-
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
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
|
|
1
3
|
au = |
1
3
|
bv = |
1
3
|
cw = |
1
3
|
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
|
|
|
= (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 + y2z2
x2 + 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 - |
æ
ç
è
|
|
u-1
u
|
+ |
1
u+p
|
|
ö
÷
ø
|
= |
p
u(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
|
|
1
un
|
+ |
1
n
|
= |
1
v(n+1)
|
+ |
1
n+1
|
= |
1
w(n+2)
|
+ |
1
n+2
|
. |
|
These yield the equations
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
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 = |
u2
n2 + n + u
|
= |
v2
n2 + 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
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
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+v+1
m+v
|
= |
m2 + m(u + v + 2) + uv + (u + v) + 1
m2 + m(u + v) + uv
|
|
| |
|
= |
m2 + (m+1)(u + v) + m(m+1) + 2m + 1
m2 + 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+1
m
|
. |
|
|
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
where au = br + 1 and bv = as + 1. Hence
|
bv = |
br + 1
u
|
s + 1 = |
brs + s + u
u
|
, |
|
so that
b = b(uv - rs) = s + u and
|
a = |
sr + ur + 1
u
|
= |
m+1-ur
u
|
= 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]
|
|
|
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) -(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(1 + t2) £ |
æ
ç
è
|
|
1-t
1 + t3
|
|
ö
÷
ø
|
(1 + t2) |
| |
| = |
1 - t + t2 - t3
1 + 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
|
|
|
= r3 sin3 q+ r3 cos3 q- rsinq+ rcosq |
| |
|
> rsinq(sin2 q- 1) + rcos3 q+ rcosq |
| |
|
= -rsinqcos2 q+ rcos3 q+ rcosq |
| |
|
= rcos2 q |
æ
ç
è
|
cosq+ |
1
cosq
|
-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- sinq
cos3 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- sinq
cos3 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
|
|
|
= |
cos2 q- sin2 q
(cosq+ sinq)2(1 - cosqsinq)
|
|
| |
|
|
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) |
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
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 - au2
bu2 - d
|
= |
c - av2
bv2 - d
|
. |
|
Then
|
|
|
= (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 = |
|
-(ad + bc) ± |
| _______________
Ö(ad - bc)2 + 4bdr2
|
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
Then
|
|
| |
|
= |
æ
ç
è
|
|
ad - bc
u + v
|
|
ö
÷
ø
|
2
|
[(u + v)2 - 4uv] |
| |
| = |
(ad - bc)2(u - v)2
(u+v)2
|
|
|
|
is a rational square, and so x is rational. There are infinitely
many possible ways of choosing u, v and each gives a different
sum u + v and so a different value of r and x. The desired
result follows.
