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Problem Set 4

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).
View solution
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.
View solution
21.
Solve the following system for real values of x and y:
 2x2 + y + 2x + y2 = 8
 Öx + Öy = 2 .
View solution

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.
View solution
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.
View solution
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.
View solution

Solutions to Problem Set 4

19.

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
 f(2u + 1) + f(2v + 1) = (3u + 1) + (3v + 1) = 3(u + v) + 2 < 3(u + v + 1) = 3(m + 1) .
Hence f(2(m + 1)) = 3(m + 1).

Suppose 2m + 3 is the sum of 2z and 2w + 1. Then z + w = m + 1 and
 f(2z) + f(2w + 1) = 3z + 3w + 1 = 3(z + w) + 1 = 3(m + 1) + 1 .
Hence f(2(m+1) + 1) = 3(m+1) + 1. The conjecture is established by induction.

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
 f(i+1) + f(j-1) = f(i) + 1 - f(j) - 2 = f(i) + f(j) - 1
while if i and j are both odd, then
 f(i+1) + f(j-1) = f(i) + 2 - f(j) - 1 = f(i) + f(j) + 1 .
Thus, f(2k+2) = f(i) + f(j) if and only if i and j are both even. In particular, f(2k+2) = f(2k) + f(2) = f(2k+1) - 1 + 3 = f(2k) + 2. We thus find that f(a+b) = f(a) + f(b) if and only if at least one of a and b is even.

20.

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
 (c - 1)(a - b) ³ 0      and        (1 - a)(c - b) ³ 0 .
If a = c = 1, then the inequalities hold for all values of b and so there is no largest value of b. If a = 1 < c, then the inequalities hold if and only if b £ a = 1, so that largest value of b is 1. If a < 1 = c, then the inequalities hold if and only if b £ c = 1, so the largest value of b is 1. Finally, if a < 1 < c, then the inequalities hold if and only if b £ a, so the largest value of b is a.

21.

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) = 2x2 + y + 2x + y2 and g(x, y) = Öx + Öy, we have that f(0, y) = 2y + 2y2 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.
 (Öx + Öy)2 = x + 2 __Öxy + y £ x + (x + y) + y = 2(x + y)
by the Arithmetic-Geometric Means Inequality. Hence
 Öx + Öy £ ______Ö2(x + y) .
Also, by the same AGM inequality,
 2x2 + y + 2x + y2 ³ 2 Ö 2x2 + y + x + y2 .
Now, using the inequality again, we find that
 x2 + y + x + y2 = (x2 + y2) + (x + y) ³ 12 (x + y)2 + (x + y)
so that
 2x2 + y + 2x + y2 ³ 21 + 1/4(x + y)2 +1/2(x + y) = 21/4[(x + y + 1)2 + 3] .
Suppose the (x, y) satisfies the system. Then
 ______Ö2(x + y) ³ 2 Þ (x + y) ³ 2
and
 14 [(x + y + 1)2 + 3] £ 3 Þ (x + y + 1)2 £ 9 Þ x + y + 1 £ 3 Þ x + y £ 2  .
Hence x + y = 2 and all inequalities are equalities. Therefore x = y = 1.

21.
Second solution. [A. Rodriguez] Wolog, we may assume that x ³ 1. Let Öx + Öy = 2; then y = (2 - Öx)2. Define
 g(x)
 = x + y2 + y + x2 = (2 - Öx)4 + x2 + x + (2 - Öx)2
 = 2x2 - 8x3/2 + 26x - 36x1/2 + 20 .
Then
 g¢(x)
 = 4x - 12x1/2 + 26 - 18x-1/2 = 2x-1/2 (2x3/2 - 6x + 13x1/2 - 9)
 = 2x-1/2(x1/2 - 1)(2x - 4x1/2 + 9) = 2x-1/2(x1/2 - 1)[2(x1/2 - 1)2 + 7] > 0
for x > 1. Hence g(x) is strictly increasing for x > 1, so that g(x) ³ g(1) = 4 for x ³ 1 with equality if and only if x = 1. Thus, if the first equation holds, then
 8 = 2x2 + y + 2x + y2 ³ 2 Ö 2g(x) Þ 16 ³ 2g(x) Þ g(x) £ 4  .
Hence g(x) = 4, so that x = 1 and y = 1. Thus, (x, y) = (1, 1) is the only solution.

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
 8 = 2x2 + y + 2x + y2 = 2u4 + 7u2 + 2 æç è 24u3 + 2u + 124u3 + 2u ö÷ ø ³ 22 (2) = 8
with equality if and only if u = 0. Since the extremes of this inequality are equal, we must have u = 0, so x = y = 1.

21.
Fourth solution. [C. Hsia] With Öx = 1 + u and Öy = 1 - u, we can write the first equation as
 24u3 + 2u + 124u3 + 2u = 21 - 7u2 - u4 .
Let z = 24u3 + 2u. We note that the quadratic z2 - 21 - 7u2 - u4 z + 1 = 0 is solvable, and so has nonnegative discriminant. Hence
 22 - 14u2 - 2u4 ³ 4 = 22 Þ -14u2 - 2u4 ³ 0Þ u = 0  .
Hence x = y = 1.

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
 f(x) + f(y) ³ 2f( 12 (x + y)) ³ 2f(1) = 4
so that x2 + x + y2 + y ³ 4. Equality occurs if and only if x = y = 1. Applying the Arithmetic-Geometric Means Inequality, we find that
 4 = 12 (2x2 + y + 2x + y2) ³ 21/2(x2 + y2 + x + y)
so that x2 + x + y2 + y £ 4. Hence x2 + x + y2 + y = 4 and so x = y = 1.

Comment. Note that 2(x2 + y2) £ (x + y)2 with equality if and only if x = y. Hence
 x2 + y2 + x + y ³ 12 (x + y)2 + (x + y) ³ 4
with equality if and only if x = y = 1. This avoids the use of the convexity of the function f.

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
 8
 = 2x2 + y + 2x + y2 ³ 2 Ö 2x2 + y + x + y2
 = 2 Ö 2(x+y)(x+y+1) - 2xy ³ 2 Ö 22·3 - 2·1 = 23 = 8
with equality if and only if x = y.

22.

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
 AC2 - AB2 = CH2 - BH2 = (CH - BH)(CH + BH) = (CH - HE)BC = CE ·BC = CE[2(AC - AB)] .
Hence AC + AB = 2CE. Also AC - AB = 1/2BC. Therefore 2AB + 1/2BC = 2CE.

 AB = AE = DE Þ 2DE + 12 BC = 2CEÞ BC = 4(CE - DE) = 4CD Þ BD = 3CD .
Conversely, suppose that BD = 3CD. Then
 BC = 4CD Þ 14 BC = CE - DE .
From the above,
 AB = CE - 14 BC = DE Þ AE = DE
 Þ ÐABD = ÐAEB = 2ÐADB .

Case (ii): Suppose ÐB = 90°. Then
 AC2 - AB2 = BC2 = 2(AC - AB)·BC ÞAC + AB = 2BC
 Þ 12 BC + AB + AB = 2BCÞ AB = 34 BC  .
 Þ BD = 34 BC Þ BD = 3CD .
 BD = 3CD Þ BD = 34 BC = ABÞ ÐADB = ÐBAD = 45° = 12 ÐABD  .

Case (iii): Suppose ÐB exceeds 90°. Let AH ^BC and E be on CH produced such that AE = AB. Then
 AC2 - CH2 = AB2 - BH2 Þ(AC - AB)(AC + AB) = CH2 - BH2 = (CH - BH)(CH + BH) = CB·CE
 Þ AC + AB = 2CE .
Also
 AC - AB = 12 BC Þ 2AB + 12 BC = 2CE Þ AB + 14 BC = CE  .

Also
Hence
 AB = ED Þ 2ED + 12 BC = 2CE ÞBC = 4(CE - DE) = 4CD Þ BD = 3CD  .
Conversely, suppose that BD = 3CD. Then BC = 4CD and ED = CE - CD = CE - 1/4BC = AB so that ED = AE and ÐEAD = ÐADE. Therefore

22.
Second solution. [R. Hoshino] Let ÐABD = 2q. By the Law of Cosines, with the usual conventions for a, b, c,
 1 - 2sin2 q
 = cos2q = c2 + 4(b-c)2 - b24c(b-c)
 = b-cc - b+c4c = 3b-5c4c (since  b ¹ c)
 Þ 3(b-c) = 6c - 8csin2 q
 Þ 3(b-c)2 sinq = c(3sinq-4sin3 q) = c sin3q
 Þ sinqc = 2 sin3 q3(b-c) .  (*)

Suppose now that D is selected so that ÐADB = q. Then, by the Law of Sines,
 sinqc = sin(180° - 3q)x = sin3qx
where x = |BD |. Comparison with (*) yields x = 1/2(3(b-c)) so 4BD = 3BC Þ BD = 3CD as desired.

On the other hand, suppose D is selected so that BD = 3CD. Then BD = 3/2(b-c). Let ÐADB = f. Then
sinf
c
= sin(180° - f- 2q)
 32 (b-c)
= sin(f+ 2q)
 32 (b-c)
.
Hence
 sin(f+ 2q)sinf = sin3qsinq
 Þ sinqsin(f+ 2q) = sin3qsinf
 Þ 12 [cos(q+ f) - cos(3q+ f)] = 12 [cos(3q- f) - cos(3q+ f)]
 Þ cos(q+ f) = cos(3q- f)
 Þ q+ f = ±(3q- f)      or    q+ f+ 3q- f = 360° .
The only viable possibility is q+ f = 3q- fÞ q = f as desired.

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
cosÐABC = 3b - 5c
4c
and    cosÐADB = 3b - c
 2 _______Ö2c(3b-c)
.
Then cos2ÐADB = (3b - 5c)/4c. Hence, either 2ÐADB = ÐABC or ÐABC + 2ÐADB = 360°. In the latter case, ÐABC + ÐADB = 360° - ÐADB > 180°, which is false. Hence ÐABC = 2ÐADB.

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,
 sin(2q+ f)3 = (1 + 2cos2q)sinf3
so that
 0
 = sinf+ 2sinfcos2q- sin(2q+ f)
 = sinf+ sinfcos2q- sin2qcosf
 = sinf+ sin(f- 2q) = 2sin(f- q) cosq  .
Since 0 £ |f- q| < 180°, we find that f = q as desired. The converse can be obtained as in the third solution.

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,
 (u + 3v)2 = (u + v)2 + (4v)2 - 8v(u + v)cosB
so that
 cosB = 8v2 - 4uv8v(u + v) = 2v - u2(u + v) .
Hence
 |AD |2 = (u+v)2 + (3v)2 - 6v(u+v)cosB = u2 + 5uv + 4v2 = (u + 4v)(u + v)   .

Since sin2 ÐABD = 1 - cos2 B = [3u(u+4v)]/[4(u+v)2], and, by the Law of Sines,
 sin2 ÐADBsin2 ÐABD = u+vu+4v ,
we have that
 sin2 ÐADB = 3u4(u + v) and    cos2 ÐADB = u + 4v4(u + v) .
Thus sin2 ÐABD = 4sin2 ÐADB cos2 ÐADB so that either ÐABD = 2ÐADB or ÐABD + 2ÐADB = 180°. The latter case would yield ÐADB = ÐBAD, so that AB = BD. This would make DABC a 3-4-5 right triangle and DABD an isosceles right triangle, whence 90° = ÐABD = 2ÐADB. The converse can be shown as in the previous solutions. The result follows.

23.

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
 |OP | = r ( [(R+r)/(R-r)] ) |OU | = [(4Rr2)/(R2 - r2)]
 |UP | = r ( [(R-r)/(R+r)] ) |AU | = [(2r)/(R+r)] Ö[(Rr)]
 |OQ | = R ( [(R+r)/(R-r)] ) |OV | = [(4R2r)/(R2 - r2)]
 |VQ | = R ( [(R-r)/(R+r)] ) |BV | = [(2R)/(R+r)] Ö[(Rr)]

The volume of the cone obtained by rotating OBV is

 13 p|BV |2 |OV | = 16 pR5 r23(R+r)3 (R-r)
and the volume of the cone obtained by rotating OAU is
 16 pR2 r53(R+r)3 (R-r)
so that the volume of the frustum obtained by rotating AUVB is
 16 pR2 r2 (R3 - r3)3 (R + r)3 (R - r) = 16 pR2 r23 (R + r)3 (R2 + Rr + r2)  .

The volume of a slice of a sphere of radius a and height h from the equatorial plane is
 p óõ h 0 (a2 - t2)dt = p[a2 h - h3/3] .
The portion of the larger sphere included within the frustum has volume
 2pR33
 - p éê ë R3 æç è R-rR+r ö÷ ø - R33 æç è R-rR+r ö÷ ø 3 ùú û
 = pR33 éê ë 2 - 3 æç è R-rR+r ö÷ ø + æç è R-rR+r ö÷ ø 3 ùú û
 = pR33 (R+r)3 [ 4r3 + 12Rr2 ] = 4pR2 r23 (R+r)3 [Rr + 3R2]
and the portion of the smaller sphere included within the frustum has volume
 2pr33 + p éê ë r3 æç è R-rR+r ö÷ ø - r33 æç è R-rR+r ö÷ ø 3 ùú û = 4pR2 r23 (R + r)3 [Rr + 3r2]   .
Hence, the portions of the sphere lying within the frustum have total volume
 4pR2 r23 (R + r)3 [3R2 + 2Rr + 3r2] .
Subtracting this from the volume of the frustum yields the volume of the ring-shaped region
 4pR2 r23 (R + r)3 [(4R2 + 4Rr + 4r2) - (3R2 + 2Rr + 3r2)] = 4pR2 r23 (R + r)3 [R2 + 2Rr + r2] = 4 pR2 r23(R + r) .

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
 pa2 h - 13 ph3  .

24.

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
 a1sina1 = a2sina2
and similarly that
 b1sinb1 = b2sinb2 and c1sing1 = c2sing2 .
Let a = ÐBAE. Then a = ÐFAC since ÐFAB = ÐEAC. Similarly, let b = ÐFBC = ÐABD and g = ÐBCE = ÐACD.

Let |AB | = c, |BC | = a, |AC | = b, |AD | = u, |BE | = v, |CF | = w. By the Law of Sines, we find that
 vsina = csinb1 and vsing = asinb2
so that
 c sinasinb1 = a singsinb2 Þ sinb1sinb2 = ca · sinasing .
Similarly
 sina1sina2 = bc · singsinb and sing1sing2 = ab · sinbsina .

Putting this altogether yields
 a1a2 · b1b2 · c1c2 = sina1sina2 · sinb1sinb2 · sing1sing2 = bc · ca · ab · singsinb · sinasing · sinbsina = 1 .
By the converse of Ceva's Theorem, the cevians AP, BQ and CR are concurrent and the result follows.

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