location:
 Solutions

139.
Let A, B, C be three pairwise orthogonal faces of a tetrahedran meeting at one of its vertices and having respective areas a, b, c. Let the face D opposite this vertex have area d. Prove that

 d2 = a2 + b2 + c2 .

Solution 1. Let the tetrahedron be bounded by the three coordinate planes in R3 and the plane with equation x/u + y/v + z/w = 1, where u, v, w are positive. The vertices of the tetrahedron are (0, 0, 0), (u, 0, 0), (0, v, 0), (0, 0, w). Let d, a, b, c be the areas of the faces opposite these respective vertices. Then the volume V of the tetrahedron is equal to

 13 au = 13 bv = 13 cw = 13 dk ,
where k is the distance from the origin to its opposite face. The foot of the perpendicular from the origin to this face is located at ((um)-1, (vm)-1. (wm)-1), where m = u-2 + v-2 + w-2, and its distance from the origin is m-1/2. Since a = 3Vu-1, b = 3Vv-1, c = 3Vw-1 and d = 3Vm1/2, the result follows.

Solution 2. [J. Chui] Let edges of lengths x, y, z be common to the respective pairs of faces of areas (b, c), (c, a), (a, b). Then 2a = yz, 2b = zx and 2c = xy. The fourth face is bounded by sides of length u = Ö[(y2 + z2)], v = Ö[(z2 + x2)] and w = Ö[(x2 + y2)]. By Heron's formula, its area d is given by the relation

 16d2
 = (u + v + w)(u + v - w)(u - v + w)(-u + v + w)
 = [(u + v)2 - w2][(w2 - (u - v)2] = [2uv + (u2 + v2 - w2)][2uv - (u2 + v2 - w2)]
 = 2u2v2 + 2v2w2 + 2w2u2 - u4 - v4 - w4
 = 2(y2 + z2)(x2 + z2) + 2(x2 + z2)(x2 + y2) +2(x2 + y2)(x2 + z2)
 - (y2 + z2)2 - (x2 + z2)2 - (x2 + y2)2
 = 4x2y2 + 4x2z2 + 4y2z2 = 16a2 + 16b2 + 16c2  ,
whence the result follows.

Solution 3. Use the notation of Solution 2. There is a plane through the edge bounding the faces of areas a and b perpendicular to the edge bounding the faces of areas c and d. Suppose it cuts the latter faces in altitudes of respective lengths u and v. Then 2c = uÖ[(x2 + y2)], whence u2(x2 + y2) = x2y2. Hence

 v2 = z2 + u2 = x2y2 + x2z2 + y2z2x2 + y2 = 4(a2 + b2 + c2)x2 + y2 ,
so that

 2d = v ______Öx2 + y2 Þ4d2 = 4(a2 + b2 + c2) ,
as desired.

Solution 4. [R. Ziman] Let a, b, c, d be vectors orthogonal to the respective faces of areas a, b, c, d that point inwards from these faces and have respective magnitudes a, b, c, d. If the vertices opposite the respective faces are x, y, z, O, then the first three are pairwise orthogonal and 2c = x×y, 2b = z× x, 2c = x×y, and 2d = (z - y)×(z - x) = - (z×x) - (y×z) - (x×y). Hence d = - (a + b + c), so that

 d2 = d·d = (a + b + c)·(a + b + c) = a2 + b2 + c2 .

140.
Angus likes to go to the movies. On Monday, standing in line, he noted that the fraction x of the line was in front of him, while 1/n of the line was behind him. On Tuesday, the same fraction x of the line was in front of him, while 1/(n+1) of the line was behind him. On Wednesday, the same fraction x of the line was in front of him, while 1/(n+2) of the line was behind him. Determine a value of n for which this is possible.

Answer. When x = 5/6, he could have 1/7 of a line of 42 behind him, 1/8 of a line of 24 behind him and 1/9 of a line of 18 behind him. When x = 11/12, he could have 1/14 of a line of 84 behind him, 1/15 of a line of 60 behind him and 1/16 of a line of 48 behind him. When x = 13/15, he could have 1/8 of a line of 120 behind him, 1/9 of a line of 45 behind him and 1/10 of a line of 30 behind him.

Solution 1. The strategy in this solution is to try to narrow down the search by considering a special case. Suppose that x = (u-1)/u for some positive integer exceeding 1. Let 1/(u+p) be the fraction of the line behind Angus. Then Angus himself represents this fraction of the line:

 1 - æç è u-1u + 1u+p ö÷ ø = pu(u+p) ,
so that there would be u(u+p)/p people in line. To make this an integer, we can arrange that u is a multiple of p. For n = u + 1, we want to get an integer for p = 1, 2, 3, and so we may take u to be any multiple of 6. Thus, we can arrange that x is any of 5/6, 11/12, 17/18, 23/24, and so on.

More generally, for u(u+1), u(u+2)/2 and u(u+3)/3 to all be integers we require that u be a multiple of 6, and so can take n = 6k + 1. On Monday, there would be 36k2 + 6k people in line with 36k2 - 1 in front and 6k behind; on Tuesday, 18k2 + 6k with 18k2 + 3k - 1 in front and 3k behind; on Wednesday, 12k2 + 6k with 12k2 + 4k - 1 and 2k behind.

Solution 2. [O. Bormashenko] On the three successive days, the total numbers numbers of people in line are un, v(n+1) and w(n+2) for some positive integers u, v and w. The fraction of the line constituted by Angus and those behind him is

 1un + 1n = 1v(n+1) + 1n+1 = 1w(n+2) + 1n+2 .
These yield the equations

 (n-v)(n+1+u) = n(n+1)
and

 (n+1-w)(n+2+v) = (n+1)(n+2) .
We need to find an integer v for which n-v divides n(n+1) and n+2+v divides (n+1)(n+2). This is equivalent to determining p, q for which p + q = 2(n+1), p < n, p divides n(n+1), q > n+2 and q divides (n+1)(n+2). The triple (n, p, q) = (7, 4, 12) works and yields (u, v, w) = (6, 3, 2). In this case, x = 5/6.

Comment 1. Solution 1 indicates how we can select x for which the amount of the line behind Angus is represented by any number of consecutive integer reciprocals. For example, in the case of x = 11/12, he could also have 1/13 of a line of 156 behind him. Another strategy might be to look at x = (u-2)/u, i.e. successively at x = 3/5, 5/7, 7/9,¼. In this case, we assume that 1/(u-p) is the line is behind him, and need to ensure that u - 2p is a positive divisor of u(u-p) for three consecutive values of p. If u is odd, we can achieve this with u any odd multiple of 15, starting with p = 1/2(u-1).

Comment 2. With the same fraction in front on two days, suppose that 1/n of a line of u people is behind the man on the first day, and 1/(n+1) of a line of v people is behind him on the second day. Then

 1u + 1n = 1v + 1n+1
so that uv = n(n+1)(u-v). This yields both (n2 + n - v)u = (n2 + n)v and (n2 + n + u)v = (n2 + n)u, leading to

 u - v = u2n2 + n + u = v2n2 + n - v .
Two immediate possibilities are (n, u, v) = (n, n+1, n) and (n, u, v) = (n, n(n+1), 1/2n(n+1)). To get some more, taking u - v = k, we get the quadratic equation

 u2 - ku - k(n2 + n) = 0
with discriminant

 D = k2 + 4(n2 + n)k = [k + 2(n2 + n)]2 - 4(n2 + n)2 ,
a pythagorean relationship when D is square and the equation has integer solutions. Select a, b, g so that gab = n2 + n and let k = g(a2 + b2 - 2ab) = g(a-b)2; this will make the discriminant D equal to a square.

Taking n = 3, for example, yields the possibilities (u, v) = (132, 11), (60, 10), (36, 9), (24, 8), (12, 6), (6, 4),(4, 3). In general, we find that (n, u, v) = (n, ga(a- b),gb(a- b)) when n2 + n = gab with a > b. It turns out that k = u - v = g(a- b)2.

141.
In how many ways can the rational 2002/2001 be written as the product of two rationals of the form (n+1)/n, where n is a positive integer?

Solution 1. We begin by proving a more general result. Let m be a positive integer, and denote by d(m) and d(m+1), the number of positive divisors of m and m+1 respectively. Suppose that

 m+1m = p+1p · q+1q ,
where p and q are positive integers exceeding m. Then (m+1)pq = m(p+1)(q+1), which reduces to (p - m)(q - m) = m(m+1). It follows that p = m + u and q = m + v, where uv = m(m+1). Hence, every representation of (m+1)/m corresponds to a factorization of m(m+1).

On the other hand, observe that, if uv = m(m+1), then

 m+u+1m+u
 · m+v+1m+v = m2 + m(u + v + 2) + uv + (u + v) + 1m2 + m(u + v) + uv
 = m2 + (m+1)(u + v) + m(m+1) + 2m + 1m2 + m(u + v) +m(m+1)
 = (m+1)2 + (m+1)(u+v) + m(m+1)m2 + m(u+v) + m(m+1)
 = (m+1)[(m+1) + (u+v) + m]m[m + (u + v) + m + 1] = m+1m .
Hence, there is a one-one correspondence between representations and pairs (u, v) of complementary factors of m(m+1). Since m and m+1 are coprime, the number of factors of m(m+1) is equal to d(m)d(m+1), and so the number of representations is equal to 1/2d(m)d(m+1).

Now consider the case that m = 2001. Since 2001 = 3 ×23×29, d(2001) = 8; since 2002 = 2 ×7 ×11 ×13, d(2002) = 16. Hence, the desired number of representations is 64.

Solution 2. [R. Ziman] Let m be an arbitrary positive integer. Then, since (m+1)/m is in lowest terms, pq must be a multiple of m. Let m+1 = uv for some positive integers u and v and m = rs for some positive integers r and s, where r is the greatest common divisor of m and p; suppose that p = br and q = as, with s being the greatest common divisor of m and q. Then, the representation must have the form

 m+1m = aubr · bvas ,
where au = br + 1 and bv = as + 1. Hence

 bv = br + 1u s + 1 = brs + s + uu ,
so that b = b(uv - rs) = s + u and

 a = sr + ur + 1u = m+1-uru = v + r .
Thus, a and b are uniquely determined. Note that we can get a representation for any pair (u, v) of complementary factors or m+1 and (r, s) of complementary factors of m, and there are d(m+1)d(m) of selecting these. However, the selections { (u, v), (r, s) } and { (v, u), (s, r) } yield the same representation, so that number of representations is 1/2d(m+1)d(m). The desired answer can now be found.

142.
Let x, y > 0 be such that x3 + y3 £ x - y. Prove that x2 + y2 £ 1.

Solution 1. [R. Barrington Leigh] We have that

 x - y ³ x3 + y3 > x3 - y3 .
Since x - y ³ x3 + y3 > 0, we can divide this inequality by x - y to obtain

 1 > x2 + xy + y2 > x2 + y2 .

Solution 2. [S.E. Lu]

 x - y
 ³ x3 + y3 > x3 > x3 - [y3 + xy(x - y)]
 = x3 - x2 y + xy2 - y3 = (x2 + y2)(x - y) ,
whereupon a division by the positive quantity x - y yields that 1 > x2 + y2.

Solution 3. [O. Bormashenko] Observe that y < x and that x3 < x3 + y3 £ x - y < x, so that 0 < y < x < 1. It follows that

 x(x + y) < 2 Þ xy(x + y) < 2xy .
(1)
The given condition can be rewritten

 (x + y)(x2 + y2) - xy(x + y) £ x - y  .
(2)
Adding inequalities (1) and (2) yields

 (x + y)(x2 + y2) < x + y ,
whence x2 + y2 < 1.

Solution 4. [R. Furmaniak] We have that

 (x - y)(1 - x2 - y2)
 = (x - y) -(x3 - x2y + xy2 - y3)
 ³ (x3 + y3) - (x3 - x2y + xy2 - y3) = 2y3 + x2y - xy2 = y(x2 - xy + 2y2)
 = y[(x - Ö2y)2 + (2Ö2 - 1)xy] ³ 0 ,
from which the result follows upon division by x - y.

Solution 5. Let y = tx. Since x > y > 0, we have that 0 < t < 1. Then x3(1 + t3) £ x(1 - t) Þx2(1 + t3) £ (1 - t). Therefore,

 x2 + y2
 = x2(1 + t2) £ æç è 1-t1 + t3 ö÷ ø (1 + t2)
 = 1 - t + t2 - t31 + t3 = 1 - t(1 - t + 2t2)1 + t3 .
Since 1 - t + 2t2, having negative discriminant, is always positive, the desired result follows.

Solution 6. [J. Chui] Suppose, if possible, that x2 + y2 = r2 > 1. We can write x = rsinq and y = rcosq for 0 £ q £ p/2. Then

 x3 + y3 - (x - y)
 = r3 sin3 q+ r3 cos3 q- rsinq+ rcosq
 > rsinq(sin2 q- 1) + rcos3 q+ rcosq
 = -rsinqcos2 q+ rcos3 q+ rcosq
 = rcos2 q æç è cosq+ 1cosq -sinq ö÷ ø
 > rcos2 q(2 - sinq) > 0 ,
contrary to hypothesis. The result follows by contradiction.

Solution 7. Let r > 0 and r2 = x2 + y2. Since x > y > 0, we can write x = rcosq and y = rsinq, where 0 < q < p/4. The given equality is equivalent to

 r2 £ cosq- sinqcos3 q+sin3 q ,
so it suffices to show that the right side does not exceed 1 to obtain the desired r2 £ 1.

Observe that

 1 - cosq- sinqcos3 q+ sin3 q
 = (cosq+ sinq)(1 - cosqsinq)- (cosq- sinq)cos3 q+ sin3 q
 = sinq(2 - cosqsinq- cos2 q)cos3 q+ sin3 q > 0  ,
from which the desired result follows.

Solution 8. Begin as in Solution 7. Then

 cosq- sinqcos3 q+ sin3 q
 = cos2 q- sin2 q(cosq+ sinq)2(1 - cosqsinq)
= cos2q
 (1 + sin2q)(1 - 12 sinq
) = cos2q
 1 + 12 sin2q(1 - sin2q)
< 1 ,
from which the result follows.

143.
A sequence whose entries are 0 and 1 has the property that, if each 0 is replaced by 01 and each 1 by 001, then the sequence remains unchanged. Thus, it starts out as 010010101001 ¼. What is the 2002th term of the sequence?

Solution. Let us define finite sequences as follows. Suppose that S1 = 0. Then, for each k ³ 2, Sk is obtained by replacing each 0 in Sk-1 by 01 and each 1 in Sk-1 by 001. Thus,

 S1 = 0;  S2 = 01;  S3 = 01001;  S4 = 010010101001; S5 = 01001010100101001010010101001; ¼
Each Sk-1 is a prefix of Sk; in fact, it can be shown that, for each k ³ 3,

 Sk = Sk-1 * Sk-2 * Sk-1 ,
where * indicates juxtaposition. The respective number of symbols in Sk for k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 is equal to 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378.

The 2002th entry in the given infinite sequence is equal to the 2002th entry in S10, which is equal to the (2002-985-408)th =(609)th entry in S9. This in turn is equal to the (609-408-169)th = (32)th entry in S8, which is equal to the (32)th entry of S6, or the third entry of S3. Hence, the desired entry is 0.

Comment. Suppose that f(n) is the position of the nth one, so that f(1) = 2 and f(2) = 5. Let g(n) be the number of zeros up to and including the nth position, and so n - g(n) is the number of ones up to and including the nth position. Then we get the two equations

 f(n) = 2g(n) + 3(n - g(n)) = 3n - g(n)
(1)

 g(f(n)) = f(n) - n .
(2)
These two can be used to determine the positions of the ones by stepping up; for example, we have f(2) = 5, g(5) = 3, f(5) = 15 - 3 = 12, g(12) = f(5) - 5 = 7, and so on. By messing around, one can arrive at the result, but it would be nice to formulate this approach in a nice clean efficient zeroing in on the answer.

144.
Let a, b, c, d be rational numbers for which bc ¹ ad. Prove that there are infinitely many rational values of x for which Ö[((a + bx)(c + dx))] is rational. Explain the situation when bc = ad.
Solution 1. We study the possibility of making c + dx = (a + bx)t2 for some rational numbers t. This would require that

 x = c - at2bt2 - d .
Since the condition bc ¹ ad prohibits b = d = 0, at least one of b and d must fail to vanish. Let us now construct our solution.

Let t be an arbitrary positive rational number for which bt2 ¹ d. Then a + bx = (bc - ad)(bt2 - d)-1 and c + dx = (bc - ad)t2(bt2 - d)-1, whence

 ____________Ö(a + bx)(c + dx) = |(bc - ad) t (bt2 - d)-1 |
is rational.

We need to show that distinct values of t deliver distinct values of x. Let u and v be two values of t for which

 c - au2bu2 - d = c - av2bv2 - d .
Then

 0
 = (c - au2)(bv2 - d) - (c - av2)(bu2 - d)
 = bc(v2 - u2) + ad(u2 - v2) = (bc - ad)(u2 - v2)  ,
so that u = v, and the result follows.

Consider the case that bc = ad. if both sides equal zero, then one of the possibilities (a, b, c, d) = (0, 0, c, d),(a, b, 0, 0), (a, 0, c, 0), (0, b, 0, d) must hold. In the first two cases, any x will serve. In the third, any value of x will serve provided that ac is a rational square, and in the fourth, provided bd is a rational square; otherwise, no x can be found. Otherwise, let c/a = d/b = s, for some nonzero rational s, so that (a + bx)(c + dx) = s(a + bx)2. If s is a rational square, any value of x will do; if s is irrational, then only x = -a/b = -c/d will work.

Solution 2. (a + bx)(c + dx) = r2 for rational r is equivalent to

 bdx2 + (ad + bc)x + (ac - r2) = 0 .
If b = d = 0, this is satisfiable by all rational x provided ac is a rational square and r2 = ac, and by no rational x otherwise. If exactly one of b and d is zero and ad + bc ¹ 0, then each positive rational value is assumed by Ö[((a + bx)(c + dx))] for a suitable value of x.

Otherwise, let bd ¹ 0. Then, given r, we have the corresponding

x =

2bd
.
If ad = bc, then this yields a rational x if and only if bd is a rational square. Let ad ¹ bc. We wish to make (ad - bc)2+ 4bdr2 = s2 for some rational s. This is equivalent to

 4bdr2 = (ad - bc)2 - s2 = (ad - bc + s)(ad - bc - s) .

Pick a pair u, v of rationals for which u + v ¹ 0 and uv = bd. We want to make

 2ur = ad - bc + s           and            2vr = ad - bc - s
so that (u + v)r = ad - bc and s = (u - v)r. Thus, let

 r = ad - bcu + v .
Then