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

90.
Let $m$ be a positive integer, and let $f\left(m\right)$ 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 $f\left(m\right)$.

Solution. [N. Sato] The value of $f\left(m\right)$ is $2m-1$. The set of $2m-2$ numbers consisting of $m-1$ zeros and $m-1$ ones does not satisfy the property; from this we can see that $n$ cannot be less than $2m-1$.
We first establish that, if $f\left(u\right)=2u-1$ and $f\left(v\right)=2v-1$, then $f\left(\mathrm{uv}\right)=2\mathrm{uv}-1$. Suppose that $2\mathrm{uv}-1$ numbers are given. Select any $2u-1$ 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 $2\mathrm{uv}-1=\left(2v-1\right)u+\left(u-1\right)$, we will have $2v-1$ sets of $u$ numbers summing to a multiple of $u$. For $1\le i\le 2v-1$, let ${\mathrm{ua}}_{i}$ be the sum of the $i$th of these $2v-1$ 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 $f\left(p\right)=2p-1$ for each prime $p$, we can use the fact that each number is a product of primes to show that $f\left(m\right)=2m-1$ for each positive integer $m$.
Let ${x}_{1}$, ${x}_{2}$, $\dots$, ${x}_{2p-1}$ be $2p-1$ integers. Wolog, we can assume that the ${x}_{i}$ have been reduced to their least non-negative residue modulo $p$ and that they are in increasing order. For $1\le i\le p-1$, 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 ${A}_{0}=\left\{0\right\}$, and, for $k\ge 1$, that ${A}_{k}$ is the set of distinct numbers $i$ with $0\le i\le p-1$ which either lie in ${A}_{k-1}$ or are congruent to $a+{y}_{k}$ for some $a$ in ${A}_{k-1}$. 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 $\left\{0,1,\dots ,p-1\right\}$.
Suppose that going from ${A}_{j-1}$ to ${A}_{j}$ yields no new elements. Since $0\in {A}_{j-1}$, ${y}_{j}\in {A}_{j}$, which means that ${y}_{j}\in {A}_{j-1}$. Then $2{y}_{j}={y}_{j}+{y}_{j}\in {A}_{j}={A}_{j-1}$, $3{y}_{j}=2{y}_{j}+{y}_{j}\in {A}_{j}={A}_{j-1}$, and so on. Thus, all multiples of ${y}_{j}$ (modulo $p$) are in ${A}_{j-1}$. As $p$ is prime, we find that ${A}_{j-1}$ must contain $\left\{0,1,\dots ,p-1\right\}$. 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 $\angle A={90}^{ˆ}$. 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 ${180}^{ˆ}$ 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 $\angle {\mathrm{QAV}}_{1}={60}^{ˆ}$. Similarly, a ${180}^{ˆ}$ rotation about $P$ takes quadrilateral $\mathrm{PBUW}$ to ${\mathrm{PAU}}_{1}{W}_{1}$ with ${\mathrm{WPW}}_{1}$ a line and $\angle {U}_{1}\mathrm{AP}={60}^{ˆ}$. Since $\angle {U}_{1}\mathrm{AP}=\angle \mathrm{PAQ}=\angle {\mathrm{QAV}}_{1}={60}^{ˆ}$, ${U}_{1}{\mathrm{AV}}_{1}$ is a line and

${U}_{1}{V}_{1}={U}_{1}A+{\mathrm{AV}}_{1}=\mathrm{UB}+\mathrm{CV}=\frac{1}{2}\mathrm{BC}=\mathrm{UV} .$

Translate $U$ and $V$ to fall on ${U}_{1}$ and ${V}_{1}$ respectively; let $W$ fall on ${W}_{2}$. Since

$\angle {W}_{1}{U}_{1}{W}_{2}=\angle {W}_{1}{U}_{1}A+\angle {W}_{2}{U}_{1}A=\angle \mathrm{WUB}+\angle \mathrm{WUV}={180}^{ˆ} ,$

$\angle {W}_{2}{V}_{1}{Z}_{1}=\angle {W}_{2}{V}_{1}A+\angle {\mathrm{AV}}_{1}{Z}_{1}=\angle \mathrm{WVU}+\angle \mathrm{CVZ}={180}^{ˆ} ,$

and

$\angle {W}_{2}=\angle {W}_{1}=\angle {Z}_{1}=\angle \mathrm{WZQ}={90}^{ˆ} ,$

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

$\begin{array}{cc}{W}_{1}{W}_{2}\hfill & ={W}_{1}{U}_{1}+{U}_{1}{W}_{2}=\mathrm{WU}+\mathrm{UW}=\mathrm{WU}+\mathrm{QZ}\hfill \\ \multicolumn{0}{c}{}& <\mathrm{UQ}=\mathrm{PV}=\mathrm{PZ}+\mathrm{ZV}=\mathrm{PZ}+\mathrm{PW}=\mathrm{PZ}+{\mathrm{PW}}_{1}={W}_{1}Z ,\hfill \end{array}$

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

$‖\mathrm{PU}‖=\sqrt{3} , ‖\mathrm{PQ}‖=2 , ‖\mathrm{QU}‖=‖\mathrm{PV}‖=\sqrt{7} .$

Since the triangles $\mathrm{PUW}$ and $\mathrm{PVU}$ are similar, $\mathrm{UW}:\mathrm{PU}=\mathrm{VU}:\mathrm{PV}$, whence $‖\mathrm{UW}‖=2\sqrt{21}/7$. Thus, $‖{W}_{1}{W}_{2}‖=4\sqrt{21}/7\ne \sqrt{7}=‖{W}_{1}Z‖$.
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

$\frac{{x}_{1}{x}_{2}\dots {x}_{n}}{\left(1+{x}_{1}\right)\left({x}_{1}+{x}_{2}\right)\dots \left({x}_{n-1}+{x}_{n}\right)\left({x}_{n}+{a}^{n+1}\right)}$

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 $1\le i\le n-1$ 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

$\begin{array}{cc}\left(1+{u}_{0}\right)\hfill & \left(1+{u}_{1}\right)\left(1+{u}_{2}\right)\dots \left(1+{u}_{n}\right)\hfill \\ \multicolumn{0}{c}{}& =1+\sum {u}_{i}+\sum {u}_{i}{u}_{j}+\dots +\sum {u}_{{i}_{1}}{u}_{{i}_{2}}\dots {u}_{{i}_{k}}+\dots +{u}_{0}{u}_{1}\dots {u}_{n} .\hfill \end{array}$

For $k=1,2,\dots ,n$, the sum $\sum {u}_{{i}_{1}}{u}_{{i}_{2}}\dots {u}_{{i}_{k}}$ adds together all the $\left(\genfrac{}{}{0}{}{n+1}{k}\right)$ $k-$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 $\left(\genfrac{}{}{0}{}{n}{k-1}\right)$, namely, to $a$ raised to the power $k\left(\genfrac{}{}{0}{}{n+1}{k}\right)$. By the arithmetic-geometric means inequality

$\sum {u}_{{i}_{1}}{u}_{{i}_{2}}\dots {u}_{{i}_{k}}\ge \left(\genfrac{}{}{0}{}{n+1}{k}\right){a}^{k} .$

Hence

$\left(1+{u}_{0}\right)\left(1+{u}_{1}\right)\dots \left(1+{u}_{n}\right)\ge 1+\left(n+1\right)a+\dots +\left(\genfrac{}{}{0}{}{n+1}{k}\right){a}^{k}+\dots {a}^{n+1}=\left(1+a\right){}^{n+1} ,$

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 $\left(1+a\right){}^{-\left(n+1\right)}$, 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) $\left\{{a}_{0},{a}_{1},\dots ,{a}_{n}\dots \right\}$ satisfying, for each positive integer $n$, the equation

$\frac{{a}_{0}}{{a}_{1}}+\frac{{a}_{0}}{{a}_{2}}+\dots +\frac{{a}_{0}}{{a}_{n}}+\frac{p}{{a}_{n+1}}=1 .$

Solution. For $n\ge 3$ we have that

$\begin{array}{cc}1=\hfill & \frac{{a}_{0}}{{a}_{1}}+\dots +\frac{{a}_{0}}{{a}_{2}}+\dots +\frac{{a}_{0}}{{a}_{n-2}}+\frac{p}{{a}_{n-1}}\hfill \\ \multicolumn{0}{c}{}& =\frac{{a}_{0}}{{a}_{1}}+\frac{{a}_{0}}{{a}_{2}}+\dots +\frac{{a}_{0}}{{a}_{n-1}}+\frac{p}{{a}_{n}}\hfill \end{array}$

whence

$\frac{p}{{a}_{n-1}}=\frac{{a}_{0}}{{a}_{n-1}}+\frac{p}{{a}_{n}} ,$

so that

${a}_{n}=\frac{{\mathrm{pa}}_{n-1}}{p-{a}_{0}} .$

Thus, for $n\ge 2$, we have that

${a}_{n}=\frac{{p}^{n-2}{a}_{2}}{\left(p-{a}_{0}\right){}^{n-2}} .$

Since $1\le p-{a}_{0}\le p-1$, $p-{a}_{0}$ 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, ${a}_{0}=p-1$ and ${a}_{n}={p}^{n-2}{a}_{2}$ for $n\ge 2$. Thus, once ${a}_{1}$ and ${a}_{2}$ are selected, then the rest of the sequence $\left\{{a}_{n}\right\}$ is determined. The remaining condition that has to be satisfied is

$1=\frac{{a}_{0}}{{a}_{1}}+\frac{p}{{a}_{2}}=\frac{p-1}{{a}_{1}}+\frac{p}{{a}_{2}} .$

This is equivalent to

$\left(p-1\right){a}_{2}+{\mathrm{pa}}_{1}={a}_{1}{a}_{2} ,$

or

$\left[{a}_{1}-\left(p-1\right)\right]\left[{a}_{2}-p\right]=p\left(p-1\right) .$

The factors ${a}_{1}-\left(p-1\right)$ and ${a}_{2}-p$ must be both negative or both positive. The former case is excluded by the fact that $\left(p-1\right)-{a}_{1}$ and $p-{a}_{2}$ are respectively less than $p-1$ and $p$. Hence, each choice of the pair $\left({a}_{1},{a}_{2}\right)$ corresponds to a choice of a pair of positive divisors of $p\left(p-1\right)$. There are $d\left(p\left(p-1\right)\right)=2d\left(p-1\right)$ such choices, where $d\left(n\right)$ is the number of positive divisors of the positive integer $n$.
Comment. When $p=5$, for example, the possibilities for $\left({a}_{1},{a}_{2}\right)$ are $\left(5,25\right)$, $\left(6,15\right)$, $\left(8,10\right)$, $\left(9,9\right)$, $\left(14,7\right)$, $\left(24,6\right)$. In general, particular choices of sequences that work are

$\left\{p-1,p,{p}^{2},{p}^{3},\dots \right\}$

$\left\{p-1,2p-1,2p-1,p\left(2p-1\right),\dots \right\}$

$\left\{p-1,{p}^{2}-1,p+1,p\left(p+1\right),\dots \right\} .$

A variant on the argument showing that the ${a}_{n}$ from some point on constituted a geometric progression started with the relation $p\left({a}_{n}-{a}_{n-1}\right)={a}_{0}{a}_{n}$ for $n\ge 3$, whence

$\frac{{a}_{n-1}}{{a}_{n}}=1-\frac{{a}_{0}}{p} .$

Thus, for $n\ge 3$, ${a}_{n+1}{a}_{n-1}={a}_{n}^{2}$, which forces $\left\{{a}_{2},{a}_{3},\dots ,\right\}$ to be a geometric progession. The common ratio must be a positive integer $r$ for which $r=p/\left(p-{a}_{0}\right)$. This forces $p-{a}_{0}$ to be equal to 1.
Quite a few solvers lost points because of poor book-keeping; 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 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 $f\left(x\right)$ be the difference between the value of the union and the volume of the intersection of the two solids. Then

$f\left(x\right)=\left\{\begin{array}{cc}\hfill \mathrm{abc}-{x}^{3}\hfill & \hfill \left(0\le x

The function decreases for $0\le x\le a$ and increases for $x\ge c$. For $b\le x\le c$,

$\begin{array}{cc}f\left(x\right)-f\left(b\right)\hfill & ={x}^{3}-2\mathrm{abx}-{b}^{3}+2{\mathrm{ab}}^{2}\hfill \\ \multicolumn{0}{c}{}& =\left(x-b\right)\left[{x}^{2}+\mathrm{bx}+{b}^{2}-2\mathrm{ab}\right]\hfill \\ \multicolumn{0}{c}{}& =\left(x-b\right)\left[\left({x}^{2}-\mathrm{ab}\right)+b\left(x-a\right)+{b}^{2}\right]\ge 0 ,\hfill \end{array}$

so that $f\left(x\right)\ge f\left(b\right)$. Hence, the minimum value of $f\left(x\right)$ must be assumed when $a\le x\le b$.
For $a\le x\le b$, $f\text{'}\left(x\right)=3{x}^{2}-4\mathrm{ax}=x\left[3x-4a\right]$, so that $f\left(x\right)$ increases for $x\ge 4a/3$ and decreases for $x\le 4a/3$. When $b\le 4a/3$, then $f\left(x\right)$ is decreasing on the closed interval $\left[a,b\right]$ and assumes its minimum for $x=b$. If $b>4a/3>a$, then $f\left(x\right)$ increases on $\left[4a/3,b\right]$ and so achieves its minimum when $x=4a/3$. Hence, the function $f\left(x\right)$ is minimized when $x=\mathrm{min}\left(b,4a/3\right)$.

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

$S=\left(\sqrt{{S}_{1}}+\sqrt{{S}_{2}}\right)\left(\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}}\right){}^{2} .$

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

$\frac{{R}_{0}}{{R}_{1}}=\frac{{R}_{2}}{{R}_{0}} \mathrm{and} \frac{{Q}_{2}}{{Q}_{1}}=\frac{{S}_{2}}{{S}_{1}}=\frac{{R}_{2}^{2}}{{R}_{1}^{2}} .$

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 $\left(1/3\right){R}_{2}\left({Q}_{2}+{S}_{2}\right)$. 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 $\left(1/3\right){R}_{0}\left({Q}_{2}-{S}_{2}\right)$.
Hence

$\frac{{Q}_{2}-{S}_{2}}{{Q}_{2}+{S}_{2}}=\frac{{R}_{2}}{{R}_{0}}=\frac{{R}_{2}}{\sqrt{{R}_{1}{R}_{2}}}=\frac{\sqrt{{R}_{2}}}{\sqrt{{R}_{1}}}=\frac{\sqrt[4]{{S}_{2}}}{\sqrt[4]{{S}_{1}}}$

$⇒{Q}_{2}\left(\sqrt[4]{{S}_{1}}-\sqrt[4]{{S}_{2}}\right)={S}_{2}\left(\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}}\right) ,$

so that

$\begin{array}{cc}S\hfill & ={Q}_{1}-{Q}_{2}=\frac{{Q}_{2}}{{S}_{2}}\left({S}_{1}-{S}_{2}\right)\hfill \\ \multicolumn{0}{c}{}& =\left[\frac{\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}}}{\sqrt[4]{{S}_{1}}-\sqrt[4]{{S}_{2}}}\right]\left[\sqrt{{S}_{1}}-\sqrt{{S}_{2}}\right]\left[\sqrt{{S}_{1}}+\sqrt{{S}_{2}}\right]\hfill \\ \multicolumn{0}{c}{}& =\left(\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}}\right){}^{2}\left(\sqrt{{S}_{1}}+\sqrt{{S}_{2}}\right) .\hfill \end{array}$

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 ${A}_{2}$),

$\frac{{h}_{1}}{{h}_{2}}=\frac{{a}_{1}}{{a}_{2}}=\sqrt{\frac{{\sigma }_{1}}{{\sigma }_{2}}} .$

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

$\begin{array}{cc}\frac{1}{2}\left({a}_{1}+{a}_{2}\right)\left({h}_{1}+{h}_{2}\right)\hfill & =\frac{1}{2}\left[{a}_{1}{h}_{1}+{a}_{2}{h}_{2}+{a}_{1}{h}_{2}+{a}_{2}{h}_{1}\right]\hfill \\ \multicolumn{0}{c}{}& =\frac{1}{2}\left[{a}_{1}{h}_{1}+{a}_{2}{h}_{2}+2\sqrt{{a}_{1}{a}_{2}{h}_{1}{h}_{2}}\right]\hfill \\ \multicolumn{0}{c}{}& =\frac{1}{2}\left[{a}_{1}{h}_{1}+{a}_{2}{h}_{2}+2{a}_{2}{h}_{2}\sqrt{{\sigma }_{1}/{\sigma }_{2}}\right] .\hfill \end{array}$

Adding the corresponding equations over all the lateral faces $C$ yields

$\sigma ={\sigma }_{1}+{\sigma }_{2}+\sqrt{{\sigma }_{1}{\sigma }_{2}}=\left(\sqrt{{\sigma }_{1}}+\sqrt{{\sigma }_{2}}\right){}^{2} .$

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 $\left({S}_{2},{S}_{0}\right)$ and $\left({S}_{0},{S}_{1}\right)$, we obtain that

$\begin{array}{cc}S\hfill & =\left(\sqrt{{S}_{1}}+\sqrt{{S}_{0}}\right){}^{2}+\left(\sqrt{{S}_{0}}+\sqrt{{S}_{1}}\right){}^{2}\hfill \\ \multicolumn{0}{c}{}& =\left(\sqrt{{S}_{1}}+\sqrt[4]{{S}_{1}{S}_{2}}\right){}^{2}+\left(\sqrt[4]{{S}_{1}{S}_{2}}+\sqrt{{S}_{2}}\right){}^{2}\hfill \\ \multicolumn{0}{c}{}& =\left(\sqrt{{S}_{1}}+\sqrt{{S}_{2}}\right)\left(\sqrt[4]{{S}_{1}}+\sqrt[4]{{S}_{2}}\right){}^{2} .\hfill \end{array}$

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