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

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 $((\mathrm{um}){}^{-1},(\mathrm{vm}){}^{-1}.(\mathrm{wm}){}^{-1})$, where $m={u}^{-2}+{v}^{-2}+{w}^{-2}$, and its distance from the origin is ${m}^{-1/2}$. Since $a=3{\mathrm{Vu}}^{-1}$, $b=3{\mathrm{Vv}}^{-1}$, $c=3{\mathrm{Vw}}^{-1}$ and $d=3{\mathrm{Vm}}^{1/2}$, the result follows.

whence the result follows.

so that

as desired.

- 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.

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
$36{k}^{2}+6k$
people in line with
$36{k}^{2}-1$ in front and
$6k$ behind;
on Tuesday,
$18{k}^{2}+6k$ with
$18{k}^{2}+3k-1$ in front and
$3k$ behind; on Wednesday,
$12{k}^{2}+6k$ with
$12{k}^{2}+4k-1$ and
$2k$ behind.

These yield the equations

and

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, $x=5/6$.

so that $\mathrm{uv}=n(n+1)(u-v)$. This yields both $({n}^{2}+n-v)u=({n}^{2}+n)v$ and $({n}^{2}+n+u)v=({n}^{2}+n)u$, leading to

Two immediate possibilities are $(n,u,v)=(n,n+1,n)$ and $(n,u,v)=(n,n(n+1),\frac{1}{2}n(n+1))$. To get some more, taking $u-v=k$, we get the quadratic equation

with discriminant

a pythagorean relationship when $\Delta $ is square and the equation has integer solutions. Select $\alpha $, $\beta $, $\gamma $ so that $\gamma \alpha \beta ={n}^{2}+n$ and let $k=\gamma ({\alpha}^{2}+{\beta}^{2}-2\alpha \beta )=\gamma (\alpha -\beta ){}^{2}$; this will make the discriminant $\Delta $ 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
$(n,u,v)=(n,\gamma \alpha (\alpha -\beta ),\gamma \beta (\alpha -\beta ))$ when
${n}^{2}+n=\gamma \alpha \beta $ with
$\alpha >\beta $. It turns out that
$k=u-v=\gamma (\alpha -\beta ){}^{2}$.

- 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?

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

On the other hand, observe that, if
$\mathrm{uv}=m(m+1)$, then

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 $\frac{1}{2}d(m)d(m+1)$.

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

where $\mathrm{au}=\mathrm{br}+1$ and $\mathrm{bv}=\mathrm{as}+1$. Hence

so that $b=b(\mathrm{uv}-\mathrm{rs})=s+u$ and

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 $\frac{1}{2}d(m+1)d(m)$. The desired answer can now be found.

- 142.
- Let $x,y>0$ be such that ${x}^{3}+{y}^{3}\le x-y$. Prove that ${x}^{2}+{y}^{2}\le 1$.

Since $x-y\ge {x}^{3}+{y}^{3}>0$, we can divide this inequality by $x-y$ to obtain

whereupon a division by the positive quantity $x-y$ yields that $1>{x}^{2}+{y}^{2}$.

The given condition can be rewritten

Adding inequalities (1) and (2) yields

whence ${x}^{2}+{y}^{2}<1$.

from which the result follows upon division by $x-y$.

Since $1-t+2{t}^{2}$, having negative discriminant, is always positive, the desired result follows.

contrary to hypothesis. The result follows by contradiction.

so it suffices to show that the right side does not exceed 1 to obtain the desired ${r}^{2}\le 1$.

Observe that

from which the desired result follows.

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\dots $. What is the $2002$th term of the sequence?

Each ${S}_{k-1}$ is a prefix of ${S}_{k}$; in fact, it can be shown that, for each $k\ge 3$,

where $*$ indicates juxtaposition. The respective number of symbols in ${S}_{k}$ 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
$2002$th entry in the given infinite sequence is equal to
the
$2002$th entry in
${S}_{10}$, which is equal to the
$(2002-985-408)$th
$=(609)$th entry in
${S}_{9}$. This in turn is
equal to the
$(609-408-169)$th
$=(32)$th entry in
${S}_{8}$,
which is equal to the
$(32)$th entry of
${S}_{6}$, or the third
entry of
${S}_{3}$. Hence, the desired entry is 0.

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 $\mathrm{bc}\ne \mathrm{ad}$. Prove that there are infinitely many rational values of $x$ for which $\sqrt{(a+\mathrm{bx})(c+\mathrm{dx})}$ is rational. Explain the situation when $\mathrm{bc}=\mathrm{ad}$.

Since the condition $\mathrm{bc}\ne \mathrm{ad}$ 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
${\mathrm{bt}}^{2}\ne d$. Then
$a+\mathrm{bx}=(\mathrm{bc}-\mathrm{ad})({\mathrm{bt}}^{2}-d){}^{-1}$ and
$c+\mathrm{dx}=(\mathrm{bc}-\mathrm{ad}){t}^{2}({\mathrm{bt}}^{2}-d){}^{-1}$, whence

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

Then

so that $u=v$, and the result follows.

Consider the case that
$\mathrm{bc}=\mathrm{ad}$. 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
$\mathrm{ac}$ is a rational square, and in the
fourth, provided
$\mathrm{bd}$ is a rational square; otherwise, no
$x$ can
be found. Otherwise, let
$c/a=d/b=s$, for some nonzero rational
$s$, so that
$(a+\mathrm{bx})(c+\mathrm{dx})=s(a+\mathrm{bx}){}^{2}$. 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.

If $b=d=0$, this is satisfiable by all rational $x$ provided $\mathrm{ac}$ is a rational square and ${r}^{2}=\mathrm{ac}$, and by no rational $x$ otherwise. If exactly one of $b$ and $d$ is zero and $\mathrm{ad}+\mathrm{bc}\ne 0$, then each positive rational value is assumed by $\sqrt{(a+\mathrm{bx})(c+\mathrm{dx})}$ for a suitable value of $x$.

Otherwise, let
$\mathrm{bd}\ne 0$. Then, given
$r$, we have the corresponding

If $\mathrm{ad}=\mathrm{bc}$, then this yields a rational $x$ if and only if $\mathrm{bd}$ is a rational square. Let $\mathrm{ad}\ne \mathrm{bc}$. We wish to make $(\mathrm{ad}-\mathrm{bc}){}^{2}+4{\mathrm{bdr}}^{2}={s}^{2}$ for some rational $s$. This is equivalent to

Pick a pair
$u,v$ of rationals for which
$u+v\ne 0$ and
$\mathrm{uv}=\mathrm{bd}$. We want to make

so that $(u+v)r=\mathrm{ad}-\mathrm{bc}$ and $s=(u-v)r$. Thus, let

Then

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.