Category Archives: Puzzles

April 23, 2025 · 9:30 am

Problem Solving

I am came across this problem and was fascinated. It’s from this book

At first I went straight to the 14-sided polygon, and tried to draw the diamonds (parallelograms), but then I thought let’s start smaller and see if there is a pattern.

Clearly a square contains 1 diamond (itself).

Pentagon

It’s not possible with a pentagon.

Hexagon

A hexagon has 6 diamonds

Septagon

I am guessing it’s not possible to fill a regular 7-sided shape with diamonds

It’s not possible with odd numbers of sides. Regular polygons with an odd number of sides have no parallel sides, so we can’t cover it with rhombi (which have opposite sides parallel).

Octagon

An octagon has 6 diamonds.

We know a decoagon has 10 diamonds (from the question)

Let’s put together what we know

$n$	$4$	$6$	$8$	$10$
Diamonds	$1$	$3$	$6$	$10$

These are the triangular numbers, so when $n=12$ the number of diamonds is $15$ , and for $n=14$ it’s $21$ .

We can work out a rule for calculating the number of diamonds given the number of sides.

Because the difference in the $n$ values is not $1$ , I am going to get $n$ and $D$ in terms of $k$ and then combine the two equations.

From the above table, $n=2k+2$

We know this rule is quadratic as the second difference is constant, hence

$D=\frac{1}{2}k^2+bk+c$

$\begin{equation*}1=\frac{1}{2}+b+c\end{equation}$

(1) $\begin{equation*}\frac{1}{2}=b+c\end{equation*}$

$\begin{equation*}3=\frac{1}{2}2^2+2b+c\end{equation}$

(2) $\begin{equation*}1=2b+c\end{equation*}$

Solve simultaneously, subtract equation $1$ from equation $2$

(3) $\begin{equation*}\frac{1}{2}=b\end{equation*}$

Substitute for $b=\frac{1}{2}$ into equation $1$

$\begin{equation*}\frac{1}{2}=\frac{1}{2}+c\end{equation}$

$c=0$ , therefore $D=\frac{1}{2}k^2+\frac{1}{2}k$

We know $n=2k+2$ hence $k=\frac{n-2}{2}$

Hence $D=\frac{1}{2}(\frac{n-2}{2})^2+\frac{1}{2}(\frac{n-2}{2})$

$D=\frac{1}{8}(n^2-4n+4)+\frac{1}{4}(n-2)=\frac{n^2}{8}-\frac{n}{2}+\frac{1}{2}+\frac{n}{4}-\frac{1}{2}=\frac{n^2}{8}-\frac{n}{4}$

$\begin{equation*}D=\frac{n^2}{8}-\frac{n}{4}\end{equation}$

Let’s test our rule for $n=14$

$\begin{equation*}D=\frac{14^2}{8}-\frac{14}{4}=\frac{49}{2}-\frac{7}{2}=\frac{42}{2}=21\end{equation}$

Interesting Sum

$S=\sum_{n=1}^\infty (tan^{-1}(\frac{2}{n^2}))$ , find $S$ .

I came across this sum in An Imaginary Tale by Nahin and I was fascinated.

Let $tan(\alpha)=n+1$ and $tan(\beta)=n-1$ .

tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}

tan(\alpha-\beta)=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}

Which means,

$\begin{equation*}S=\sum_{n=1}^\infty(tan^{-1}(n+1)-tan^{-1}(n-1))\end{equation}$

Let’s try a few partial sums

$S_4=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)$

$S_4=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(4)+tan^{-1}(5)$

$S_6=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)+tan^{-1}(6)-tan^{-1}(4)+tan^{-1}(7)-tan^{-1}(5)$

$S_6=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(6)+tan^{-1}(7)$

Hence, $S_N=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(N)+tan^{-1}(N+1)$

$S_N=-\frac{\pi}{4}-0+tan^{-1}(N)+tan^{-1}(N+1)$

What happens as $N\rightarrow \infty$ ?

$\lim\limits_{N\to \infty}\ S_N=-\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{3\pi}{4}$

Because we know $tan(\frac{\pi}{2})$ is undefined.

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Filed under Identities, Interesting Mathematics, Puzzles, Sequences, Trigonometry

Tagged as limits, sum, trigonometric identities

December 10, 2024 · 6:00 am

Puzzle Page 2

If $x^2-3x+1=0$ , then find $x^5+\frac{1}{x^5}$ .

My first thought was to solve for $x$ , but it doesn’t factorise easily, and I didn’t want to find the fifth power of an expression involving surds $(x=\frac{3\pm \sqrt{5}}{2})$ , there must be an easier way.

Because $x\neq0$ , we can divide by $x$

$\begin{equation*}x-3+\frac{1}{x}=0\end{equation}$

Hence

(1) $\begin{equation*}x+\frac{1}{x}=3\end{equation*}$

What is the expansion of $(x+\frac{1}{x})^5$ ?

Using the binomial expansion theorem

$\begin{equation*}(x+\frac{1}{x})^5=x^5+5x^4(\frac{1}{x})+10x^3(\frac{1}{x^2})+10x^2(\frac{1}{x^3})+5x(\frac{1}{x^4})+\frac{1}{x^5}\end{equation}$

$\begin{equation*}(x+\frac{1}{x})^5=x^5+\frac{1}{x^5}+5(x^3+\frac{1}{x^3})+10(x+\frac{1}{x})\end{equation}$

Therefore

(2) $\begin{equation*}x^5+\frac{1}{x^5}=(x+\frac{1}{x})^5-5(x^3+\frac{1}{x^3})-10(x+\frac{1}{x})\end{equation*}$

Let’s do it again for $x^3+\frac{1}{x^3}$

$\begin{equation*}(x+\frac{1}{x})^3=x^3+3x^2(\frac{1}{x})+3x(\frac{1}{x^2})+\frac{1}{x^3}\end{equation}$

(3) $\begin{equation*}x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3(x+\frac{1}{x})\end{equation*}$

Substitute $3$ into $2$

$\begin{equation*}x^5+\frac{1}{x^5}=(x+\frac{1}{x})^5-5((x+\frac{1}{x})^3-3(x+\frac{1}{x}))-10(x+\frac{1}{x})\end{equation}$

Remember $x+\frac{1}{x}=3$

Therefore

$\begin{equation*}x^5+\frac{1}{x^5}=3^5-5(3^3)+15\times3-10\times3\end{equation}$

$\begin{equation*}x^5+\frac{1}{x^5}=243-135+45-30=123\end{equation}$

This would be a good extension question for students learning the binomial expansion theorem. We also use this technique for trigonometric identities using complex numbers.