# SOLUTIONS 

## QUESTION 1

## Solution 1.

Let $\mathcal{S}$ denote the given sum. Then

$$
\begin{aligned}
\mathcal{S} & =\sum_{n=1}^{1994}(-1)^{n}\left(\frac{n}{(n-1) !}+\frac{n+1}{n !}\right) \\
& =\sum_{n=0}^{1993}(-1)^{n+1} \frac{n+1}{n !}+\sum_{n=1}^{1994}(-1)^{n} \frac{n+1}{n !} \\
& =-1+\frac{1995}{1994 !}
\end{aligned}
$$

## Solution 2.

For positive integers $k$, define

$$
\mathcal{S}(k)=\sum_{n=1}^{k}(-1)^{n} \frac{n^{2}+n+1}{n !}
$$

We prove by induction on $k$ that

$$
\text { (*) } \quad \mathcal{S}(k)=-1+(-1)^{k} \frac{k+1}{k !}
$$

The given sum is the case when $k=1994$. For $k=1, \mathcal{S}(1)=-3=-1-\frac{2}{1 !}$. Suppose $\left({ }^{*}\right)$ holds for some $k \geq 1$, then

$$
\begin{aligned}
\mathcal{S}(k & +1)=\mathcal{S}(k)+(-1)^{k+1} \frac{(k+1)^{2}+(k+1)+1}{(k+1) !} \\
& =-1+(-1)^{k} \frac{k+1}{k !}+(-1)^{k+1}\left(\frac{k+1}{k !}+\frac{k+2}{(k+1) !}\right) \\
& =-1+(-1)^{k+1} \frac{k+2}{(k+1) !}
\end{aligned}
$$

completing the induction.

## QUESTION 2

## Solution 1.

Fix a positive integer $n$. Let $a=(\sqrt{2}-1)^{n}$ and $b=(\sqrt{2}+1)^{n}$. Then clearly $a b=1$. Let $c=(b+a) / 2$ and $d=(b-a) / 2$. If $n$ is even, $n=2 k$, then from the Binomial Theorem we get

$$
\begin{aligned}
c= & \frac{1}{2} \sum_{i=0}^{n}\left(\begin{array}{c}
n \\
i
\end{array}\right)\left(\sqrt{2}^{n-i}+(-1)^{i} \sqrt{2}^{n-i}\right) \\
& =\sum_{j=0}^{k}\left(\begin{array}{c}
2 k \\
2 j
\end{array}\right) \sqrt{2}^{2 k-2 j} \\
& =\sum_{j=0}^{k}\left(\begin{array}{c}
2 k \\
2 j
\end{array}\right) 2^{k-j}
\end{aligned}
$$

and

$$
\begin{aligned}
\frac{d}{\sqrt{2}} & =\frac{1}{\sqrt{2}} \sum_{i=0}^{n}\left(\begin{array}{c}
n \\
i
\end{array}\right)\left(\sqrt{2}^{n-i}-(-1)^{i} \sqrt{2}^{n-i}\right) \\
& =\frac{2}{\sqrt{2}} \sum_{j=0}^{k-1}\left(\begin{array}{c}
2 k \\
2 j+1
\end{array}\right) \sqrt{2}^{2 k-2 j-1} \\
& =\sum_{j=0}^{k-1}\left(\begin{array}{c}
2 k \\
2 j+1
\end{array}\right) 2^{k-j}
\end{aligned}
$$

showing that $c$ and $\frac{d}{\sqrt{2}}$ are both positive integers. Similarly, when $n$ is odd we see that $\frac{c}{\sqrt{2}}$ and $d$ are both positive integers. In either case, $c^{2}$ and $d^{2}$ are both integers. Notes that

$$
c^{2}-d^{2}=\frac{1}{4}\left((b+a)^{2}-(b-a)^{2}\right)=a b=1 .
$$

Hence if we let $m=c^{2}$, then $m-1=c^{2}-1=d^{2}$ and $a=c-d=\sqrt{m}-\sqrt{m-1}$.

## Solution 2.

Let $m$ and $n$ be positive integers. Observe that

$$
(\sqrt{2}-1)^{n}(\sqrt{2}+1)^{n}=1=(\sqrt{m}-\sqrt{m-1})(\sqrt{m}+\sqrt{m-1})
$$

and so

$$
\text { (*) } \quad(\sqrt{2}-1)^{n}=\sqrt{m}-\sqrt{m-1} \text { if and only if }(\sqrt{2}+1)^{n}=\sqrt{m}+\sqrt{m-1} \text {. }
$$

Assuming $m$ and $n$ satisfy (*), then adding the two equivalent equations we get $2 \sqrt{m}=$ $(\sqrt{2}-1)^{n}+(\sqrt{2}+1)^{n}$ whence:

$$
(* *) \quad m=\frac{1}{4}\left[(\sqrt{2}-1)^{2 n}+2+(\sqrt{2}+1)^{2 n}\right]
$$

Now we show that the steps above are reversible and that $m$ defined by $\left({ }^{* *}\right)$ is a positive integer. From $\left({ }^{* *}\right.$ ) one sees easily that

$$
\sqrt{m}=\frac{1}{2}\left[(\sqrt{2}-1)^{n}+(\sqrt{2}+1)^{n}\right] \text { and } \sqrt{m-1}=\frac{1}{2}\left[(\sqrt{2}+1)^{n}-(\sqrt{2}-1)^{n}\right]
$$

and so $\sqrt{m}-\sqrt{m-1}=(\sqrt{2}-1)^{n}$ as required. Finally, from the Binomial Theorem,

$$
\begin{aligned}
(\sqrt{2} & -1)^{2 n}+(\sqrt{2}+1)^{2 n}= \\
& =\sum_{k=0}^{2 n}\left(\begin{array}{c}
2 n \\
k
\end{array}\right)\left[(-1)^{k} 2^{(2 n-k) / 2}+2^{(2 n-k) / 2}\right] \\
& =\sum_{\ell=0}^{n}\left(\begin{array}{c}
2 n \\
2 \ell
\end{array}\right) 2^{n-\ell+1}
\end{aligned}
$$

which is congruent to 2 modulo 4 since $2^{n-\ell+1} \equiv 0(\bmod 4)$ for all $\ell=0,1,2, \cdots, n-1$. Therefore, $(\sqrt{2}-1)^{2 n}+2+(\sqrt{2}+1)^{2 n}$ is a multiple of 4 , as required.

## SOLUTIONS (Cont'd)

## Solution 3.

We show by induction that

$$
\text { (*) } \quad(\sqrt{2}-1)^{n}=\left\{\begin{array}{l}
a \sqrt{2}-b \text { where } 2 a^{2}=b^{2}+1 \text { if } n \text { is odd } \\
a-b \sqrt{2} \text { where } a^{2}=2 b^{2}+1 \text { if } n \text { is even }
\end{array}\right. \text {. }
$$

Thus $m=2 a^{2}$ when $n$ is odd and $m=a^{2}$ when $n$ is even and the problem is solved.

The induction is as follows:

$$
\begin{aligned}
& (\sqrt{2}-1)^{1}=1 \sqrt{2}-1 \text { where } 2\left(1^{2}\right)=1^{2}+1 \\
& (\sqrt{2}-1)^{2}=3-2 \sqrt{2} \text { where } 3^{2}=2\left(2^{2}\right)+1
\end{aligned}
$$

Assume (*) holds for some $n \geq 1, n$ odd. Then

$$
\begin{aligned}
(\sqrt{2} & -1)^{n+1} \\
& =(a \sqrt{2}-b)(\sqrt{2}-1) \text { where } 2 a^{2}=b^{2}+1 \\
& =(2 a+b)-(a+b) \sqrt{2} \\
& =A-B \sqrt{2} \text { where } A=2 a+b, B=a+b .
\end{aligned}
$$

Moreover, $A^{2}=2 a^{2}+4 a b+b^{2}+2 a^{2}=2 a^{2}+4 a b+2 b^{2}+1=2 B^{2}+1$.

Assume (*) holds for some $n \geq 2, n$ even. Then

$$
\begin{aligned}
(\sqrt{2} & -1)^{n+1} \\
& =(a-b \sqrt{2})(\sqrt{2}-1) \text { where } a^{2}=2 b^{2}+1 \\
& =(a+b) \sqrt{2}-(a+2 b) \\
& =A \sqrt{2}-B \text { where } A \approx a+b, B=a+2 b .
\end{aligned}
$$

Moreover, $2 A^{2}=2 a^{2}+4 a b+2 b^{2}=a^{2}+4 a b+4 b^{2}+a^{2}-2 b^{2}=B^{2}+1$.

## Solution 4.

From $(\sqrt{2}-1)^{1}=\sqrt{2}-1,(\sqrt{2}-1)^{2}=3-2 \sqrt{2},(\sqrt{2}-1)^{3}=5 \sqrt{2}-7,(\sqrt{2}-1)^{4}=17-12 \sqrt{2}$, etc, we conjecture that

$$
(*) \quad(\sqrt{2}-1)^{n}=s_{n} \sqrt{2}+t_{n}
$$

where $s_{1}=1, t_{1}=1, s_{n+1}=(-1)^{n}\left(\left|s_{n}\right|+\left|t_{n}\right|\right), t_{n+1}=(-1)^{n+1}\left(2\left|s_{n}\right|+\left|t_{n}\right|\right)$.

Note that $s_{n}$ is positive (negative) if $n$ is odd (even) and $t_{n}$ is negative (positive) if $n$ is odd (even).

We now show by induction that $\left(^{*}\right)$ holds and that each $s_{n} \sqrt{2}+t_{n}$ of the form $\sqrt{m}-\sqrt{m-1}$ for some $m$.

It is easily verified that $\left({ }^{*}\right)$ is correct for $n=1$ and 2 . Assume $\left({ }^{*}\right)$ holds for some $n \geq 2$. Then

$$
(\sqrt{2}-1)^{n+1}=\left(s_{n} \sqrt{2}+t_{n}\right)(\sqrt{2}-1)=\left(t_{n}-s_{n}\right) \sqrt{2}+\left(2 s_{n}-t_{n}\right)
$$

If $n$ is odd, then

$$
\begin{aligned}
& t_{n}-s_{n}=-\left(\left|t_{n}\right|+\left|s_{n}\right|\right)=s_{n+1} \\
& 2 s_{n}-t_{n}=2\left|s_{n}\right|+\left|t_{n}\right|=t_{n+1}
\end{aligned}
$$

If $n$ is even, then

$$
\begin{aligned}
& t_{n}-s_{n}=\left|t_{n}\right|+\left|s_{n}\right|=s_{n+1} \\
& 2 s_{n}-t_{n}=-2\left|s_{n}\right|-\left|t_{n}\right|=t_{n+1}
\end{aligned}
$$

We have shown that $\left(^{*}\right)$ is correct for all $n$.

Observe now that $\left(s_{n+1} \sqrt{2}\right)^{2}-t_{n+1}^{2}=2\left(s_{n}^{2}-2 s_{n} t_{n}+t_{n}^{2}\right)-\left(4 s_{n}^{2}-4 s_{n} t_{n}+t_{n}^{2}\right)=-2 s_{n}^{2}+t_{n}^{2}$ $=-\left(\left(s_{n} \sqrt{2}\right)^{2}-t_{n}^{2}\right)$. Since $\left(s_{1} \sqrt{2}\right)^{2}-t_{1}^{2}=1$, it follows that $\left(s_{n} \sqrt{2}\right)^{2}-t_{n}^{2}=(-1)^{n+1}$ for all $n$. To complete the proof it suffices to take $m=\left(s_{n} \sqrt{2}\right)^{2}, m-1=t_{n}^{2}$ when $n$ is odd and $m=t_{n}^{2}, m-1=\left(s_{n} \sqrt{2}\right)^{2}$ when $n$ is even.

## QUESTION 3

First observe that if two neighbours have the same response on the $n^{\text {th }}$ vote, then they both will respond the same way on the $(n+1)^{\text {th }}$ vote. Moreover, neither will ever change his response after the $n^{\text {th }}$ vote.

Let $A_{n}$ be the set of men who agree with at least one of their neighbours on the $n^{\text {th }}$ vote. The previous paragraph says that $A_{n} \subset A_{n+1}$ for every $n \geq 1$. Moreover, we will be done if we can show that $A_{n}$ contains all 25 men for some $n$.

Since there are an odd number of men at the table, it is not pssible that every man disagrees with both of his neighbours on the first vote. Therefore $A_{1}$ contains at least two men. And since $A_{n} \subset A_{n+1}$ for every $n$, there exists a $T<25$ such that $A_{T}=A_{T+1}$. Suppose that $A_{T}$ does not contain all 25 men; we shall use this to derive a contradiction. Since $A_{T}$ is not empty, there must exist two neighbours, whom we shall call $x$ and $y$, such that $x \in A_{T}$ and $y \notin A_{T}$. Since $x \in A_{T}$, he will respond the same way on the $T^{\text {th }}$ and $(T+1)^{\text {th }}$ votes. But $y \notin A_{T}$, so $y$ 's response on the $T^{\text {th }}$ vote differs from $x$ 's response. In fact, we know that $y$ disagrees with both of his neighbours on the $T^{\text {th }}$ vote, and so he will change his response on the $(T+1)^{t h}$ vote. Therefore, on the $(T+1)^{\text {th }}$ vote, $y$ responds the same way as does $x$. This implies that $y \in A_{T+1}$. But $y \notin A_{T}$, which contradicts the fact that $A_{T}=A_{T+1}$. Therefore we conclude that $A_{T}$ contains all 25 men, and we are done.

## QUESTION 4

There are three cases to be considered:

Case 1: If $P$ is outside $\Omega$ (see figures I, II, and III), then since $\angle A U B=\angle A V B=\pi / 2$, we have

$$
\cos (\angle A P B)=\frac{P U}{P B}=\frac{P V}{P A}=\sqrt{\frac{P U}{P A} \cdot \frac{P V}{P B}}=\sqrt{s t}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_99b8b91ebd335becafa4g-6.jpg?height=556&width=485&top_left_y=1809&top_left_x=83)

Figure I

![](https://cdn.mathpix.com/cropped/2024_04_17_99b8b91ebd335becafa4g-6.jpg?height=545&width=1054&top_left_y=1804&top_left_x=606)

Figure II
Figure III

## SOLUTIONS (Cont'd)

Case 2: If $P$ is on $\Omega$ (see figure IV), then

$$
P=U=V \Rightarrow P U=P V=0 \Rightarrow s=t=0 \text {. }
$$

Since $\angle A P B=\pi / 2, \cos (\angle A P B)=0=\sqrt{s t}$ holds again.

![](https://cdn.mathpix.com/cropped/2024_04_17_99b8b91ebd335becafa4g-7.jpg?height=445&width=485&top_left_y=753&top_left_x=455)

Figure IV

![](https://cdn.mathpix.com/cropped/2024_04_17_99b8b91ebd335becafa4g-7.jpg?height=435&width=485&top_left_y=774&top_left_x=1063)

Figure $\mathrm{V}$

Case 3: If $P$ is inside $\Omega$ (figure $\mathrm{V}$ ), then

$$
\cos (\angle A P B)=\cos (\pi-\angle A P V)=-\cos (\angle A P V)=-\frac{P V}{P A}
$$

and

$$
\cos (\angle A P B)=\cos (\pi-\angle B P U)=-\cos (\angle B P U)=-\frac{P U}{P B}
$$

Therefore $\cos (\angle A P B)=-\sqrt{\frac{P U}{P A} \cdot \frac{P V}{P B}}=-\sqrt{s t}$.

## SOLUTIONS (Cont'd)

## QUESTION 5

## Solution 1.

100 degree angle grestion? bit haden?

From $A$ draw a lkine $\ell$ parallel to $B C$. Extend $D F$ and $D E$ to meet $\ell$ at $P$ and $Q$ respectively (See Figure I). Then from similar triangles, we have

$$
\frac{A P}{B D}=\frac{A F}{F B} \text { qne } \frac{A Q}{C D}=\frac{A E}{E C}
$$

or

$$
A P=\frac{A F}{F B} \cdot B D \text { and } A Q=\frac{A E}{E C} \cdot C D
$$

By Ceva's Theorem, $\frac{A F}{F B} \cdot \frac{B D}{D C} \cdot \frac{C E}{E A}=1$ and thus

$$
\frac{A F}{F B} \cdot B D=\frac{A E}{E C} \cdot C D
$$

From (1) and (2) we get $A P=A Q$ and hence $\triangle A D P \simeq \triangle A D Q$ from which $\angle E D H=$ $\angle F D H$ follows.

![](https://cdn.mathpix.com/cropped/2024_04_17_99b8b91ebd335becafa4g-8.jpg?height=461&width=660&top_left_y=1492&top_left_x=648)

Figure I

## Solution 2.

Use cartesian coordinates, with $D$ at $(0,0), A=(0, a), B=(-b, 0), C=(c, 0)$. Let $H=(0, h), E=(u, v)$ and $F=(-r, s)$ where $a, b, c, h, u, v, r, s$ are all positive (See Figure II).

It clearly suffices to show that $\frac{v}{u}=\frac{s}{\tau}$. Since $E C$ and $A C$ have the same slope, we have $\frac{v}{u-c}=\frac{a}{-c}$. Similarly, since $E B$ and $H B$ have the same slope, $\frac{v}{u+b}=\frac{k}{b}$. Thus

$$
\frac{v}{a}=\frac{u-c}{-c}=\frac{-u}{c}+1
$$

and

$$
\frac{v}{h}=\frac{u+b}{b}=\frac{u}{b}+1
$$

(2)-(1) we get $v\left(\frac{1}{h}-\frac{1}{a}\right)=u\left(\frac{1}{b}+\frac{1}{c}\right)$ and thus

$$
\frac{v}{u}=\frac{\frac{1}{b}+\frac{1}{c}}{\frac{1}{h}-\frac{1}{a}}=\frac{a h(b+c)}{b c(a-h)} .
$$

With $u, v, b$ and $c$ replaced by $-r, s,-c$ and $-b$ respectively, we have, by a similar argument that

$$
\frac{s}{-r}=\frac{a h(-c-b)}{b c(a-h)} \text { or } \frac{s}{r}=\frac{a h(b+c)}{b c(a-h)} \text {. }
$$

Therefore, $\frac{v}{u}=\frac{s}{\tau}$ as desired.

![](https://cdn.mathpix.com/cropped/2024_04_17_99b8b91ebd335becafa4g-9.jpg?height=470&width=724&top_left_y=1679&top_left_x=582)

Figure II