# Solutions 

Note: The points to be awarded for each part of the solution are indicated on the right side.
Problem 1.

$$
\begin{gathered}
1=\frac{1 \times 2}{2} \\
1+\frac{1}{3}=\frac{2 \times 2}{3} \\
1+\frac{1}{3}+\frac{1}{6}+\ldots+\frac{1}{r_{n}}=\frac{n<2}{n+1}
\end{gathered}
$$

which is easily shown by induction.
(up to 3 points)
Now 5 is the sum of the reciprocals of these numbers where the last, $1993006=$
$1996 \times 1997$ $\frac{1996 \times 1997}{2}=8996$. Thus we have

$$
\begin{gathered}
S=\frac{1}{2}\left(\frac{2}{1}+\frac{3}{2}+\ldots+\frac{1997}{1996}\right) \\
=\frac{1}{2}\left(1996+\left(1+\frac{1}{2}+\ldots+\frac{1}{1996}\right)\right)
\end{gathered}
$$

(ap to 3 points)

$$
\therefore \frac{1}{2}(1996+6)
$$

$$
=1001
$$

P'roblem 2. Note that $2^{n+2}=2\left(2^{n-1}+1\right)$ so that $n$ is of the form 2 r with r odd. We will consider two cases.
i) $n=2 p$ with $p$ prime. $2 p \mid 2^{2 p}+2$, implies that $p!2^{2 n-1}+1$ and hence, hence $p \mid 2^{40-2}-1$. On the
![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-1.jpg?height=82&width=1678&top_left_y=1935&top_left_x=229) follows that $p \mid 2^{d}-1$. But $d \mid p-1$ and $d \mid 4 p-2=4(p-1)+2$. Fence $d \mid 2$ and since $p-1,4 p-2$ are even $d=2$. Then $p=3$ and $n=6<100$.
(up to 2 points)
ii) $n=2 p q$ where $p, q$ are odd primes, $p<q$ and $p q<\frac{1997}{2}$. Now $n!2 r+2$ impites that $p i 2^{n-1}$ +1 and therefore that $\mathrm{p}^{\prime} 2^{2 \mathrm{~m}-2} . .1=2^{4 \mathrm{~m}-2}-1$. Once again by Femat's theorern we have pl $2^{n-1}$ 1 which implies that $\mathrm{p}-1 / 4 \mathrm{pq}$-2. The same holds tole for q so that

$$
q-1 \mid 4 p q-2
$$

Both $p-1$ and $q-1$ are thus multiples of 2 but not of 4 su that $p \equiv q \equiv 3$ (mod 4).
( points)
laking $p=3$, we huve $4 p q-2=12 q-2$. Now from (1) we have

$$
12=\frac{12 q-12}{q-1}<\frac{12 q-2}{q-1}=\frac{12(q-1)+10}{q-1}=12+\frac{10}{q-1} \leq 1
$$

if $q \geq 11$, and clearly $\frac{12 q-?}{q-1}=13$ if $q=11$. But this gives $n=2(3)(11)=66<100$. Furthermore $(p, \varphi)=:(3,7)$ does not salisty (1).

Taking $p=7$ we observe that $4 \mathrm{pq}-2=28 \mathrm{q}-2$, and from (1) we have

$$
28<\frac{28 q-2}{q-1}=\frac{28(q-1)+26}{q-1}=28+\frac{26}{q-1} \leq 2
$$

if $q \geq 27$ and clearly $\frac{28 q-2}{q-1}=29$ if $q=27$. But 27 is not prime and the cases $(p, q)=(7$. 11), (7,19) and (7,23) do not satisfy (1).

Taking $p=11$, then $4 p q-2=44 q-2$, and

$$
44<\frac{44 q-2}{y-1} \text { and } \frac{44 q-2}{q-1} \leq 45 \text { if } q \geq 43 .
$$

Now clearly $\frac{44 q-2}{q-1}=45$ when $q=43$. In this case we have $n:=2 p q=2(11)(43)=946$.
Furthermore, $\frac{2^{44 x}+2}{946}$ is incieed an integer. The cases $(p, q)=(11,19),(11,23)$ and 111,31$)$ do not satisfy (1).
(2 points)
[Aditionally for completeness, if $p=19$ then $4 p y-2=76 q-2$ and $76<\frac{76 q-2}{4-1}=77$ if $q$ $\geqq 75$. Now 75 is not prime and for the cases $(p, 4)=(19,23),(19,31),(19,43)$ and $(19,47), q$ -1 is not a divisor of $74=2 \times 37$.

Similarly, if $p=23$ then $4 \mathrm{pq}-2=92 \mathrm{q}-2$ and $92<\frac{92 q-2}{q-\Gamma} \leq 93$ if $q \geq 91$ and $\frac{92 q-2}{q-1}=93$ if $q=91$. But 91 is not prime and of the cases $(p, q)=(23,31),(23,43)$, when q $=31$ all of the conditions are satisfied. But, $n=2 p q=1426$ is not a solution because $\frac{2^{1426}+2}{1+20}$
is not an integer.

No other pairs of F, q yield numbers within the required range.]
(1) point)

## Problem 3.

![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-3.jpg?height=864&width=1050&top_left_y=690&top_left_x=565)
$\angle A L D=\frac{1}{2}(\overparen{M C}+\widehat{A B})$
![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-3.jpg?height=147&width=386&top_left_y=1114&top_left_x=1718)
$\angle \triangle N M$

It is known (see Geometry Revisited) or easily derivable that

$$
m_{a}^{2}=(A L)^{2}=b c\left(1-\left(\frac{a}{b+c}\right)^{2}\right)
$$

![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-3.jpg?height=275&width=570&top_left_y=1559&top_left_x=1520)
(1 point)
From $\triangle$ ADL a MAN we have

$$
\begin{gathered}
\frac{A D}{A L}=\frac{4 M}{M N} \Rightarrow \\
h_{a} \cdot 2 R=A L \cdot A M=m_{a} \cdot M a \\
h_{a}=\frac{2(A B C)}{a}
\end{gathered}
$$

$$
\begin{gathered}
\frac{2(A B C)}{a} \cdot 2 R=m_{a} M_{a} \\
\frac{\frac{u b c}{d} \cdot 4 R}{a}=m_{a} M_{a} \\
b c=m_{a} M_{u}
\end{gathered}
$$

So that

$$
l_{a}=\frac{m_{a}^{2}}{m_{a} I_{a}}=1-(b+c)^{2}
$$

with similar expressions for $l_{b}$ and $l_{c}$.
(2 points)
$\therefore$ Given that $\sin A=\frac{t^{2}}{2 k}$, etc. the expression we are working with becomes

$$
\begin{aligned}
& \frac{l_{a}}{\sin ^{2} \cdot A}+\frac{l_{b}}{\sin ^{2} A}+\frac{l_{b}}{\sin ^{2} C^{2}}=\frac{4 R^{2}}{a^{2}}\left(1-\left(\frac{b^{2}}{b+c}\right)^{2}\right)+\frac{4 R^{2}}{b^{2}}\left(1-\left(\frac{b}{a+c}\right)^{2}\right)+\frac{4 R^{2}}{c^{2}}\left(1-\left(\frac{i}{a-b}\right)^{2}\right) \\
& =4 R^{2}\left[\left(\frac{1}{a^{2}}-\frac{1}{(b+c)^{2}}\right)+\left(\frac{1}{b^{2}}-\frac{1}{(a+c)^{2}}\right)+\left(\frac{1}{c^{2}}-\frac{1}{(a+b)^{2}}\right)\right] \\
& \geq 4 R^{2}\left[\frac{1}{a^{2}}-\frac{1}{4 b c}+\frac{1}{b^{2}}-\frac{1}{4 a c}+\frac{1}{c^{2}}-\frac{1}{4 a b}\right] \\
& =2 \mathrm{R}^{2}\left[\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)+\left(\frac{1}{a^{2}}+\frac{1}{c^{2}}\right)+\left(\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)-\frac{1}{2 a b}-\frac{1}{2 a c}-\frac{1}{2 b c}\right] \\
& 2\left[\frac{2}{a b}+\frac{2}{a c}+\frac{2}{b c}-\frac{1}{2 a b}-\frac{1}{2 a c}-\frac{1}{2 b c}\right] \\
& =2 R^{2}\left[2 \frac{3}{2 b}+\frac{3}{2 a c}+\frac{3}{2 b c}\right] \\
& =3 R^{2}\left[\frac{1}{a b}+\frac{1}{a c}+\frac{1}{b c}\right]=3 R^{2}\left[\frac{a+b+c}{a b c}\right]
\end{aligned}
$$

But $a b c=f R(.4 B C)$ so that this last expression becomes

$$
\frac{3 R(a+b+c)}{4(A B C)}=\frac{3 R \cdot 2 s}{4 s r}=3 \cdot \frac{\beta}{2 r}=3
$$

since $R \geq 2 r$. All of the inequalitics are equalities iff $a=b=c$.
(1 point)

## Problem 4.

- (a) Consider the gequence of triangles on the plane 4 . $42,4, A_{2}, 4,4,4,4,4, \ldots$ It is easy to see that any pair of them are similar. Let's prove that triangles $A_{1} A_{3} A_{5}$ and $A_{3} A$; are similar.
Triangles $A_{2} A_{3} A_{4}$ and $A_{4} A_{4} A_{6}$ are similar and their altitudes are $A_{4} A_{5}$ and $A_{6}, A_{0}$, then

$$
\frac{A_{2} A_{3}}{A_{4}}=\frac{A_{4} A_{5}}{A_{6} A_{7}} .
$$

Triangles $A A_{*} A_{s}$ and $A_{s} A_{*} A$ - are similar. then

$$
\frac{A+A_{5}}{A_{0} A_{7}}=\frac{A 3 A s}{A_{5} A_{7}}
$$

Now we can conclude that

$$
\frac{A_{1} A_{3}}{A_{3} A_{s}}=\frac{A_{3} A_{s}}{A_{4}}
$$

and triangles $A, A: A$ and $A ; A: A-$ are similar.
![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-5.jpg?height=63&width=1644&top_left_y=1389&top_left_x=227) $\triangle \quad A \cdot A_{s} A_{1}=\triangle \quad A-A_{3} P=\triangle \quad A_{A} A_{1}$, so triangle $A-A, P$ has a righr angle at $P$ and lines $A_{1} A_{5}$ and $A A^{A}$ - are perpendicular. In the same way lines $A A_{2}$. and $A$ are perpendicular and lines $A, A y$ and $A_{-1} A_{11}$ are perpendicular, hence $A_{1}, A_{3}, A_{9}$ are collinear and $A_{-}, A_{1 i} . A_{j}$ are collincar. It follows that triangle $A_{l A} A_{2} A_{;}$and $A_{0} A_{10} A_{l \mid}$ are homothetic and the center of homothety is P . Moreover, all
![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-5.jpg?height=56&width=1630&top_left_y=1639&top_left_x=231) an interior point to any of these triangles and there is no other point distinct from $P$ that is interior to any of these triangles. So this is the point we are looking for.

$$
\text { (up to }+ \text { points) }
$$

(b) Since $\triangle \quad A_{1} P A_{3} 90^{\prime \prime}$ then $P$ lies on the circle with diameter $A_{i}, A_{3}$. Let $A_{i} A_{3}=1, A_{2} A_{2}=s, A_{3} A_{3}$ $=r$ and let $A_{i} A_{2} A_{3}$ be clockwise. Triangles $A_{1} A_{2} A_{3}$ and $A_{3} A_{+} A_{5}$ are similar, thus $A_{2} A_{5}: r=s: 1$, and so $A_{2} A_{5}=r s$. Besides $A_{3} A_{y}=r \sqrt{1+s^{2}}$ (Pythagoras), and area of triangle $A_{1} A_{2} A_{3}=$ $\frac{1}{2} r \sqrt{1+s^{2}} \cdot \sqrt{1+s^{2}}=\frac{1}{2} s \cdot 1$. Thus $r=\frac{s}{1+s^{2}}$. By the arithmetic-geometric mean $\frac{s}{1+s^{2}} \leq \frac{1}{2}$, thus $r \leq \frac{1}{4}$ and the set of all possible values of $r$ consists of two real intervals $\left[-\frac{1}{2}, 0\right)$ and $\left(0, \frac{1}{2}\right]$. $\triangle A_{2} A_{1} P$ takes the maximum value when $r= \pm$ thus the locus of $P$
consists of two continuous arcs from the circle with dianeter $4, A_{i}$ with two extreme positions corresponding to $r=-\frac{1}{2}$ and $r=\frac{1}{2}$.
(up to 3 points)
![](https://cdn.mathpix.com/cropped/2024_11_22_a2786f8a9f2e9957a62dg-6.jpg?height=1558&width=920&top_left_y=390&top_left_x=597)

## Problem 5.

A redistribution can be written as $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ where $x_{1}$ denotes the number of objects transferred from $A_{i}$ to $A_{i+1}$. Our objective is to minimize the function
$F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{i=1}^{n}\left|x_{1}\right|$
After redistribution we should have at each $A_{i}, a_{i}-x_{1}+x_{i-1}=N$ for $i \in\{1,2, \ldots, n\}$ where $x_{0}$ means $x_{n}$.
(1 point)
Solving this system of linear equations we obtain:
$x_{i}=x_{1}-\left[(i-1) N-a_{2}-a_{3}-\ldots-a_{i}\right]$
for $i \in\{1,2, \ldots, n\}$.
Hence

$$
\begin{aligned}
F\left(x_{1}, x_{2}, \ldots, x_{n}\right)= & \left|x_{1}\right|+\left|x_{1}-\left(N-a_{2}\right)\right|+\left|x_{1}-2 N-a_{2}-a_{3}\right| \\
& +\ldots+\left|x_{1}-\left[(n-1) N-a_{2}-a_{3}-\ldots-a_{n}\right]\right|
\end{aligned}
$$

Basically the problem reduces to find the minimum of $F(x)=\sum_{i=1}^{n}\left|x-\alpha_{i}\right|$ where $\alpha_{i}=(i-1) N-\sum_{j=2}^{i} a_{j}$.
(up to. 3 points)

First rearrange $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ in non decreasing order. Collecting terms which are equal to one another we write the ordered sequence $\beta_{1}<\beta_{2}<\cdots<\beta_{m}$, each $\beta_{i}$ occurs $k_{i}$ times in the family $\left\{\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}\right\}$. Thus $k_{1}+k_{2}+\cdots+k_{m}=n$.

Consider the intervals $\left(-\infty, \beta_{1}\right],\left[\begin{array}{l}\beta_{1}, \beta_{2}\end{array}\right], \cdots,\left[\begin{array}{cc}\beta_{m-1}, \beta_{m}\end{array}\right],\left[\begin{array}{l}\left.\beta_{m}, \infty\right) \\ \end{array}\right.$ the graph of $F(x)=\sum_{i=1}^{n}\left|x-\alpha_{i}\right|=\sum_{i=1}^{m} k_{i}\left|x-\beta_{i}\right|$ is a continuos piece wise linear graph define in the following way:

$$
F(x)=\left\{\begin{array}{c}
k_{1}\left(\beta_{1}-x\right)+k_{2}\left(\beta_{2}-x\right)+\cdots+k_{m}\left(\beta_{m}-x\right) \text { if } \mathrm{x} \in\left(-\infty, \beta_{1}\right] \\
k_{1}\left(x-\beta_{1}\right)+k_{2}\left(\beta_{2}-x\right)+\cdots+k_{m}\left(\beta_{m}-x\right) \text { if } \mathrm{x} \in\left[\beta_{1}, \beta_{2}\right] \\
\vdots \\
\left.k_{1}\left(x-\beta_{1}\right)+k_{2}\left(x-\beta_{2}\right)+\cdots+k_{m}\right)\left(x-\beta_{m}\right) \text { if } \mathrm{x} \in\left[\beta_{m}, \infty\right)
\end{array}\right.
$$

The slopes of each line segment on each interval are respectively: $S_{0}=-k_{1}-k_{2}-k_{3}-\cdots-k_{m}$
$S_{1}=k_{1}-k_{2}-k_{3}-\cdots-k_{m}$
$S_{2}=k_{1}+k_{2}-k_{3}-\cdots-k_{m}$
$S_{m}=k_{1}+k_{2}+k_{3}+\cdots+k_{m}$
Note that this sequence of increasing numbers goes from a negative to a positive number, hence for some $t \geq 1$ there is an

$$
S_{t}=0 \text { or } S_{t-1}<0<S_{t}
$$

In the first case the minimum occurs at $x=\beta_{t}$ or $\beta_{t+1}$ and in the second case the minimum occurs at $x=\beta_{t}$
(Up to 7 points)
We can rephrase the computations above in terms of $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ rather than $\beta_{1}, \beta_{2}, \cdots, \beta_{m}$. After rearranging the $\alpha^{\prime} s$ in non decreasing order, pick $x=\alpha \quad$ if $n$ is odd and take $x=\alpha$ or $\alpha \quad$ if $n$ is even.

$$
\frac{n+1}{2} \quad \frac{n}{2} \quad \frac{n}{2}+1
$$

If no justification is given for the choice of $x$, give up to 4 points.