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

${d}^{2}={a}^{2}+{b}^{2}+{c}^{2} .$

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 $\left(0,0,0\right)$, $\left(u,0,0\right)$, $\left(0,v,0\right)$, $\left(0,0,w\right)$. 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

$\frac{1}{3}\mathrm{au}=\frac{1}{3}\mathrm{bv}=\frac{1}{3}\mathrm{cw}=\frac{1}{3}\mathrm{dk} ,$

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 $\left(\left(\mathrm{um}\right){}^{-1},\left(\mathrm{vm}\right){}^{-1}.\left(\mathrm{wm}\right){}^{-1}\right)$, 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 $\left(b,c\right)$, $\left(c,a\right)$, $\left(a,b\right)$. 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

$\begin{array}{cc}16{d}^{2}\hfill & =\left(u+v+w\right)\left(u+v-w\right)\left(u-v+w\right)\left(-u+v+w\right)\hfill \\ \multicolumn{0}{c}{}& =\left[\left(u+v\right){}^{2}-{w}^{2}\right]\left[\left({w}^{2}-\left(u-v\right){}^{2}\right]=\left[2\mathrm{uv}+\left({u}^{2}+{v}^{2}-{w}^{2}\right)\right]\left[2\mathrm{uv}-\left({u}^{2}+{v}^{2}-{w}^{2}\right)\right]\hfill \\ \multicolumn{0}{c}{}& =2{u}^{2}{v}^{2}+2{v}^{2}{w}^{2}+2{w}^{2}{u}^{2}-{u}^{4}-{v}^{4}-{w}^{4}\hfill \\ \multicolumn{0}{c}{}& =2\left({y}^{2}+{z}^{2}\right)\left({x}^{2}+{z}^{2}\right)+2\left({x}^{2}+{z}^{2}\right)\left({x}^{2}+{y}^{2}\right)+2\left({x}^{2}+{y}^{2}\right)\left({x}^{2}+{z}^{2}\right)\hfill \\ \multicolumn{0}{c}{}& -\left({y}^{2}+{z}^{2}\right){}^{2}-\left({x}^{2}+{z}^{2}\right){}^{2}-\left({x}^{2}+{y}^{2}\right){}^{2}\hfill \\ \multicolumn{0}{c}{}& =4{x}^{2}{y}^{2}+4{x}^{2}{z}^{2}+4{y}^{2}{z}^{2}=16{a}^{2}+16{b}^{2}+16{c}^{2} ,\hfill \end{array}$

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 ${u}^{2}\left({x}^{2}+{y}^{2}\right)={x}^{2}{y}^{2}$. Hence

${v}^{2}={z}^{2}+{u}^{2}=\frac{{x}^{2}{y}^{2}+{x}^{2}{z}^{2}+{y}^{2}{z}^{2}}{{x}^{2}+{y}^{2}}=\frac{4\left({a}^{2}+{b}^{2}+{c}^{2}\right)}{{x}^{2}+{y}^{2}} ,$

so that

$2d=v\sqrt{{x}^{2}+{y}^{2}}⇒4{d}^{2}=4\left({a}^{2}+{b}^{2}+{c}^{2}\right) ,$

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

${d}^{2}=d·d=\left(a+b+c\right)·\left(a+b+c\right)={a}^{2}+{b}^{2}+{c}^{2} .$

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 $1/\left(n+1\right)$ of the line was behind him. On Wednesday, the same fraction $x$ of the line was in front of him, while $1/\left(n+2\right)$ 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 $x=\left(u-1\right)/u$ for some positive integer exceeding 1. Let $1/\left(u+p\right)$ be the fraction of the line behind Angus. Then Angus himself represents this fraction of the line:

$1-\left(\frac{u-1}{u}+\frac{1}{u+p}\right)=\frac{p}{u\left(u+p\right)} ,$

so that there would be $u\left(u+p\right)/p$ 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 $u\left(u+1\right)$, $u\left(u+2\right)/2$ and $u\left(u+3\right)/3$ 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.
Solution 2. [O. Bormashenko] On the three successive days, the total numbers numbers of people in line are $\mathrm{un}$, $v\left(n+1\right)$ and $w\left(n+2\right)$ for some positive integers $u$, $v$ and $w$. The fraction of the line constituted by Angus and those behind him is

$\frac{1}{\mathrm{un}}+\frac{1}{n}=\frac{1}{v\left(n+1\right)}+\frac{1}{n+1}=\frac{1}{w\left(n+2\right)}+\frac{1}{n+2} .$

These yield the equations

$\left(n-v\right)\left(n+1+u\right)=n\left(n+1\right)$

and

$\left(n+1-w\right)\left(n+2+v\right)=\left(n+1\right)\left(n+2\right) .$

We need to find an integer $v$ for which $n-v$ divides $n\left(n+1\right)$ and $n+2+v$ divides $\left(n+1\right)\left(n+2\right)$. This is equivalent to determining $p,q$ for which $p+q=2\left(n+1\right)$, $p, $p$ divides $n\left(n+1\right)$, $q>n+2$ and $q$ divides $\left(n+1\right)\left(n+2\right)$. The triple $\left(n,p,q\right)=\left(7,4,12\right)$ works and yields $\left(u,v,w\right)=\left(6,3,2\right)$. 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 $x=\left(u-2\right)/u$, i.e. successively at $x=3/5,5/7,7/9,\dots$. In this case, we assume that $1/\left(u-p\right)$ is the line is behind him, and need to ensure that $u-2p$ is a positive divisor of $u\left(u-p\right)$ for three consecutive values of $p$. If $u$ is odd, we can achieve this with $u$ any odd multiple of 15, starting with $p=\frac{1}{2}\left(u-1\right)$.
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 $1/\left(n+1\right)$ 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 $\mathrm{uv}=n\left(n+1\right)\left(u-v\right)$. This yields both $\left({n}^{2}+n-v\right)u=\left({n}^{2}+n\right)v$ and $\left({n}^{2}+n+u\right)v=\left({n}^{2}+n\right)u$, leading to

$u-v=\frac{{u}^{2}}{{n}^{2}+n+u}=\frac{{v}^{2}}{{n}^{2}+n-v} .$

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

${u}^{2}-\mathrm{ku}-k\left({n}^{2}+n\right)=0$

with discriminant

$\Delta ={k}^{2}+4\left({n}^{2}+n\right)k=\left[k+2\left({n}^{2}+n\right)\right]{}^{2}-4\left({n}^{2}+n\right){}^{2} ,$

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 $k=\gamma \left({\alpha }^{2}+{\beta }^{2}-2\alpha \beta \right)=\gamma \left(\alpha -\beta \right){}^{2}$; this will make the discriminant $\Delta$ equal to a square.
Taking $n=3$, for example, yields the possibilities $\left(u,v\right)=\left(132,11\right)$, (60, 10), (36, 9), (24, 8), (12, 6), (6, 4),(4, 3). In general, we find that $\left(n,u,v\right)=\left(n,\gamma \alpha \left(\alpha -\beta \right),\gamma \beta \left(\alpha -\beta \right)\right)$ when ${n}^{2}+n=\gamma \alpha \beta$ with $\alpha >\beta$. It turns out that $k=u-v=\gamma \left(\alpha -\beta \right){}^{2}$.

141.
In how many ways can the rational $2002/2001$ be written as the product of two rationals of the form $\left(n+1\right)/n$, 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 $d\left(m\right)$ and $d\left(m+1\right)$, the number of positive divisors of $m$ and $m+1$ respectively. Suppose that

$\frac{m+1}{m}=\frac{p+1}{p}·\frac{q+1}{q} ,$

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

$\begin{array}{cc}\frac{m+u+1}{m+u}\hfill & ·\frac{m+v+1}{m+v}=\frac{{m}^{2}+m\left(u+v+2\right)+\mathrm{uv}+\left(u+v\right)+1}{{m}^{2}+m\left(u+v\right)+\mathrm{uv}}\hfill \\ \multicolumn{0}{c}{}& =\frac{{m}^{2}+\left(m+1\right)\left(u+v\right)+m\left(m+1\right)+2m+1}{{m}^{2}+m\left(u+v\right)+m\left(m+1\right)}\hfill \\ \multicolumn{0}{c}{}& =\frac{\left(m+1\right){}^{2}+\left(m+1\right)\left(u+v\right)+m\left(m+1\right)}{{m}^{2}+m\left(u+v\right)+m\left(m+1\right)}\hfill \\ \multicolumn{0}{c}{}& =\frac{\left(m+1\right)\left[\left(m+1\right)+\left(u+v\right)+m\right]}{m\left[m+\left(u+v\right)+m+1\right]}=\frac{m+1}{m} .\hfill \end{array}$

Hence, there is a one-one correspondence between representations and pairs $\left(u,v\right)$ of complementary factors of $m\left(m+1\right)$. Since $m$ and $m+1$ are coprime, the number of factors of $m\left(m+1\right)$ is equal to $d\left(m\right)d\left(m+1\right)$, and so the number of representations is equal to $\frac{1}{2}d\left(m\right)d\left(m+1\right)$.
Now consider the case that $m=2001$. Since $2001=3×23×29$, $d\left(2001\right)=8$; since $2002=2×7×11×13$, $d\left(2002\right)=16$. Hence, the desired number of representations is 64.
Solution 2. [R. Ziman] Let $m$ be an arbitrary positive integer. Then, since $\left(m+1\right)/m$ 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

$\frac{m+1}{m}=\frac{\mathrm{au}}{\mathrm{br}}·\frac{\mathrm{bv}}{\mathrm{as}} ,$

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

$\mathrm{bv}=\frac{\mathrm{br}+1}{u}s+1=\frac{\mathrm{brs}+s+u}{u} ,$

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

$a=\frac{\mathrm{sr}+\mathrm{ur}+1}{u}=\frac{m+1-\mathrm{ur}}{u}=v+r .$

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

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

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

$x-y\ge {x}^{3}+{y}^{3}>{x}^{3}-{y}^{3} .$

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

$1>{x}^{2}+\mathrm{xy}+{y}^{2}>{x}^{2}+{y}^{2} .$

Solution 2. [S.E. Lu]

$\begin{array}{cc}x-y\hfill & \ge {x}^{3}+{y}^{3}>{x}^{3}>{x}^{3}-\left[{y}^{3}+\mathrm{xy}\left(x-y\right)\right]\hfill \\ \multicolumn{0}{c}{}& ={x}^{3}-{x}^{2}y+{\mathrm{xy}}^{2}-{y}^{3}=\left({x}^{2}+{y}^{2}\right)\left(x-y\right) ,\hfill \end{array}$

whereupon a division by the positive quantity $x-y$ yields that $1>{x}^{2}+{y}^{2}$.
Solution 3. [O. Bormashenko] Observe that $y and that ${x}^{3}<{x}^{3}+{y}^{3}\le x-y, so that $0. It follows that

$x\left(x+y\right)<2⇒\mathrm{xy}\left(x+y\right)<2\mathrm{xy} .\text{ }\left(1\right)$

The given condition can be rewritten

$\left(x+y\right)\left({x}^{2}+{y}^{2}\right)-\mathrm{xy}\left(x+y\right)\le x-y .\text{ }\left(2\right)$

Adding inequalities (1) and (2) yields

$\left(x+y\right)\left({x}^{2}+{y}^{2}\right)

whence ${x}^{2}+{y}^{2}<1$.
Solution 4. [R. Furmaniak] We have that

$\begin{array}{cc}\left(x-y\right)\left(1-{x}^{2}-{y}^{2}\right)\hfill & =\left(x-y\right)-\left({x}^{3}-{x}^{2}y+{\mathrm{xy}}^{2}-{y}^{3}\right)\hfill \\ \multicolumn{0}{c}{}& \ge \left({x}^{3}+{y}^{3}\right)-\left({x}^{3}-{x}^{2}y+{\mathrm{xy}}^{2}-{y}^{3}\right)=2{y}^{3}+{x}^{2}y-{\mathrm{xy}}^{2}=y\left({x}^{2}-\mathrm{xy}+2{y}^{2}\right)\hfill \\ \multicolumn{0}{c}{}& =y\left[\left(x-\sqrt{2}y\right){}^{2}+\left(2\sqrt{2}-1\right)\mathrm{xy}\right]\ge 0 ,\hfill \end{array}$

from which the result follows upon division by $x-y$.
Solution 5. Let $y=\mathrm{tx}$. Since $x>y>0$, we have that $0. Then ${x}^{3}\left(1+{t}^{3}\right)\le x\left(1-t\right)⇒{x}^{2}\left(1+{t}^{3}\right)\le \left(1-t\right)$. Therefore,

$\begin{array}{cc}{x}^{2}+{y}^{2}\hfill & ={x}^{2}\left(1+{t}^{2}\right)\le \left(\frac{1-t}{1+{t}^{3}}\right)\left(1+{t}^{2}\right)\hfill \\ \multicolumn{0}{c}{}& =\frac{1-t+{t}^{2}-{t}^{3}}{1+{t}^{3}}=1-\frac{t\left(1-t+2{t}^{2}\right)}{1+{t}^{3}} .\hfill \end{array}$

Since $1-t+2{t}^{2}$, 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

$\begin{array}{cc}{x}^{3}+{y}^{3}-\left(x-y\right)\hfill & ={r}^{3}\mathrm{sin}{}^{3}\theta +{r}^{3}\mathrm{cos}{}^{3}\theta -r\mathrm{sin}\theta +r\mathrm{cos}\theta \hfill \\ \multicolumn{0}{c}{}& >r\mathrm{sin}\theta \left(\mathrm{sin}{}^{2}\theta -1\right)+r\mathrm{cos}{}^{3}\theta +r\mathrm{cos}\theta \hfill \\ \multicolumn{0}{c}{}& =-r\mathrm{sin}\theta \mathrm{cos}{}^{2}\theta +r\mathrm{cos}{}^{3}\theta +r\mathrm{cos}\theta \hfill \\ \multicolumn{0}{c}{}& =r\mathrm{cos}{}^{2}\theta \left(\mathrm{cos}\theta +\frac{1}{\mathrm{cos}\theta }-\mathrm{sin}\theta \right)\hfill \\ \multicolumn{0}{c}{}& >r\mathrm{cos}{}^{2}\theta \left(2-\mathrm{sin}\theta \right)>0 ,\hfill \end{array}$

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

${r}^{2}\le \frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta } ,$

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

$\begin{array}{cc}1-\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta }\hfill & =\frac{\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right)\left(1-\mathrm{cos}\theta \mathrm{sin}\theta \right)-\left(\mathrm{cos}\theta -\mathrm{sin}\theta \right)}{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta }\hfill \\ \multicolumn{0}{c}{}& =\frac{\mathrm{sin}\theta \left(2-\mathrm{cos}\theta \mathrm{sin}\theta -\mathrm{cos}{}^{2}\theta \right)}{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta }>0 ,\hfill \end{array}$

from which the desired result follows.
Solution 8. Begin as in Solution 7. Then

$\begin{array}{cc}\frac{\mathrm{cos}\theta -\mathrm{sin}\theta }{\mathrm{cos}{}^{3}\theta +\mathrm{sin}{}^{3}\theta }\hfill & =\frac{\mathrm{cos}{}^{2}\theta -\mathrm{sin}{}^{2}\theta }{\left(\mathrm{cos}\theta +\mathrm{sin}\theta \right){}^{2}\left(1-\mathrm{cos}\theta \mathrm{sin}\theta \right)}\hfill \\ \multicolumn{0}{c}{}& =\frac{\mathrm{cos}2\theta }{\left(1+\mathrm{sin}2\theta \right)\left(1-\frac{1}{2}\mathrm{sin}\theta }\right)=\frac{\mathrm{cos}2\theta }{1+\frac{1}{2}\mathrm{sin}2\theta \left(1-\mathrm{sin}2\theta \right)}<1 ,\hfill \end{array}$

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 ${S}_{k-1}$ by $01$ and each $1$ in ${S}_{k-1}$ by $001$. Thus,

${S}_{1}=0; {S}_{2}=01; {S}_{3}=01001; {S}_{4}=010010101001; {S}_{5}=01001010100101001010010101001;\dots$

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

${S}_{k}={S}_{k-1}*{S}_{k-2}*{S}_{k-1} ,$

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 $\left(2002-985-408\right)$th $=\left(609\right)$th entry in ${S}_{9}$. This in turn is equal to the $\left(609-408-169\right)$th $=\left(32\right)$th entry in ${S}_{8}$, which is equal to the $\left(32\right)$th entry of ${S}_{6}$, or the third entry of ${S}_{3}$. Hence, the desired entry is 0.
Comment. Suppose that $f\left(n\right)$ is the position of the $n$th one, so that $f\left(1\right)=2$ and $f\left(2\right)=5$. Let $g\left(n\right)$ be the number of zeros up to and including the $n$th position, and so $n-g\left(n\right)$ is the number of ones up to and including the $n$th position. Then we get the two equations

$f\left(n\right)=2g\left(n\right)+3\left(n-g\left(n\right)\right)=3n-g\left(n\right)\text{ }\left(1\right)$

$g\left(f\left(n\right)\right)=f\left(n\right)-n .\text{ }\left(2\right)$

These two can be used to determine the positions of the ones by stepping up; for example, we have $f\left(2\right)=5$, $g\left(5\right)=3$, $f\left(5\right)=15-3=12$, $g\left(12\right)=f\left(5\right)-5=7$, 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 $\sqrt{\left(a+\mathrm{bx}\right)\left(c+\mathrm{dx}\right)}$ is rational. Explain the situation when $\mathrm{bc}=\mathrm{ad}$.
Solution 1. We study the possibility of making $c+\mathrm{dx}=\left(a+\mathrm{bx}\right){t}^{2}$ for some rational numbers $t$. This would require that

$x=\frac{c-{\mathrm{at}}^{2}}{{\mathrm{bt}}^{2}-d} .$

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 $a+\mathrm{bx}=\left(\mathrm{bc}-\mathrm{ad}\right)\left({\mathrm{bt}}^{2}-d\right){}^{-1}$ and $c+\mathrm{dx}=\left(\mathrm{bc}-\mathrm{ad}\right){t}^{2}\left({\mathrm{bt}}^{2}-d\right){}^{-1}$, whence

$\sqrt{\left(a+\mathrm{bx}\right)\left(c+\mathrm{dx}\right)}=‖\left(\mathrm{bc}-\mathrm{ad}\right)t\left({\mathrm{bt}}^{2}-d\right){}^{-1}‖$

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

$\frac{c-{\mathrm{au}}^{2}}{{\mathrm{bu}}^{2}-d}=\frac{c-{\mathrm{av}}^{2}}{{\mathrm{bv}}^{2}-d} .$

Then

$\begin{array}{cc}0\hfill & =\left(c-{\mathrm{au}}^{2}\right)\left({\mathrm{bv}}^{2}-d\right)-\left(c-{\mathrm{av}}^{2}\right)\left({\mathrm{bu}}^{2}-d\right)\hfill \\ \multicolumn{0}{c}{}& =\mathrm{bc}\left({v}^{2}-{u}^{2}\right)+\mathrm{ad}\left({u}^{2}-{v}^{2}\right)=\left(\mathrm{bc}-\mathrm{ad}\right)\left({u}^{2}-{v}^{2}\right) ,\hfill \end{array}$

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 $\left(a,b,c,d\right)=\left(0,0,c,d\right),\left(a,b,0,0\right),\left(a,0,c,0\right),\left(0,b,0,d\right)$ 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 $\left(a+\mathrm{bx}\right)\left(c+\mathrm{dx}\right)=s\left(a+\mathrm{bx}\right){}^{2}$. 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. $\left(a+\mathrm{bx}\right)\left(c+\mathrm{dx}\right)={r}^{2}$ for rational $r$ is equivalent to

${\mathrm{bdx}}^{2}+\left(\mathrm{ad}+\mathrm{bc}\right)x+\left(\mathrm{ac}-{r}^{2}\right)=0 .$

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 $\sqrt{\left(a+\mathrm{bx}\right)\left(c+\mathrm{dx}\right)}$ for a suitable value of $x$.
Otherwise, let $\mathrm{bd}\ne 0$. Then, given $r$, we have the corresponding

$x=\frac{-\left(\mathrm{ad}+\mathrm{bc}\right)±\sqrt{\left(\mathrm{ad}-\mathrm{bc}\right){}^{2}+4{\mathrm{bdr}}^{2}}}{2\mathrm{bd}} .$

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 $\left(\mathrm{ad}-\mathrm{bc}\right){}^{2}+4{\mathrm{bdr}}^{2}={s}^{2}$ for some rational $s$. This is equivalent to

$4{\mathrm{bdr}}^{2}=\left(\mathrm{ad}-\mathrm{bc}\right){}^{2}-{s}^{2}=\left(\mathrm{ad}-\mathrm{bc}+s\right)\left(\mathrm{ad}-\mathrm{bc}-s\right) .$

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

$2\mathrm{ur}=\mathrm{ad}-\mathrm{bc}+s \mathrm{and} 2\mathrm{vr}=\mathrm{ad}-\mathrm{bc}-s$

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

$r=\frac{\mathrm{ad}-\mathrm{bc}}{u+v} .$

Then

$\begin{array}{cc}\left(\mathrm{ad}-\mathrm{bc}\right){}^{2}+4{\mathrm{bdr}}^{2}\hfill & =\left(\mathrm{ad}-\mathrm{bc}\right){}^{2}+4{\mathrm{uvr}}^{2}\hfill \\ \multicolumn{0}{c}{}& =\left(\frac{\mathrm{ad}-\mathrm{bc}}{u+v}{\right)}^{2}\left[\left(u+v\right){}^{2}-4\mathrm{uv}\right]\hfill \\ \multicolumn{0}{c}{}& =\frac{\left(\mathrm{ad}-\mathrm{bc}\right){}^{2}\left(u-v\right){}^{2}}{\left(u+v\right){}^{2}}\hfill \end{array}$

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.
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