We begin with an old problem that no one managed to solve.

90.

Let
$m$ be a positive integer, and let
$f(m)$ be the
smallest value of
$n$ for which the following statement is true:


given any set of
$n$ integers, it is always possible
to find a subset of
$m$ integers whose sum is divisible by
$m$


Determine
$f(m)$.
Solution. [N. Sato] The value of
$f(m)$ is
$2m1$.
The set of
$2m2$ numbers consisting of
$m1$ zeros and
$m1$ ones does not satisfy the property; from this we can
see that
$n$ cannot be less than
$2m1$.
We first establish that, if
$f(u)=2u1$ and
$f(v)=2v1$,
then
$f(\mathrm{uv})=2\mathrm{uv}1$. Suppose that
$2\mathrm{uv}1$ numbers are given.
Select any
$2u1$ at random. By hypothesis, there exists a
$u$subset whose sum is divisible by
$u$; remove these
$u$ elements.
Continue removing
$u$subsets in this manner until there are fewer than
$u$ numbers remaining. Since
$2\mathrm{uv}1=(2v1)u+(u1)$, we will
have
$2v1$ sets of
$u$ numbers summing to a multiple of
$u$.
For
$1\le i\le 2v1$, let
${\mathrm{ua}}_{i}$ be the sum of the
$i$th of these
$2v1$ sets. We can choose exactly
$v$ of the
${a}_{i}$ whose sum is
divisible by
$v$. The
$v$
$u$sets corresponding to these form
the desired
$\mathrm{uv}$ elements whose sum is divisible by
$\mathrm{uv}$.
Thus, if we can show that
$f(p)=2p1$ for each prime
$p$,
we can use the fact that each number is a product of primes to
show that
$f(m)=2m1$ for each positive integer
$m$.
Let
${x}_{1}$,
${x}_{2}$,
$\dots $,
${x}_{2p1}$ be
$2p1$ integers.
Wolog, we can assume that the
${x}_{i}$ have been reduced to their least
nonnegative residue modulo
$p$ and that they are in increasing
order. For
$1\le i\le p1$, let
${y}_{i}={x}_{p+i}{x}_{i}$;
we have that
${y}_{i}\ge 0$. If
${y}_{i}=0$ for some
$i$, then
${x}_{i+1}=\dots ={x}_{p+i}$, in which case
${x}_{i+1}+\dots +{x}_{p+i}$ is a multiple of
$p$ and we have achieved
our goal. Henceforth, assume that
${y}_{i}>0$ for all
$i$
Let
$s={x}_{1}+{x}_{2}+\dots +{x}_{p}$. Replacing
${x}_{i}$ by
${x}_{p+i}$ in this
sum is equivalent to adding
${y}_{i}$. We wish to show that there is
a set of the
${y}_{i}$ whose sum is congruent to
$s$ modulo
$p$; this
would indicate which of the first
$p$
${x}_{i}$ to replace to get a
sum which is a multiple of
$p$.
Suppose that
${A}_{0}=\{0\}$, and, for
$k\ge 1$, that
${A}_{k}$ is the set of distinct numbers
$i$ with
$0\le i\le p1$
which either lie in
${A}_{k1}$ or are congruent to
$a+{y}_{k}$ for
some
$a$ in
${A}_{k1}$. Note that the elements of each
${A}_{k}$ is
equal to 0 or congruent (modulo
$p$) to a sum of distinct
${y}_{i}$.
We claim that the number of elements in
${A}_{k}$ must increase by at least one for every
$k$ until
${A}_{k}$
is equal to
$\{0,1,\dots ,p1\}$.
Suppose that going from
${A}_{j1}$ to
${A}_{j}$ yields
no new elements. Since
$0\in {A}_{j1}$,
${y}_{j}\in {A}_{j}$, which means that
${y}_{j}\in {A}_{j1}$. Then
$2{y}_{j}={y}_{j}+{y}_{j}\in {A}_{j}={A}_{j1}$,
$3{y}_{j}=2{y}_{j}+{y}_{j}\in {A}_{j}={A}_{j1}$, and so on. Thus, all multiples of
${y}_{j}$ (modulo
$p$) are in
${A}_{j1}$. As
$p$ is prime, we find that
${A}_{j1}$ must contain
$\{0,1,\dots ,p1\}$. We deduce that some
sum of the
${y}_{i}$ is congruent to
$s$ modulo
$p$ and obtain the desired
result.

145.

Let
$\mathrm{ABC}$ be a right triangle with
$\angle A={90}^{\u02c6}$.
Let
$P$ be a point on the hypotenuse
$\mathrm{BC}$, and let
$Q$ and
$R$
be the respective feet of the perpendiculars from
$P$ to
$\mathrm{AC}$ and
$\mathrm{AB}$. For what position of
$P$ is the length of
$\mathrm{QR}$ minimum?
Solution.
$\mathrm{PQAR}$, being a quadrilateral with right angles
at
$A$,
$Q$ and
$R$, is a rectangle. Therefore, its diagonals
$\mathrm{QR}$ and
$\mathrm{AP}$ are equal. The length of
$\mathrm{QR}$ is minimized when
the length of
$\mathrm{AP}$ is minimized, and this occurs when
$P$ is the
foot of the perpendicular from
$A$ to
$\mathrm{BC}$.
Comment.
$P$ must be chosen so that
$\mathrm{PB}:\mathrm{PC}={\mathrm{AB}}^{2}:{\mathrm{AC}}^{2}$.

146.

Suppose that
$\mathrm{ABC}$ is an equilateral triangle.
Let
$P$ and
$Q$ be the respective midpoint of
$\mathrm{AB}$ and
$\mathrm{AC}$,
and let
$U$ and
$V$ be points on the side
$\mathrm{BC}$ with
$4\mathrm{BU}=4\mathrm{VC}=\mathrm{BC}$ and
$2\mathrm{UV}=\mathrm{BC}$. Suppose that
$\mathrm{PV}$ is joined and
that
$W$ is the foot of the perpendicular from
$U$ to
$\mathrm{PV}$ and
that
$Z$ is the foot of the perpendicular from
$Q$ to
$\mathrm{PV}$.


Explain how that four polygons
$\mathrm{APZQ}$,
$\mathrm{BUWP}$,
$\mathrm{CQZV}$ and
$\mathrm{UVW}$ can be rearranged to form a rectangle. Is this rectangle
a square?
Solution. Consider a
${180}^{\u02c6}$ rotation about
$Q$ so that
$C$ falls on
$A$,
$Z$ falls on
${Z}_{1}$ and
$V$ falls on
${V}_{1}$. The
quadrilateral
$\mathrm{QZVC}$ goes to
${\mathrm{QZ}}_{1}{V}_{1}A$,
${\mathrm{ZQZ}}_{1}$ is a line and
$\angle {\mathrm{QAV}}_{1}={60}^{\u02c6}$. Similarly, a
${180}^{\u02c6}$ rotation
about
$P$ takes quadrilateral
$\mathrm{PBUW}$ to
${\mathrm{PAU}}_{1}{W}_{1}$ with
${\mathrm{WPW}}_{1}$ a
line and
$\angle {U}_{1}\mathrm{AP}={60}^{\u02c6}$. Since
$\angle {U}_{1}\mathrm{AP}=\angle \mathrm{PAQ}=\angle {\mathrm{QAV}}_{1}={60}^{\u02c6}$,
${U}_{1}{\mathrm{AV}}_{1}$ is a line and
${U}_{1}{V}_{1}={U}_{1}A+{\mathrm{AV}}_{1}=\mathrm{UB}+\mathrm{CV}=\frac{1}{2}\mathrm{BC}=\mathrm{UV}\hspace{1em}.$
Translate
$U$ and
$V$ to fall on
${U}_{1}$ and
${V}_{1}$ respectively;
let
$W$ fall on
${W}_{2}$. Since
$\angle {W}_{1}{U}_{1}{W}_{2}=\angle {W}_{1}{U}_{1}A+\angle {W}_{2}{U}_{1}A=\angle \mathrm{WUB}+\angle \mathrm{WUV}={180}^{\u02c6}\hspace{1em},$
$\angle {W}_{2}{V}_{1}{Z}_{1}=\angle {W}_{2}{V}_{1}A+\angle {\mathrm{AV}}_{1}{Z}_{1}=\angle \mathrm{WVU}+\angle \mathrm{CVZ}={180}^{\u02c6}\hspace{1em},$
and
$\angle {W}_{2}=\angle {W}_{1}=\angle {Z}_{1}=\angle \mathrm{WZQ}={90}^{\u02c6}\hspace{1em},$
it follows that
${Z}_{1}{W}_{2}{W}_{1}Z$ is a rectangle composed of isometric
images of
$\mathrm{APZQ}$,
$\mathrm{BUWP}$,
$\mathrm{CQZV}$ and
$\mathrm{UVW}$.
Since
$\mathrm{PU}$ and
$\mathrm{QV}$ are both parallel to the median from
$A$ to
$\mathrm{BC}$, we have that
$\mathrm{PQVU}$ is a rectangle for which
$\mathrm{PU}<\mathrm{PB}=\mathrm{PQ}$. Thus,
$\mathrm{PQVU}$ is not a square and so its diagonals
$\mathrm{PV}$ and
$\mathrm{QU}$ do not intersect at right angles. It follows that
$W$ and
$Z$ do not lie on
$\mathrm{QU}$ and so must be distinct.
Since
$\mathrm{PZQ}$ and
$\mathrm{VWU}$ are right triangles with
$\angle \mathrm{QPZ}=\angle \mathrm{UVW}$ and
$\mathrm{PQ}=\mathrm{VU}$, they must be congruent, so that
$\mathrm{PZ}=\mathrm{VW}$,
$\mathrm{PW}=\mathrm{ZV}$ and
$\mathrm{UW}=\mathrm{QZ}$. Since
$\begin{array}{cc}{W}_{1}{W}_{2}\hfill & ={W}_{1}{U}_{1}+{U}_{1}{W}_{2}=\mathrm{WU}+\mathrm{UW}=\mathrm{WU}+\mathrm{QZ}\hfill \\ \multicolumn{0}{c}{}& <\mathrm{UQ}=\mathrm{PV}=\mathrm{PZ}+\mathrm{ZV}=\mathrm{PZ}+\mathrm{PW}=\mathrm{PZ}+{\mathrm{PW}}_{1}={W}_{1}Z\hspace{1em},\hfill \end{array}$
the adjacent sides of
${Z}_{1}{W}_{2}{W}_{1}Z$ are unequal, and so the rectangle
is not square.
Comment. The inequality of the adjacent sides of the rectangle
can be obtained also by making measurements. Take 4 as the length
of a side of triangle
$\mathrm{ABC}$. Then
$\Vert \mathrm{PU}\Vert =\sqrt{3}\hspace{1em},\hspace{1em}\hspace{1em}\hspace{1em}\Vert \mathrm{PQ}\Vert =2\hspace{1em},\hspace{1em}\hspace{1em}\hspace{1em}\Vert \mathrm{QU}\Vert =\Vert \mathrm{PV}\Vert =\sqrt{7}\hspace{1em}.$
Since the triangles
$\mathrm{PUW}$ and
$\mathrm{PVU}$ are similar,
$\mathrm{UW}:\mathrm{PU}=\mathrm{VU}:\mathrm{PV}$, whence
$\Vert \mathrm{UW}\Vert =2\sqrt{21}/7$. Thus,
$\Vert {W}_{1}{W}_{2}\Vert =4\sqrt{21}/7\ne \sqrt{7}=\Vert {W}_{1}Z\Vert $.
One can also use the fact that the areas of the triangle and rectangle
are equal. The area of the triangle is
$4\sqrt{3}$. It just needs to
be verified that one of the sides of the rectangle is not equal to
the square root of this.

147.

Let
$a>0$ and let
$n$ be a positive integer.
Determine the maximum value of
$\frac{{x}_{1}{x}_{2}\dots {x}_{n}}{(1+{x}_{1})({x}_{1}+{x}_{2})\dots ({x}_{n1}+{x}_{n})({x}_{n}+{a}^{n+1})}$
subject to the constraint that
${x}_{1},{x}_{2},\dots ,{x}_{n}>0$.
Solution. Let
${u}_{0}={x}_{1}$,
${u}_{i}={x}_{i+1}/{x}_{i}$ for
$1\le i\le n1$ and
${u}_{n}={a}^{n+1}/{x}_{n}$. Observe that
${u}_{0}{u}_{1}\dots {u}_{n}={a}^{n+1}$. The quantity in the problem
is the reciprocal of
$\begin{array}{cc}(1+{u}_{0})\hfill & (1+{u}_{1})(1+{u}_{2})\dots (1+{u}_{n})\hfill \\ \multicolumn{0}{c}{}& =1+\sum {u}_{i}+\sum {u}_{i}{u}_{j}+\dots +\sum {u}_{{i}_{1}}{u}_{{i}_{2}}\dots {u}_{{i}_{k}}+\dots +{u}_{0}{u}_{1}\dots {u}_{n}\hspace{1em}.\hfill \end{array}$
For
$k=1,2,\dots ,n$, the sum
$\sum {u}_{{i}_{1}}{u}_{{i}_{2}}\dots {u}_{{i}_{k}}$
adds together all the
$\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)$
$k$fold products of the
${u}_{i}$; the product of all the terms in this sum is equal to
${a}^{n+1}$ raised to the power
$\left(\genfrac{}{}{0ex}{}{n}{k1}\right)$, namely, to
$a$ raised to the power
$k\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)$. By the arithmeticgeometric
means inequality
$\sum {u}_{{i}_{1}}{u}_{{i}_{2}}\dots {u}_{{i}_{k}}\ge \left(\genfrac{}{}{0ex}{}{n+1}{k}\right){a}^{k}\hspace{1em}.$
Hence
$(1+{u}_{0})(1+{u}_{1})\dots (1+{u}_{n})\ge 1+(n+1)a+\dots +\left(\genfrac{}{}{0ex}{}{n+1}{k}\right){a}^{k}+\dots {a}^{n+1}=(1+a){}^{n+1}\hspace{1em},$
with equality if and only if
${u}_{0}={u}_{1}=\dots ={u}_{n}=a$.
If follows from this that the quantity in the problem has maximum
value of
$(1+a){}^{(n+1)}$, with equality if and only if
${x}_{i}={a}_{i}$ for
$1\le i\le n$.
Comment. Some of you tried the following strategy. If any two
of the
${u}_{i}$ were unequal, they showed that a larger value could
be obtained for the given expression by replacing each of these by
another value. They then deduced that the maximum occurred when
all the
${u}_{i}$ were equal. There is a subtle difficulty here.
What has really been proved is that, if there is a maximum,
it can occur only when the
${u}_{i}$ are equal. However, it begs the
question of the existence of a maximum. To appreciate the point,
consider the following argument that
$1$ is the largest postive
integer. We note that, given any integer
$n$ exceeding 1, we can find
another integer that exceeds
$n$, namely
${n}^{2}$. Thus, no integer
exceeding 1 can be the largest positive integer. Therefore, 1 itself
must be the largest.
Some of you tried a similar approach with the
${x}_{i}$, and showed that
for a maximum, one must have all the
${x}_{i}$ equal to 1. However, they
neglected to build in the relationship between
${x}_{n}$ and
${a}_{n+1}$,
which of course cannot be equal if all the
${x}_{i}$ are 1 and
$a\ne 1$.
This leaves open the possibility of making the given expression larger
by bettering the relationship between the
${x}_{i}$ and
$a$ and possibly
allowing inequalities of the variables.

148.

For a given prime number
$p$, find the number of
distinct sequences of natural numbers (positive integers)
$\{{a}_{0},{a}_{1},\dots ,{a}_{n}\dots \}$ satisfying, for each
positive integer
$n$, the equation
$\frac{{a}_{0}}{{a}_{1}}+\frac{{a}_{0}}{{a}_{2}}+\dots +\frac{{a}_{0}}{{a}_{n}}+\frac{p}{{a}_{n+1}}=1\hspace{1em}.$
Solution. For
$n\ge 3$ we have that
$\begin{array}{cc}1=\hfill & \frac{{a}_{0}}{{a}_{1}}+\dots +\frac{{a}_{0}}{{a}_{2}}+\dots +\frac{{a}_{0}}{{a}_{n2}}+\frac{p}{{a}_{n1}}\hfill \\ \multicolumn{0}{c}{}& =\frac{{a}_{0}}{{a}_{1}}+\frac{{a}_{0}}{{a}_{2}}+\dots +\frac{{a}_{0}}{{a}_{n1}}+\frac{p}{{a}_{n}}\hfill \end{array}$
whence
$\frac{p}{{a}_{n1}}=\frac{{a}_{0}}{{a}_{n1}}+\frac{p}{{a}_{n}}\hspace{1em},$
so that
${a}_{n}=\frac{{\mathrm{pa}}_{n1}}{p{a}_{0}}\hspace{1em}.$
Thus, for
$n\ge 2$, we have that
${a}_{n}=\frac{{p}^{n2}{a}_{2}}{(p{a}_{0}){}^{n2}}\hspace{1em}.$
Since
$1\le p{a}_{0}\le p1$,
$p{a}_{0}$ and
$p$ are coprime.
It follows that, either
$p{a}_{0}$ must divide
${a}_{2}$ to an arbitrarily
high power (impossible!) or
$p{a}_{0}=1$.
Therefore,
${a}_{0}=p1$ and
${a}_{n}={p}^{n2}{a}_{2}$ for
$n\ge 2$.
Thus, once
${a}_{1}$ and
${a}_{2}$ are selected, then the rest of the
sequence
$\{{a}_{n}\}$ is determined. The remaining condition that
has to be satisfied is
$1=\frac{{a}_{0}}{{a}_{1}}+\frac{p}{{a}_{2}}=\frac{p1}{{a}_{1}}+\frac{p}{{a}_{2}}\hspace{1em}.$
This is equivalent to
$(p1){a}_{2}+{\mathrm{pa}}_{1}={a}_{1}{a}_{2}\hspace{1em},$
or
$[{a}_{1}(p1)][{a}_{2}p]=p(p1)\hspace{1em}.$
The factors
${a}_{1}(p1)$ and
${a}_{2}p$ must be both negative or
both positive. The former case is excluded by the fact that
$(p1){a}_{1}$ and
$p{a}_{2}$ are respectively less than
$p1$ and
$p$. Hence, each choice of the pair
$({a}_{1},{a}_{2})$ corresponds to
a choice of a pair of positive divisors of
$p(p1)$. There are
$d(p(p1))=2d(p1)$ such choices, where
$d(n)$ is the number of
positive divisors of the positive integer
$n$.
Comment. When
$p=5$, for example, the possibilities for
$({a}_{1},{a}_{2})$ are
$(5,25)$,
$(6,15)$,
$(8,10)$,
$(9,9)$,
$(14,7)$,
$(24,6)$. In general, particular choices of
sequences that work are
$\{p1,p,{p}^{2},{p}^{3},\dots \}$
$\{p1,2p1,2p1,p(2p1),\dots \}$
$\{p1,{p}^{2}1,p+1,p(p+1),\dots \}\hspace{1em}.$
A variant on the argument showing that the
${a}_{n}$ from some point
on constituted a geometric progression started with the relation
$p({a}_{n}{a}_{n1})={a}_{0}{a}_{n}$ for
$n\ge 3$, whence
$\frac{{a}_{n1}}{{a}_{n}}=1\frac{{a}_{0}}{p}\hspace{1em}.$
Thus, for
$n\ge 3$,
${a}_{n+1}{a}_{n1}={a}_{n}^{2}$, which forces
$\{{a}_{2},{a}_{3},\dots ,\}$ to be a geometric progession. The
common ratio must be a positive integer
$r$ for which
$r=p/(p{a}_{0})$. This forces
$p{a}_{0}$ to be equal to 1.
Quite a few solvers lost points because of poor bookkeeping;
they did not identify the correct place at which the geometric
progression began. It is often a good idea to write out the
first few equations of a general relation explicitly in order
to avoid this type of confusion. You must learn to pay attention
to details and check work carefully; otherwise, you may find yourself
settling for a score on a competition less than you really deserve
on the basis of ability.

149.

Consider a cube concentric with a parallelepiped
(rectangular box) with sides
$a<b<c$ and faces parallel
to that of the cube. Find the side length of the cube for which
the difference between the volume of the union and the volume of the
intersection of the cube and parallelepiped is minimum.
Solution. Let
$x$ be the length of the side of the cube and
let
$f(x)$ be the difference between the value of the union and the
volume of the intersection of the two solids. Then
$f(x)=\{\begin{array}{cc}\hfill \mathrm{abc}{x}^{3}\hfill & \hfill (0\le x<a)\hfill \\ \multicolumn{0}{c}{\mathrm{abc}+(xa){x}^{2}{\mathrm{ax}}^{2}=\mathrm{abc}+{x}^{3}2{\mathrm{ax}}^{2}}& \hfill (a\le x<b)\hfill \\ \multicolumn{0}{c}{{x}^{3}+\mathrm{ab}(cx)\mathrm{abx}=\mathrm{abc}+{x}^{3}2\mathrm{abx}}& \hfill (b\le x<c)\hfill \\ \multicolumn{0}{c}{{x}^{3}\mathrm{abc}}& \hfill (c\le x)\hfill \end{array}$
The function decreases for
$0\le x\le a$ and increases for
$x\ge c$. For
$b\le x\le c$,
$\begin{array}{cc}f(x)f(b)\hfill & ={x}^{3}2\mathrm{abx}{b}^{3}+2{\mathrm{ab}}^{2}\hfill \\ \multicolumn{0}{c}{}& =(xb)[{x}^{2}+\mathrm{bx}+{b}^{2}2\mathrm{ab}]\hfill \\ \multicolumn{0}{c}{}& =(xb)[({x}^{2}\mathrm{ab})+b(xa)+{b}^{2}]\ge 0\hspace{1em},\hfill \end{array}$
so that
$f(x)\ge f(b)$. Hence, the minimum value of
$f(x)$ must
be assumed when
$a\le x\le b$.
For
$a\le x\le b$,
$f\text{'}(x)=3{x}^{2}4\mathrm{ax}=x[3x4a]$,
so that
$f(x)$ increases for
$x\ge 4a/3$ and decreases for
$x\le 4a/3$. When
$b\le 4a/3$, then
$f(x)$ is decreasing on
the closed interval
$[a,b]$ and assumes its minimum for
$x=b$.
If
$b>4a/3>a$, then
$f(x)$ increases on
$[4a/3,b]$ and
so achieves its minimum when
$x=4a/3$. Hence, the function
$f(x)$ is minimized when
$x=\mathrm{min}(b,4a/3)$.

150.

The area of the bases of a truncated pyramid are equal
to
${S}_{1}$ and
${S}_{2}$ and the total area of the lateral surface is
$S$. Prove that, if there is a plane parallel to each of the bases
that partitions the truncated
pyramid into two truncated pyramids within
each of which a sphere can be inscribed, then
$S=(\sqrt{{S}_{1}}+\sqrt{{S}_{2}})(\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}}){}^{2}\hspace{1em}.$
Solution 1. Let
${M}_{1}$ be the larger base of the truncated
pyramid with area
${S}_{1}$, and
${M}_{2}$ the smaller base with area
${S}_{2}$.
Let
${P}_{1}$ be the entire pyramid with base
${M}_{1}$ of which the truncated
pyramid is a part. Let
${M}_{0}$ be the base parallel to
${M}_{1}$ and
${M}_{2}$ described in the problem, and let its area be
${S}_{0}$. Let
${P}_{0}$ be the pyramid with base
${M}_{0}$ and
${P}_{2}$ the pyramid with
base
${M}_{2}$.
The inscribed sphere bounded by
${M}_{0}$ and
${M}_{1}$ is determined by
the condition that it touches
${M}_{1}$ and the lateral faces of the
pyramid; thus, it is the inscribed sphere of the pyramid
${P}_{1}$ with
base
${M}_{1}$; let its radius be
${R}_{1}$. The inscribed sphere bounded by
${M}_{2}$ and
${M}_{0}$ is the inscribed sphere of the pyramid
${P}_{0}$ with base
${M}_{0}$; let its radius be
${R}_{0}$. Finally, let the inscribed sphere of
the pyramid with base
${M}_{2}$ have radius
${R}_{2}$.
Suppose
${Q}_{2}$ is the lateral area of pyramid
${P}_{2}$ and
${Q}_{1}$ the
lateral area of pyramid
${P}_{1}$. Thus,
$S={Q}_{1}{Q}_{2}$.
There is a dilation with factor
${R}_{0}/{R}_{1}$ that takes pyramid
${P}_{1}$
to
${P}_{0}$; since it takes the inscribed sphere of
${P}_{1}$ to that
of
${P}_{0}$, it takes the base
${M}_{1}$ to
${M}_{0}$ and the base
${M}_{0}$ to
${M}_{2}$. Hence, this dilation takes
${P}_{0}$ to
${P}_{2}$. The dilation
composed with itself takes
${P}_{1}$ to
${P}_{2}$. Therefore
$\frac{{R}_{0}}{{R}_{1}}=\frac{{R}_{2}}{{R}_{0}}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\mathrm{and}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\frac{{Q}_{2}}{{Q}_{1}}=\frac{{S}_{2}}{{S}_{1}}=\frac{{R}_{2}^{2}}{{R}_{1}^{2}}\hspace{1em}.$
Consider the volume of
${P}_{2}$. Since
${P}_{2}$ is the union of pyramids of
height
${R}_{2}$ and with bases the lateral faces of
${P}_{2}$ and
${M}_{2}$, its
volume is
$(1/3){R}_{2}({Q}_{2}+{S}_{2})$. However, we can find the volume of
${P}_{2}$ another way.
${P}_{2}$ can be realized as the union of pyramids
whose bases are its lateral faces and whose apexes are the centre of
the inscribed sphere with radius
${R}_{0}$ with the removal of the
pyramid of base
${M}_{2}$ and apex at the centre of the same sphere.
Thus, the volume is also equal to
$(1/3){R}_{0}({Q}_{2}{S}_{2})$.
Hence
$\frac{{Q}_{2}{S}_{2}}{{Q}_{2}+{S}_{2}}=\frac{{R}_{2}}{{R}_{0}}=\frac{{R}_{2}}{\sqrt{{R}_{1}{R}_{2}}}=\frac{\sqrt{{R}_{2}}}{\sqrt{{R}_{1}}}=\frac{\sqrt[4]{{S}_{2}}}{\sqrt[4]{{S}_{1}}}$
$\Rightarrow {Q}_{2}(\sqrt[4]{{S}_{1}}\sqrt[4]{{S}_{2}})={S}_{2}(\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}})\hspace{1em},$
so that
$\begin{array}{cc}S\hfill & ={Q}_{1}{Q}_{2}=\frac{{Q}_{2}}{{S}_{2}}({S}_{1}{S}_{2})\hfill \\ \multicolumn{0}{c}{}& =[\frac{\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}}}{\sqrt[4]{{S}_{1}}\sqrt[4]{{S}_{2}}}][\sqrt{{S}_{1}}\sqrt{{S}_{2}}][\sqrt{{S}_{1}}+\sqrt{{S}_{2}}]\hfill \\ \multicolumn{0}{c}{}& =(\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}}){}^{2}(\sqrt{{S}_{1}}+\sqrt{{S}_{2}})\hspace{1em}.\hfill \end{array}$
Solution 2. [S. En Lu] Consider an arbitrary truncated pyramid
with bases
${A}_{1}$ and
${A}_{2}$ of respective areas
${\sigma}_{1}$ and
${\sigma}_{2}$, in which a sphere
$\Gamma $ of centre
$O$ is inscribed.
Let the lateral area be
$\sigma $. Suppose that
$C$ is a lateral face
and that
$\Gamma $ touches
${A}_{1}$,
${A}_{2}$ and
$C$ in the respective points
${P}_{1}$,
${P}_{2}$ and
$Q$.
$C$ is a trapezoid with sides of lengths
${a}_{1}$ and
${a}_{2}$ incident
with the respective bases
${A}_{1}$ and
${A}_{2}$; let
${h}_{1}$ and
${h}_{2}$ be
the respective lengths of the altitudes
of triangles with apexes
${P}_{1}$ and
${P}_{2}$
and bases bordering on
$C$. By similarity (of
${A}_{1}$ and
${A}_{2}$),
$\frac{{h}_{1}}{{h}_{2}}=\frac{{a}_{1}}{{a}_{2}}=\sqrt{\frac{{\sigma}_{1}}{{\sigma}_{2}}}\hspace{1em}.$
The plane that contains these altitudes passes through
${P}_{1}{P}_{2}$
(a diameter of
$\Gamma $) as well as
$Q$, the point on
$C$ nearest to
the centre of
$\Gamma $. Since the height of
$C$ is
${a}_{1}+{a}_{2}$
[why?], its area is
$\begin{array}{cc}\frac{1}{2}({a}_{1}+{a}_{2})({h}_{1}+{h}_{2})\hfill & =\frac{1}{2}[{a}_{1}{h}_{1}+{a}_{2}{h}_{2}+{a}_{1}{h}_{2}+{a}_{2}{h}_{1}]\hfill \\ \multicolumn{0}{c}{}& =\frac{1}{2}[{a}_{1}{h}_{1}+{a}_{2}{h}_{2}+2\sqrt{{a}_{1}{a}_{2}{h}_{1}{h}_{2}}]\hfill \\ \multicolumn{0}{c}{}& =\frac{1}{2}[{a}_{1}{h}_{1}+{a}_{2}{h}_{2}+2{a}_{2}{h}_{2}\sqrt{{\sigma}_{1}/{\sigma}_{2}}]\hspace{1em}.\hfill \end{array}$
Adding the corresponding equations over all the lateral faces
$C$ yields
$\sigma ={\sigma}_{1}+{\sigma}_{2}+\sqrt{{\sigma}_{1}{\sigma}_{2}}=(\sqrt{{\sigma}_{1}}+\sqrt{{\sigma}_{2}}){}^{2}\hspace{1em}.$
With
${S}_{0}$ defined as in Solution 1, we have that
${S}_{1}/{S}_{0}={S}_{0}/{S}_{2}$, so that
${S}_{0}=\sqrt{{S}_{1}{S}_{2}}$. Using the results of
the first paragraph applied to the truncated pyramids of bases
$({S}_{2},{S}_{0})$ and
$({S}_{0},{S}_{1})$, we obtain that
$\begin{array}{cc}S\hfill & =(\sqrt{{S}_{1}}+\sqrt{{S}_{0}}){}^{2}+(\sqrt{{S}_{0}}+\sqrt{{S}_{1}}){}^{2}\hfill \\ \multicolumn{0}{c}{}& =(\sqrt{{S}_{1}}+\sqrt[4]{{S}_{1}{S}_{2}}){}^{2}+(\sqrt[4]{{S}_{1}{S}_{2}}+\sqrt{{S}_{2}}){}^{2}\hfill \\ \multicolumn{0}{c}{}& =(\sqrt{{S}_{1}}+\sqrt{{S}_{2}})(\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}}){}^{2}\hspace{1em}.\hfill \end{array}$