
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
${d}^{2}={a}^{2}+{b}^{2}+{c}^{2}\hspace{1em}.$
Solution 1. Let the tetrahedron be bounded by the
three coordinate planes in
$R{}^{3}$ and the plane
with equation
$\frac{x}{u}+\frac{y}{v}+\frac{z}{w}=1$,
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
$\frac{1}{3}\mathrm{au}=\frac{1}{3}\mathrm{bv}=\frac{1}{3}\mathrm{cw}=\frac{1}{3}\mathrm{dk}\hspace{1em},$
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.
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
$2a=\mathrm{yz}$,
$2b=\mathrm{zx}$ and
$2c=\mathrm{xy}$.
The fourth face is bounded by sides of length
$u=\sqrt{{y}^{2}+{z}^{2}}$,
$v=\sqrt{{z}^{2}+{x}^{2}}$ and
$w=\sqrt{{x}^{2}+{y}^{2}}$. By Heron's
formula, its area
$d$ is given by the relation
$\begin{array}{cc}16{d}^{2}\hfill & =(u+v+w)(u+vw)(uv+w)(u+v+w)\hfill \\ \multicolumn{0}{c}{}& =[(u+v){}^{2}{w}^{2}][({w}^{2}(uv){}^{2}]=[2\mathrm{uv}+({u}^{2}+{v}^{2}{w}^{2})][2\mathrm{uv}({u}^{2}+{v}^{2}{w}^{2})]\hfill \\ \multicolumn{0}{c}{}& =2{u}^{2}{v}^{2}+2{v}^{2}{w}^{2}+2{w}^{2}{u}^{2}{u}^{4}{v}^{4}{w}^{4}\hfill \\ \multicolumn{0}{c}{}& =2({y}^{2}+{z}^{2})({x}^{2}+{z}^{2})+2({x}^{2}+{z}^{2})({x}^{2}+{y}^{2})+2({x}^{2}+{y}^{2})({x}^{2}+{z}^{2})\hfill \\ \multicolumn{0}{c}{}& \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}({y}^{2}+{z}^{2}){}^{2}({x}^{2}+{z}^{2}){}^{2}({x}^{2}+{y}^{2}){}^{2}\hfill \\ \multicolumn{0}{c}{}& =4{x}^{2}{y}^{2}+4{x}^{2}{z}^{2}+4{y}^{2}{z}^{2}=16{a}^{2}+16{b}^{2}+16{c}^{2}\hspace{1em},\hfill \end{array}$
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
$2c=u\sqrt{{x}^{2}+{y}^{2}}$, whence
${u}^{2}({x}^{2}+{y}^{2})={x}^{2}{y}^{2}$.
Hence
${v}^{2}={z}^{2}+{u}^{2}=\frac{{x}^{2}{y}^{2}+{x}^{2}{z}^{2}+{y}^{2}{z}^{2}}{{x}^{2}+{y}^{2}}=\frac{4({a}^{2}+{b}^{2}+{c}^{2})}{{x}^{2}+{y}^{2}}\hspace{1em},$
so that
$2d=v\sqrt{{x}^{2}+{y}^{2}}\Rightarrow 4{d}^{2}=4({a}^{2}+{b}^{2}+{c}^{2})\hspace{1em},$
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
$\times $y, 2b = z
$\times $
x, 2c = x
$\times $y, and
2d = (z  y)
$\times $(z  x) =
$$ (z
$\times $x)
$$ (y
$\times $z)
$$ (x
$\times $y). Hence
d =
$$ (a + b + c), so that
${d}^{2}=d\xb7d=(a+b+c)\xb7(a+b+c)={a}^{2}+{b}^{2}+{c}^{2}\hspace{1em}.$

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
$x=(u1)/u$ 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:
$1(\frac{u1}{u}+\frac{1}{u+p})=\frac{p}{u(u+p)}\hspace{1em},$
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
$18{k}^{2}+3k1$ in front and
$3k$ behind; on Wednesday,
$12{k}^{2}+6k$ with
$12{k}^{2}+4k1$ and
$2k$ behind.
Solution 2. [O. Bormashenko] On the three successive days,
the total numbers numbers of people in line are
$\mathrm{un}$,
$v(n+1)$ 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
$\frac{1}{\mathrm{un}}+\frac{1}{n}=\frac{1}{v(n+1)}+\frac{1}{n+1}=\frac{1}{w(n+2)}+\frac{1}{n+2}\hspace{1em}.$
These yield the equations
$(nv)(n+1+u)=n(n+1)$
and
$(n+1w)(n+2+v)=(n+1)(n+2)\hspace{1em}.$
We need to find an integer
$v$ for which
$nv$ 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
$x=(u2)/u$, i.e. successively at
$x=3/5,5/7,7/9,\dots $. In this case, we assume that
$1/(up)$ is the
line is behind him, and need to ensure that
$u2p$ is a positive divisor
of
$u(up)$ for three consecutive values of
$p$. If
$u$ is odd, we can
achieve this with
$u$ any odd multiple of 15, starting with
$p=\frac{1}{2}(u1)$.
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
$\frac{1}{u}+\frac{1}{n}=\frac{1}{v}+\frac{1}{n+1}$
so that
$\mathrm{uv}=n(n+1)(uv)$. This yields both
$({n}^{2}+nv)u=({n}^{2}+n)v$ and
$({n}^{2}+n+u)v=({n}^{2}+n)u$,
leading to
$uv=\frac{{u}^{2}}{{n}^{2}+n+u}=\frac{{v}^{2}}{{n}^{2}+nv}\hspace{1em}.$
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
$uv=k$, we get the quadratic equation
${u}^{2}\mathrm{ku}k({n}^{2}+n)=0$
with discriminant
$\Delta ={k}^{2}+4({n}^{2}+n)k=[k+2({n}^{2}+n)]{}^{2}4({n}^{2}+n){}^{2}\hspace{1em},$
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
$k=uv=\gamma (\alpha \beta ){}^{2}$.

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?
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
$\frac{m+1}{m}=\frac{p+1}{p}\xb7\frac{q+1}{q}\hspace{1em},$
where
$p$ and
$q$ are positive integers exceeding
$m$. Then
$(m+1)\mathrm{pq}=m(p+1)(q+1)$, which reduces to
$(pm)(qm)=m(m+1)$. 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
$\begin{array}{cc}\frac{m+u+1}{m+u}\hfill & \xb7\frac{m+v+1}{m+v}=\frac{{m}^{2}+m(u+v+2)+\mathrm{uv}+(u+v)+1}{{m}^{2}+m(u+v)+\mathrm{uv}}\hfill \\ \multicolumn{0}{c}{}& =\frac{{m}^{2}+(m+1)(u+v)+m(m+1)+2m+1}{{m}^{2}+m(u+v)+m(m+1)}\hfill \\ \multicolumn{0}{c}{}& =\frac{(m+1){}^{2}+(m+1)(u+v)+m(m+1)}{{m}^{2}+m(u+v)+m(m+1)}\hfill \\ \multicolumn{0}{c}{}& =\frac{(m+1)[(m+1)+(u+v)+m]}{m[m+(u+v)+m+1]}=\frac{m+1}{m}\hspace{1em}.\hfill \end{array}$
Hence, there is a oneone 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.
Solution 2. [R. Ziman] Let
$m$ be an arbitrary positive
integer. Then, since
$(m+1)/m$ is in lowest terms,
$\mathrm{pq}$
must be a multiple of
$m$. Let
$m+1=\mathrm{uv}$ for some positive
integers
$u$ and
$v$ and
$m=\mathrm{rs}$ for some positive integers
$r$ and
$s$, where
$r$ is the greatest common divisor of
$m$
and
$p$; suppose that
$p=\mathrm{br}$ and
$q=\mathrm{as}$, with
$s$ being
the greatest common divisor of
$m$ and
$q$. Then, the
representation must have the form
$\frac{m+1}{m}=\frac{\mathrm{au}}{\mathrm{br}}\xb7\frac{\mathrm{bv}}{\mathrm{as}}\hspace{1em},$
where
$\mathrm{au}=\mathrm{br}+1$ and
$\mathrm{bv}=\mathrm{as}+1$. Hence
$\mathrm{bv}=\frac{\mathrm{br}+1}{u}s+1=\frac{\mathrm{brs}+s+u}{u}\hspace{1em},$
so that
$b=b(\mathrm{uv}\mathrm{rs})=s+u$ and
$a=\frac{\mathrm{sr}+\mathrm{ur}+1}{u}=\frac{m+1\mathrm{ur}}{u}=v+r\hspace{1em}.$
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
${x}^{3}+{y}^{3}\le xy$.
Prove that
${x}^{2}+{y}^{2}\le 1$.
Solution 1. [R. Barrington Leigh] We have that
$xy\ge {x}^{3}+{y}^{3}>{x}^{3}{y}^{3}\hspace{1em}.$
Since
$xy\ge {x}^{3}+{y}^{3}>0$, we can divide this inequality
by
$xy$ to obtain
$1>{x}^{2}+\mathrm{xy}+{y}^{2}>{x}^{2}+{y}^{2}\hspace{1em}.$
Solution 2. [S.E. Lu]
$\begin{array}{cc}xy\hfill & \ge {x}^{3}+{y}^{3}>{x}^{3}>{x}^{3}[{y}^{3}+\mathrm{xy}(xy)]\hfill \\ \multicolumn{0}{c}{}& ={x}^{3}{x}^{2}y+{\mathrm{xy}}^{2}{y}^{3}=({x}^{2}+{y}^{2})(xy)\hspace{1em},\hfill \end{array}$
whereupon a division by the positive quantity
$xy$ yields that
$1>{x}^{2}+{y}^{2}$.
Solution 3. [O. Bormashenko] Observe that
$y<x$ and that
${x}^{3}<{x}^{3}+{y}^{3}\le xy<x$, so that
$0<y<x<1$. It
follows that
$x(x+y)<2\Rightarrow \mathrm{xy}(x+y)<2\mathrm{xy}\hspace{1em}.\text{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(1)$
The given condition can be rewritten
$(x+y)({x}^{2}+{y}^{2})\mathrm{xy}(x+y)\le xy\hspace{1em}.\text{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(2)$
Adding inequalities (1) and (2) yields
$(x+y)({x}^{2}+{y}^{2})<x+y\hspace{1em},$
whence
${x}^{2}+{y}^{2}<1$.
Solution 4. [R. Furmaniak] We have that
$\begin{array}{cc}(xy)(1{x}^{2}{y}^{2})\hfill & =(xy)({x}^{3}{x}^{2}y+{\mathrm{xy}}^{2}{y}^{3})\hfill \\ \multicolumn{0}{c}{}& \ge ({x}^{3}+{y}^{3})({x}^{3}{x}^{2}y+{\mathrm{xy}}^{2}{y}^{3})=2{y}^{3}+{x}^{2}y{\mathrm{xy}}^{2}=y({x}^{2}\mathrm{xy}+2{y}^{2})\hfill \\ \multicolumn{0}{c}{}& =y[(x\sqrt{2}y){}^{2}+(2\sqrt{2}1)\mathrm{xy}]\ge 0\hspace{1em},\hfill \end{array}$
from which the result follows upon division by
$xy$.
Solution 5. Let
$y=\mathrm{tx}$. Since
$x>y>0$, we have that
$0<t<1$. Then
${x}^{3}(1+{t}^{3})\le x(1t)\Rightarrow {x}^{2}(1+{t}^{3})\le (1t)$. Therefore,
$\begin{array}{cc}{x}^{2}+{y}^{2}\hfill & ={x}^{2}(1+{t}^{2})\le (\frac{1t}{1+{t}^{3}})(1+{t}^{2})\hfill \\ \multicolumn{0}{c}{}& =\frac{1t+{t}^{2}{t}^{3}}{1+{t}^{3}}=1\frac{t(1t+2{t}^{2})}{1+{t}^{3}}\hspace{1em}.\hfill \end{array}$
Since
$1t+2{t}^{2}$, having negative discriminant, is always
positive, the desired result follows.
Solution 6. [J. Chui] Suppose, if possible, that
${x}^{2}+{y}^{2}={r}^{2}>1$. We can write
$x=r\mathrm{sin}\theta $ and
$y=r\mathrm{cos}\theta $
for
$0\le \theta \le \pi /2$. Then
$\begin{array}{cc}{x}^{3}+{y}^{3}(xy)\hfill & ={r}^{3}\mathrm{sin}{}^{3}\theta +{r}^{3}\mathrm{cos}{}^{3}\theta r\mathrm{sin}\theta +r\mathrm{cos}\theta \hfill \\ \multicolumn{0}{c}{}& >r\mathrm{sin}\theta (\mathrm{sin}{}^{2}\theta 1)+r\mathrm{cos}{}^{3}\theta +r\mathrm{cos}\theta \hfill \\ \multicolumn{0}{c}{}& =r\mathrm{sin}\theta \mathrm{cos}{}^{2}\theta +r\mathrm{cos}{}^{3}\theta +r\mathrm{cos}\theta \hfill \\ \multicolumn{0}{c}{}& =r\mathrm{cos}{}^{2}\theta (\mathrm{cos}\theta +\frac{1}{\mathrm{cos}\theta}\mathrm{sin}\theta )\hfill \\ \multicolumn{0}{c}{}& >r\mathrm{cos}{}^{2}\theta (2\mathrm{sin}\theta )>0\hspace{1em},\hfill \end{array}$
contrary to hypothesis. The result follows by contradiction.
Solution 7. Let
$r>0$ and
${r}^{2}={x}^{2}+{y}^{2}$. Since
$x>y>0$, we can write
$x=r\mathrm{cos}\theta $ and
$y=r\mathrm{sin}\theta $, where
$0<\theta <\pi /4$. The given equality is equivalent to
${r}^{2}\le \frac{\mathrm{cos}\theta \mathrm{sin}\theta}{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta}\hspace{1em},$
so it suffices to show that the right side does not exceed 1
to obtain the desired
${r}^{2}\le 1$.
Observe that
$\begin{array}{cc}1\frac{\mathrm{cos}\theta \mathrm{sin}\theta}{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta}\hfill & =\frac{(\mathrm{cos}\theta +\mathrm{sin}\theta )(1\mathrm{cos}\theta \mathrm{sin}\theta )(\mathrm{cos}\theta \mathrm{sin}\theta )}{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta}\hfill \\ \multicolumn{0}{c}{}& =\frac{\mathrm{sin}\theta (2\mathrm{cos}\theta \mathrm{sin}\theta \mathrm{cos}{}^{2}\theta )}{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta}>0\hspace{1em},\hfill \end{array}$
from which the desired result follows.
Solution 8. Begin as in Solution 7. Then
$\begin{array}{cc}\frac{\mathrm{cos}\theta \mathrm{sin}\theta}{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta}\hfill & =\frac{\mathrm{cos}{}^{2}\theta \mathrm{sin}{}^{2}\theta}{(\mathrm{cos}\theta +\mathrm{sin}\theta ){}^{2}(1\mathrm{cos}\theta \mathrm{sin}\theta )}\hfill \\ \multicolumn{0}{c}{}& =\frac{\mathrm{cos}2\theta}{(1+\mathrm{sin}2\theta )(1\frac{1}{2}\mathrm{sin}\theta})=\frac{\mathrm{cos}2\theta}{1+\frac{1}{2}\mathrm{sin}2\theta (1\mathrm{sin}2\theta )}<1\hspace{1em},\hfill \end{array}$
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?
Solution. Let us define finite sequences as follows.
Suppose that
${S}_{1}=0$. Then, for each
$k\ge 2$,
${S}_{k}$ is
obtained by replacing each
$0$ in
${S}_{k1}$ by
$01$ and
each
$1$ in
${S}_{k1}$ by
$001$. Thus,
${S}_{1}=0;\hspace{1em}\hspace{1em}{S}_{2}=01;\hspace{1em}\hspace{1em}{S}_{3}=01001;\hspace{1em}\hspace{1em}{S}_{4}=010010101001;\hspace{1em}{S}_{5}=01001010100101001010010101001;\dots $
Each
${S}_{k1}$ is a prefix of
${S}_{k}$; in fact, it can be shown that,
for each
$k\ge 3$,
${S}_{k}={S}_{k1}*{S}_{k2}*{S}_{k1}\hspace{1em},$
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
$(2002985408)$th
$=(609)$th entry in
${S}_{9}$. This in turn is
equal to the
$(609408169)$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.
Comment. Suppose that
$f(n)$ is the position of the
$n$th
one, so that
$f(1)=2$ and
$f(2)=5$. Let
$g(n)$ be the number
of zeros up to and including the
$n$th position, and so
$ng(n)$ is the number of ones up to and including the
$n$th
position. Then we get the two equations
$f(n)=2g(n)+3(ng(n))=3ng(n)\text{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(1)$
$g(f(n))=f(n)n\hspace{1em}.\text{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}(2)$
These two can be used to determine the positions of the ones by
stepping up; for example, we have
$f(2)=5$,
$g(5)=3$,
$f(5)=153=12$,
$g(12)=f(5)5=7$, 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
$\mathrm{bc}=\mathrm{ad}$.
Solution 1. We study the possibility of making
$c+\mathrm{dx}=(a+\mathrm{bx}){t}^{2}$ for some rational numbers
$t$. This would require
that
$x=\frac{c{\mathrm{at}}^{2}}{{\mathrm{bt}}^{2}d}\hspace{1em}.$
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
$\sqrt{(a+\mathrm{bx})(c+\mathrm{dx})}=\Vert (\mathrm{bc}\mathrm{ad})t({\mathrm{bt}}^{2}d){}^{1}\Vert $
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
$\frac{c{\mathrm{au}}^{2}}{{\mathrm{bu}}^{2}d}=\frac{c{\mathrm{av}}^{2}}{{\mathrm{bv}}^{2}d}\hspace{1em}.$
Then
$\begin{array}{cc}0\hfill & =(c{\mathrm{au}}^{2})({\mathrm{bv}}^{2}d)(c{\mathrm{av}}^{2})({\mathrm{bu}}^{2}d)\hfill \\ \multicolumn{0}{c}{}& =\mathrm{bc}({v}^{2}{u}^{2})+\mathrm{ad}({u}^{2}{v}^{2})=(\mathrm{bc}\mathrm{ad})({u}^{2}{v}^{2})\hspace{1em},\hfill \end{array}$
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.
Solution 2.
$(a+\mathrm{bx})(c+\mathrm{dx})={r}^{2}$ for rational
$r$ is
equivalent to
${\mathrm{bdx}}^{2}+(\mathrm{ad}+\mathrm{bc})x+(\mathrm{ac}{r}^{2})=0\hspace{1em}.$
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
$x=\frac{(\mathrm{ad}+\mathrm{bc})\pm \sqrt{(\mathrm{ad}\mathrm{bc}){}^{2}+4{\mathrm{bdr}}^{2}}}{2\mathrm{bd}}\hspace{1em}.$
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
$4{\mathrm{bdr}}^{2}=(\mathrm{ad}\mathrm{bc}){}^{2}{s}^{2}=(\mathrm{ad}\mathrm{bc}+s)(\mathrm{ad}\mathrm{bc}s)\hspace{1em}.$
Pick a pair
$u,v$ of rationals for which
$u+v\ne 0$ and
$\mathrm{uv}=\mathrm{bd}$. We want to make
$2\mathrm{ur}=\mathrm{ad}\mathrm{bc}+s\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\mathrm{and}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}2\mathrm{vr}=\mathrm{ad}\mathrm{bc}s$
so that
$(u+v)r=\mathrm{ad}\mathrm{bc}$ and
$s=(uv)r$.
Thus, let
$r=\frac{\mathrm{ad}\mathrm{bc}}{u+v}\hspace{1em}.$
Then
$\begin{array}{cc}(\mathrm{ad}\mathrm{bc}){}^{2}+4{\mathrm{bdr}}^{2}\hfill & =(\mathrm{ad}\mathrm{bc}){}^{2}+4{\mathrm{uvr}}^{2}\hfill \\ \multicolumn{0}{c}{}& =(\frac{\mathrm{ad}\mathrm{bc}}{u+v}{)}^{2}[(u+v){}^{2}4\mathrm{uv}]\hfill \\ \multicolumn{0}{c}{}& =\frac{(\mathrm{ad}\mathrm{bc}){}^{2}(uv){}^{2}}{(u+v){}^{2}}\hfill \end{array}$
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.