- 133.
- Prove that, if $a$, $b$, $c$, $d$ are real numbers, $b\ne c$, both sides of the equation are defined, and

then each side of the equation is equal to

Give two essentially different examples of quadruples $(a,b,c,d)$, not in geometric progression, for which the conditions are satisfied. What happens when $b=c$?

Note that the quantity is square brackets vanishes when $b=c$, so that $(b-c)$ should be a factor of it. Indeed, we have

Since $b\ne c$, we find that the given condition is equivalent to

or

It follows that

as desired. Note that, in the event that $a-b-c+d=0$, we must have $\mathrm{ad}-\mathrm{bc}=0$, as well. (Explain!)

A generic example is
$(a,b,c,d)=({r}^{k-1}\pm 1,{r}^{k}\pm 1,{r}^{k+1}\pm 1,{r}^{k+2}\pm 1)$, where
$r\ne 0,1$ and
$k$ is arbitrary.

Suppose that
$b=c$. If
$a\ne b$ and
$d\ne b$, then both sides
of the datum reduce to
$b$, and the condition is a tautology.
However, the final fraction need not be equal to
$b$ in this case:
an example is
$(a,b,c,d)=(2,1,1,3)$. On the other hand,
suppose that
$a=b=c\ne d$. The one side of the datum is
undefined, while we find that

If $a\ne b=c=d$, then we have a similar result. Finally, if $a=b=c=d$, then all fractions are undefined.

Then

and

Hence

Note that $b=k\&lrArr;c=k$. Since $b\ne c$, then both $b$ and $c$ must differ from $k$. Hence

If $a-b-c+d=0$, then $\mathrm{ad}-\mathrm{bc}=0$ and the expression in the conclusion is undefined. Otherwise,

and the result follows.

The given conditions imply that
$a-k$,
$b-k$,
$c-k$,
$d-k$ are
in geometric progresion. Conversely, pick
$u$ arbitrary and
$r\ne 0,1$, and let
$(a,b,c,d)=(k+u,k+\mathrm{ur},k+{\mathrm{ur}}^{2},k+{\mathrm{ur}}^{3})$
to obtain a generic example.

The given condition is equivalent to

from which $(\mathrm{ac}-{b}^{2})+(\mathrm{bd}-{c}^{2})=\mathrm{ad}-\mathrm{bc}$, and we can proceed as in Solution 1.

Very few full marks were given for this problem, as solvers were
not careful about details. Whenever an expression appears in a
denominator or you cancel a factor out of a product, you must
consider the possibility that it might vanish. Most people ignored
this possibility. In addition, the analysis of the situation when
$b=c$ was not at all thorough.

- 134.
- Suppose that

Prove that

Of course, if any of ${x}^{2}$, ${y}^{2}$, ${z}^{2}$ is equal to 1, then the conclusion involves undefined quantities. Give the proper conclusion in this situation. Provide two essentially different numerical examples.

On the other hand, suppose that
${x}^{2}=1$ and
$a\ne 0$. Then
$y\pm z=0$, so that

whence ${z}^{2}={y}^{2}={x}^{2}=1$. Since $a\ne 0$, it is not possible for both $b$ and $c$ to vanish. Let $b\ne 0$. Then

so that $x=-\mathrm{yz}$. In this case, all the expressions in the conclusion are undefined.

We now suppose that none of
${x}^{2}$,
${y}^{2}$,
${z}^{2}$ is equal to 1.
>From
$b=\mathrm{xc}+\mathrm{za}$ and
$\mathrm{xc}=\mathrm{xya}+{x}^{2}b$, we deduce that

Since ${x}^{2}\ne 1$, we have that $\mathrm{xy}+z\ne 0$, and so

>From $c=\mathrm{ya}+\mathrm{xb}$ and $\mathrm{xb}={x}^{2}c+\mathrm{xza}$, we deduce that

whence

>From $a=\mathrm{zb}+\mathrm{yc}$ and $\mathrm{zb}=\mathrm{xzc}+{z}^{2}a$, we deduce that

whence

>From $a=\mathrm{zb}+\mathrm{yc}$ and $\mathrm{yc}={y}^{2}a+\mathrm{xyb}$, we deduce that

whence

The result now follows.

so that

Similarly

Also, since
$(\mathrm{xy}+z)a=(1-{x}^{2})b$ and
$(\mathrm{xz}+y)a=(1-{x}^{2})c$, we
have that

Similarly, $b({x}^{2}+{y}^{2}+{z}^{2}+2\mathrm{xyz})=b$ and $c({x}^{2}+{y}^{2}+{z}^{2}+2\mathrm{xyz})=c$. For each solution of the given system for which not all of $a,b,c$ vanish, we must have ${x}^{2}+{y}^{2}+{z}^{2}+2\mathrm{xyz}=1$.

Since
$\mathrm{xc}=b-\mathrm{za}$ and
$\mathrm{xb}=c-\mathrm{ya}$, it follows that
${b}^{2}-\mathrm{zab}={c}^{2}-\mathrm{yac}$, so that

Hence

Also,

If $a,b,c$ are the sides of a triangle $\mathrm{ABC}$, then we have

[Can you prove this directly?]

Here are some examples of sextuples
$(a,b,c;x,y,z)$:
$(a,b,c;\mathrm{cos}A,\mathrm{cos}B,\mathrm{cos}C)$,
$(1,3,6;\frac{11}{9},\frac{7}{3},-\frac{13}{3})$,
$(2,5,9;\frac{17}{15},\frac{5}{3},-\frac{13}{5})$,
$(-5,5,0;4,4,-1)$.

- 135.
- For the positive integer $n$, let $p(n)=k$ if $n$ is divisible by ${2}^{k}$ but not by ${2}^{k+1}$. Let ${x}_{0}=0$ and define ${x}_{n}$ for $n\ge 1$ recursively by

Prove that every nonnegative rational number occurs exactly once in the sequence $\{{x}_{0},{x}_{1},{x}_{2},\dots ,{x}_{n},\dots \}$.

Since $({x}_{0},{x}_{1},{x}_{2})=(0,1,\frac{1}{2})$, this equation holds for the lowest value of $k$. We use an induction argument. Suppose $k\ge 1$ and that ${x}_{2k-1}={x}_{k-1}+1$. Then

and

so that ${x}_{2k+1}={x}_{k}+1$. The desired recursions follow.

>From the two recursions, we see that
${x}_{n}>1$ when
$n$ is odd and exceeds
1 and
${x}_{n}<1$ when
$n$ is even and exceeds 0. Thus, the values
${x}_{0}=0$ and
${x}_{1}=1$ are assumed only once. Suppose, if possible,
that some rational is assumed twice. Let
$r$ be the smallest index
for which
${x}_{r}={x}_{s}$ for some
$s>r\ge 2$. Then
$r$ and
$s$
must have the same parity and it follows from the recursions that
${x}_{\lfloor r/2\rfloor}={x}_{\lfloor s/2\rfloor}$, contradicting
the minimality of
$r$. Hence, no value of
${x}_{n}$ is assumed more
than once.

It is straightforward to
see that, for each non-negative integer
$m$,

Also, suppose that ${x}_{u}{x}_{v}=1$. Then

so that ${x}_{2u+1}{x}_{2v}=1$.

Let
$m$ be a positive integer. We show that, for
$0\le k\le {2}^{m}-1$,
$i={2}^{m}+k$ and
$j={2}^{m+1}-(k+1)$,
${x}_{i}{x}_{j}=1$. This is clearly true for
$m=1$, since
${x}_{2}{x}_{3}=1$.
Suppose that it has been established for some particular value of
$m\ge 1$. We show that it holds for
$m$ replaced by
$m+1$.

Let
$0\le l\le {2}^{m+1}-1$,
$i={2}^{m+1}+l={2}^{m+2}-({2}^{m+1}-l)$ and
$j={2}^{m+2}-(l+1)={2}^{m+1}+({2}^{m+1}-l-1)$.
Since
$i$ and
$j$ have opposite parity and can be put into either
form, we may suppose wolog that
$l=2r$ is even, whereupon
$i=2v$ is even and
$j=2u+1$ is odd.
Thus
$v={2}^{m}+r$ and
$u={2}^{m}-(r+1)$, so that by
the induction hypothesis,
${x}_{u}{x}_{v}=1$, whence it follows that
${x}_{i}{x}_{j}={x}_{2v}{x}_{2u+1}=1$. The desired result follows by
induction.

This shows that the set
$S$ of positive rationals of the form
${x}_{n}$
contains 0, 1 and is closed under each of the operations
$x\to x+1$ and
$x\to 1/x$. Thus
$S$
contains all nonnegative integers. Suppose we have shown that
$S$ contains all positive rationals with denominators not exceeding
$s$. Consider a rational
$p/(s+1)$ with
$1\le p\le s$.
By the induction hypothesis,
$(s+1)/p\in S$, so that
$p/(s+1)\in S$.
Hence, for each nonnegative integer
$t$,

and so we can conclude that every positive rational with denominator $s+1$ belongs to $S$. Hence, by induction, we see that $S$ contains every rational.

- 136.
- Prove that, if in a semicircle of radius 1, five points $A$, $B$, $C$, $D$, $E$ are taken in consecutive order, then

and

Hence

Similarly, ${c}^{2}+{d}^{2}+{u}^{2}+\mathrm{cdu}=4$. Adding and using ${u}^{2}+{v}^{2}=4$, we obtain that

Since triangles $\mathrm{ABC}$ and $\mathrm{CDE}$ are obtuse with the largest angles at $B$ and $D$ respectively, $c=\Vert \mathrm{CD}\Vert <\Vert \mathrm{CE}\Vert =v$ and $a<u$, so that

- 137.
- Can an arbitrary convex quadrilateral be decomposed by a polygonal line into two parts, each of whose diameters is less than the diameter of the given quadrilateral?

- 138.
- (a) A room contains ten people. Among any three. there are two (mutual) acquaintances. Prove that there are four people, any two of whom are acquainted.

- (b) Does the assertion hold if ``ten'' is replaced by ``nine''?

We first show that, if any individual, say
$A$,
is acquainted with at least six people,
$B,C,D,E,F,G$, then
the conclusion follows. Suppose,
$B$, say, does not know
$C,D,E$,
then each pair of
$A,C,D,E$ must be acquainted. On the other
hand, if
$B$ knows each of
$C,D,E$, then, as two of
$C,D,E$
know each other, say
$C$ and
$D$, then each pair of
$A,B,C,D$
must be acquainted. Since any person acquainted with
$A$
must either be acquainted with or not be acquainted with three
other people acquainted with
$A$, the result follows.

(a) When there are ten people, each person either is not acquainted
with four of the others (who then make a set of four each pair of
which are acquaintances) or must be acquainted with at least
six of the others, in which case
we get the result by the previous paragraph.

(b) The assertion holds for nine people. In this case,
each person either is not
acquainted with four of the others or must be acquainted with
at least five of the others. Suppose that each person is acquainted
with at least five of the others. Then, adding together the
number of acquaintances of each of the nine people, we get a sum
of at least
$9\times 5=45$. But, each pair of acquaintances is
counted twice, so the sum must be even and so be at least 46.
But then there must be someone with at least six acquaintances, and
we can obtain the desired result.