v + w, |AC | = u + w and |AB | = u + v. By hypothesis, v + w = 2(w - v), so that w = 3v.

Suppose that BD = 3CD. Then BC = 4CD, whence |CD | = v. Hence |BD | = 3v. By the Law of Cosines,

Since sin² ÐABD = 1 - cos² B = [3u(u+4v)]/[4(u+v)²], and, by the Law of Sines,

23.

First solution. The configuration described is obtained by rotating this configuration about the line passing through the centres P and Q of the spheres.

From a consideration of similar triangles and pythagoras theorem, we find that

The volume of the cone obtained by rotating OBV is

The volume of a slice of a sphere of radius a and height h from the equatorial plane is

Comment. The volume of a slice of a sphere of radius a and height h from the equatorial plane can be obtained from the volume of a right circular cone and a cylinder using the method of Cavalieri. The area of a cross-section of the slice at height t from the equator is p(a² - t²) = pa² - pt². The term pa² represents the cross-section of a cylinder of radius a and height h while pt² represents the area of the cross section of a cone of base radius h at distance t from the vertex. Thus the area of the each cross-section of the cylinder is the sum of the areas of the corresponding cross-sections of the spherical slice and cone. Cavalieri's principle says that the volumes of the solids bear the same relation. Thus the volume of the spherical slice is

24.

First solution. Suppose AD splits BC into segments of lengths a₁ and a₂ as in the diagram and angle BDC into subangles a₁ and a₂, and let b₁, b₂, b₁, b₂, c₁, c₂, g₁, g₂ be similarly defined as in the diagram.

Applying the Law of Sines to DBPD and DCPD, we find that

Let |AB | = c, |BC | = a, |AC | = b, |AD | = u, |BE | = v, |CF | = w. By the Law of Sines, we find that

Putting this altogether yields