Category Archives: Puzzles

June 10, 2025 · 1:54 am

Square Root Puzzle

Is it possible to find three numbers, $a, b, c$ , none of which is zero or a perfect square, such that
$\sqrt{a}+\sqrt{b}=\sqrt{c}$

Can You Solve These – David Wells

As $a, b$ and $c$ can’t be perfect squares, let $a=d\times e^2, b=f\times g^2$ and $c=h\times k^2$ where $d, e, f, g, h$ and $k$ are real numbers.

Hence $\sqrt{a}=e\sqrt{d}, \sqrt{b}=g\sqrt{f}$ and $\sqrt{c}=k\sqrt{h}$ .

$\begin{equation*}\sqrt{a}+\sqrt{b}=\sqrt{c}\end{equation}$

$\begin{equation*}e\sqrt{d}+g\sqrt{f}=k\sqrt{h}\end{equation}$

For the above equation to be possible $d, f$ and $h$ must simplify to the same surd. Because we are looking for one set of numbers, let $d=f=h$ .

$\begin{equation*}e\sqrt{d}+g\sqrt{d}=k\sqrt{d}\end{equation}$

$\begin{equation*}e+g=k\end{equation}$

Let’s think of some numbers that might work…

$1+2=3$ or $2+3=5$ , etc.

Let’s try $e=1, g=2,$ and $k=3$

We now have $a=d, b=4d,$ and $c=9d$

As $a$ can’t be a square number, $d$ can’t be a square number.

Try $d=2$

$\begin{equation*}\sqrt{2}+\sqrt{8}=\sqrt{18}\end{equation}$

$LHS=\sqrt{2}+2\sqrt{2}$

$LHS=3\sqrt{2}$

$LHS=\sqrt{9\times 2}$

$LHS=\sqrt{18}$

$LHS=RHS$

One set of possible numbers are $2, 8,$ and $18$ .

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Filed under Algebra, Interesting Mathematics, Puzzles

May 15, 2025 · 10:01 am

Perfect Squares

Find all of the positive integers that make the following expression a perfect square.

(1) $\begin{equation*}(x-10)(x+14)\end{equation*}$

Let

$\begin{equation*}(x-10)(x+14)=n^2\end{equation}$

where $n$ is an integer.

Expand and simplify

$\begin{equation*}x^2+4x-140=n^2\end{equation}$

$\begin{equation*}x^2-4x-n^2=140\end{equation}$

Complete the square

$\begin{equation*}(x+2)^2-4-n^2=140\end{equation}$

$\begin{equation*}(x+2)^2-n^2=144\end{equation}$

Factorise (using difference of perfect squares)

$\begin{equation*}(x+2-n)(x+2+n)=144\end{equation}$

Find all of the factors of $144$

$(1,144), (2, 72), (3, 48), (4, 36), (6, 24), (8, 18), (9, 16), (12, 12)$

First pair,

$\begin{equation*}x+2-n=1 \tag {1} \end{equation}$

$\begin{equation*}x+2+n=144 \tag {2} \end{equation}$

$2x=141$

$x$ must be an integer.

I then used a spreadsheet

Hence the integers that make $(x-10)(x+14)$ are perfect square are, $10, 11, 13, 18,$ and $35$ .

Let’s try another one,

$(x-6)(x+14)$

(2) $\begin{equation*}(x-6)(x+14)=n^2\end{equation*}$

$\begin{equation*}(x^2+8x-84=n^2\end{equation}$

$\begin{equation*}(x+4)^2-n^2=100\end{equation}$

$\begin{equation*}(x+4-n)(x+4+n)=100\end{equation}$

Factors of 100,

$(1, 100), (2, 50), (4, 25), (5, 20), (10, 10)$

So the possible integers are $6$ and $22$ .

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Filed under Algebra, Arithmetic, Divisibility, Interesting Mathematics, Puzzles, Quadratic, Solving Equations

Tagged as perfect squares

May 3, 2025 · 3:34 am

Geometry Puzzle

Another puzzle from this book

Two ladders are propped up vertically in a narrow passageway between two vertical buildings. The ends of the ladders are 8 metres and 4 metres above the pavement.
Find the height above the ground, $T$ ,

$\Delta ABC \sim \Delta TEC$ and $\Delta DCB \sim \Delta TEB$ (Angle Angle Similarity)

Hence,

(1) $\begin{equation*}\frac{h}{8}=\frac{x}{x+y}\end{equation*}$

(2) $\begin{equation*}\frac{h}{4}=\frac{y}{x+y}\end{equation*}$

From equation $1$ $h=\frac{8x}{x+y}$ and from $2$ $h=\frac{4y}{x+y}$

Hence, $\frac{8x}{x+y}=\frac{4y}{x+y}$

Therefore, $8x=4y$ and $y=2x$

From equation $1$

$\begin{equation*}h=\frac{8x}{3x}=\frac{8}{3}\end{equation}$

Hence $T$ is $2 \frac{2}{3}$ m above the ground.

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Filed under Geometry, Puzzles, Similarity, Simplifying fractions, Solving Equations

Tagged as Puzzle

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 $n=14$

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

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Filed under Area, Geometry, Interesting Mathematics, Puzzles, Quadratics

Tagged as Finding a quadratic rule, regular polygons, rhombus, tessellations

December 16, 2024 · 9:19 am

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.

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Filed under Algebra, Binomial Expansion Theorem, Puzzles

December 3, 2024 · 6:54 am