-
139.
-
Let
,
,
be three pairwise orthogonal
faces of a tetrahedran meeting at one of its vertices and
having respective areas
,
,
. Let the face
opposite this vertex have area
. Prove that
Solution 1. Let the tetrahedron be bounded by the
three coordinate planes in
and the plane
with equation
,
where
are positive. The vertices of the tetrahedron
are
,
,
,
.
Let
,
,
,
be the areas of the faces opposite these
respective vertices. Then the volume
of the tetrahedron is
equal to
where
is the distance from the origin to its opposite face.
The foot of the perpendicular from the origin to this face is
located at
, where
, and its distance from the
origin is
. Since
,
,
and
, the result follows.
Solution 2. [J. Chui] Let edges of lengths
,
,
be common to the respective pairs of faces of areas
,
,
. Then
,
and
.
The fourth face is bounded by sides of length
,
and
. By Heron's
formula, its area
is given by the relation
whence the result follows.
Solution 3. Use the notation of Solution 2.
There is a plane through the edge bounding the
faces of areas
and
perpendicular to the edge bounding the
faces of areas
and
. Suppose it cuts the latter faces in
altitudes of respective lengths
and
. Then
, whence
.
Hence
so that
as desired.
Solution 4. [R. Ziman] Let a, b, c,
d be vectors orthogonal to the respective faces of areas
,
,
,
that point inwards from these faces and have
respective magnitudes
,
,
,
. 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
-
140.
-
Angus likes to go to the movies. On Monday,
standing in line, he noted that the fraction
of the
line was in front of him, while
of the line was behind
him. On Tuesday, the same fraction
of the line was
in front of him, while
of the line was behind him.
On Wednesday, the same fraction
of the line was in front
of him, while
of the line was behind him.
Determine a value of
for which this is possible.
Answer. When
, 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
, 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
,
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
for some positive integer exceeding 1.
Let
be the fraction of the line behind Angus.
Then Angus himself represents this fraction of the line:
so that there would be
people in line. To make this
an integer, we can arrange that
is a multiple of
.
For
, we want to get an integer for
,
and so we may take
to be any multiple of 6. Thus, we can
arrange that
is any of 5/6, 11/12, 17/18, 23/24, and so on.
More generally, for
,
and
to all
be integers we require that
be a multiple of 6, and so can
take
. On Monday, there would be
people in line with
in front and
behind;
on Tuesday,
with
in front and
behind; on Wednesday,
with
and
behind.
Solution 2. [O. Bormashenko] On the three successive days,
the total numbers numbers of people in line are
,
and
for some positive integers
,
and
. The fraction
of the line constituted by Angus and those behind him is
These yield the equations
and
We need to find an integer
for which
divides
and
divides
. This is equivalent to determining
for which
,
,
divides
,
and
divides
. The triple
works and yields
. In this case,
.
Comment 1. Solution 1 indicates how we can select
for which the amount of the line behind Angus is represented by
any number of consecutive integer reciprocals. For example, in
the case of
, he could also have
of a line of
156 behind him. Another strategy might be to look at
, i.e. successively at
. In this case, we assume that
is the
line is behind him, and need to ensure that
is a positive divisor
of
for three consecutive values of
. If
is odd, we can
achieve this with
any odd multiple of 15, starting with
.
Comment 2. With the same fraction in front on two days,
suppose that
of a line of
people is behind the
man on the first day, and
of a line of
people
is behind him on the second day. Then
so that
. This yields both
and
,
leading to
Two immediate possibilities are
and
. To get some
more, taking
, we get the quadratic equation
with discriminant
a pythagorean relationship when
is square and the
equation has integer solutions. Select
,
,
so
that
and let
; this will make the discriminant
equal to a square.
Taking
, for example, yields the possibilities
, (60, 10), (36, 9), (24, 8), (12, 6),
(6, 4),(4, 3). In general, we find that
when
with
. It turns out that
.
-
141.
-
In how many ways can the rational
be written as the product of two rationals of the form
, where
is a positive integer?
Solution 1. We begin by proving a more general result.
Let
be a positive integer, and denote by
and
, the number of positive divisors of
and
respectively. Suppose that
where
and
are positive integers exceeding
. Then
, which reduces to
. It follows that
and
, where
. Hence, every representation of
corresponds to a factorization of
.
On the other hand, observe that, if
, then
Hence, there is a one-one correspondence between representations
and pairs
of complementary factors of
.
Since
and
are coprime, the number of factors of
is equal to
, and so the number of
representations is equal to
.
Now consider the case that
. Since
,
; since
,
. Hence, the desired number of representations
is 64.
Solution 2. [R. Ziman] Let
be an arbitrary positive
integer. Then, since
is in lowest terms,
must be a multiple of
. Let
for some positive
integers
and
and
for some positive integers
and
, where
is the greatest common divisor of
and
; suppose that
and
, with
being
the greatest common divisor of
and
. Then, the
representation must have the form
where
and
. Hence
so that
and
Thus,
and
are uniquely determined. Note that we can
get a representation for any pair
of complementary factors
or
and
of complementary factors of
, and there
are
of selecting these. However, the selections
and
yield the
same representation, so that number of representations is
. The desired answer can now be found.
-
142.
-
Let
be such that
.
Prove that
.
Solution 1. [R. Barrington Leigh] We have that
Since
, we can divide this inequality
by
to obtain
Solution 2. [S.E. Lu]
whereupon a division by the positive quantity
yields that
.
Solution 3. [O. Bormashenko] Observe that
and that
, so that
. It
follows that
The given condition can be rewritten
Adding inequalities (1) and (2) yields
whence
.
Solution 4. [R. Furmaniak] We have that
from which the result follows upon division by
.
Solution 5. Let
. Since
, we have that
. Then
. Therefore,
Since
, having negative discriminant, is always
positive, the desired result follows.
Solution 6. [J. Chui] Suppose, if possible, that
. We can write
and
for
. Then
contrary to hypothesis. The result follows by contradiction.
Solution 7. Let
and
. Since
, we can write
and
, where
. The given equality is equivalent to
so it suffices to show that the right side does not exceed 1
to obtain the desired
.
Observe that
from which the desired result follows.
Solution 8. Begin as in Solution 7. Then
from which the result follows.
-
143.
-
A sequence whose entries are
and
has the
property that, if each
is replaced by
and each
by
, then the sequence remains unchanged. Thus, it starts
out as
. What is the
th term of the
sequence?
Solution. Let us define finite sequences as follows.
Suppose that
. Then, for each
,
is
obtained by replacing each
in
by
and
each
in
by
. Thus,
Each
is a prefix of
; in fact, it can be shown that,
for each
,
where
indicates juxtaposition. The respective number of
symbols in
for
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
is equal to 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378.
The
th entry in the given infinite sequence is equal to
the
th entry in
, which is equal to the
th
th entry in
. This in turn is
equal to the
th
th entry in
,
which is equal to the
th entry of
, or the third
entry of
. Hence, the desired entry is 0.
Comment. Suppose that
is the position of the
th
one, so that
and
. Let
be the number
of zeros up to and including the
th position, and so
is the number of ones up to and including the
th
position. Then we get the two equations
These two can be used to determine the positions of the ones by
stepping up; for example, we have
,
,
,
, 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
,
,
,
be rational numbers for which
. Prove that there are infinitely many rational values
of
for which
is rational. Explain the
situation when
.
Solution 1. We study the possibility of making
for some rational numbers
. This would require
that
Since the condition
prohibits
, at least
one of
and
must fail to vanish. Let us now construct our
solution.
Let
be an arbitrary positive rational number for which
. Then
and
, whence
is rational.
We need to show that distinct values of
deliver distinct values of
. Let
and
be two values of
for which
Then
so that
, and the result follows.
Consider the case that
. if both sides equal zero, then
one of the possibilities
must hold. In the
first two cases, any
will serve. In the third, any value of
will serve provided that
is a rational square, and in the
fourth, provided
is a rational square; otherwise, no
can
be found. Otherwise, let
, for some nonzero rational
, so that
. If
is a rational
square, any value of
will do; if
is irrational, then only
will work.
Solution 2.
for rational
is
equivalent to
If
, this is satisfiable by all rational
provided
is a rational square and
, and by no rational
otherwise.
If exactly one of
and
is zero and
,
then each positive rational value is assumed by
for a suitable value of
.
Otherwise, let
. Then, given
, we have the corresponding
If
, then this yields a rational
if and only if
is
a rational square. Let
. We wish to make
for some rational
. This is equivalent to
Pick a pair
of rationals for which
and
. We want to make
so that
and
.
Thus, let
Then
is a rational square, and so
is rational. There are infinitely
many possible ways of choosing
and each gives a different
sum
and so a different value of
and
. The desired
result follows.