Racquel Sanderson

January 8, 2026 · 9:25 am

Love & Math – Edward Frenkel

A friend left this behind when they moved countries (knowing that I liked maths). This was my second attempt at reading it.

Here’s the blurb …

What if you had to take an art class in which you were only taught how to paint a fence? What if you were never shown the paintings of van Gogh and Picasso, weren’t even told they existed? Alas, this is how math is taught, and so for most of us it becomes the intellectual equivalent of watching paint dry.

In Love and Math , renowned mathematician Edward Frenkel reveals a side of math we’ve never seen, suffused with all the beauty and elegance of a work of art. In this heartfelt and passionate book, Frenkel shows that mathematics, far from occupying a specialist niche, goes to the heart of all matter, uniting us across cultures, time, and space.

Love and Math tells two intertwined stories: of the wonders of mathematics and of one young man’s journey learning and living it. Having braved a discriminatory educational system to become one of the twenty-first century’s leading mathematicians, Frenkel now works on one of the biggest ideas to come out of math in the last 50 years: the Langlands Program. Considered by many to be a Grand Unified Theory of mathematics, the Langlands Program enables researchers to translate findings from one field to another so that they can solve problems, such as Fermat’s last theorem, that had seemed intractable before.

At its core, Love and Math is a story about accessing a new way of thinking, which can enrich our lives and empower us to better understand the world and our place in it. It is an invitation to discover the magic hidden universe of mathematics.

I believe the aim of this book was to show non-maths people the beauty and joy of mathematics. However, I think it would be hard going if you didn’t have a mathematics background. This is high level mathematics (at one stage he mentions that there are probably about 12 people in the world who understand what he is discussing). Having said that, I liked it. I particularly liked the personal aspects of the narrative – his life experiences, and the people he met and with whom he worked.

A review

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Filed under Book Review

Tagged as edward frenkel, frenkel edward, Love and math

January 2, 2026 · 7:08 am

Geometry Puzzle (finding a fraction of an area)

Geometry Puzzles in Felt Tip: A Compilation of puzzles from 2018 – Catriona Shearer

Band $1$ and $3$ have the same area.

We want to find the area of the shaded segment.

As the dots are equally spaced, the sector’s angle is $\frac{\pi}{2} = (\frac{2\pi}{12}\times 3)$

Remember the area of a segment is $A=\frac{1}{2}r^2(\theta-sin(\theta))$ where the angle measurement is in radians.

(1) $\begin{equation*}A=\frac{1}{2}r^2(\frac{\pi}{2}-sin(\frac{\pi}{2}))=\frac{1}{2}r^2(\frac{\pi}{2}-1))=\frac{\pi r^2}{4}-\frac{r^2}{2}\end{equation*}$

We want to find the area of the shaded segment.

As the dots are equally spaced, the sector’s angle is $\frac{2\pi}{12} = (\frac{\pi}{6})$

(2) $\begin{equation*}A=\frac{1}{2}r^2(\frac{\pi}{6}-sin(\frac{\pi}{6}))=\frac{\pi r^2}{12}-\frac{r^2}{4}\end{equation*}$

The area of band $1$ is equation $1 -$ equation $2$ .

(3) $\begin{equation*}\frac{\pi r^2}{4}-\frac{r^2}{2}-(\frac{\pi r^2}{12}-\frac{r^2}{4})=\frac{\pi r^2}{6}-\frac{r^2}{4}\end{equation*}$

Band 2 consists of two congruent triangles and two congruent sectors.

$\begin{equation*}\theta=\frac{2\pi}{12}\times 5=\frac{5\pi}{6}, \alpha=\frac{\pi}{6}\end{equation}$

(4) $\begin{equation*}A=2(\frac{1}{2}r^2sin(\frac{5\pi}{6}))+2(\frac{1}{2}r^2\frac{\pi}{6})=\frac{r^2}{2}+\frac{r^2 \pi}{6}\end{equation*}$

Hence the shaded area is $2(\frac{\pi r^2}{6}-\frac{r^2}{4})+\frac{r^2}{2}+\frac{r^2 \pi}{6}=\frac{\pi r^2}{2}$

The area of the circle is $\pi r^2$

Hence the fraction of the shaded area is $=\frac{\frac{\pi r^2}{2}}{\pi r^2}=\frac{1}{2}$

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Filed under Area, Area of Triangles (Sine), Finding an area, Geometry, Puzzles, Simplifying fractions

Tagged as catriona shearer, finding an area, geometry, geometry puzzle

December 24, 2025 · 6:28 am

Christmas Cartesian Co-ordinates

I found a co-ordinate puzzle here.

Finished image

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Filed under Co-ordinate Geometry

Tagged as Christmas Maths, Coordinate picture, Rudolf

December 19, 2025 · 8:10 am

Algebra Time Question

This question is from Challenging Problems in Algebra

It’s the type of question students hate – “Who talks like that?”

Let $t$ be the number of hours from noon.

$\begin{equation*}\frac{t}{8}+6-\frac{t}{4}=t\end{equation}$

$\begin{equation*}6=\frac{9t}{8}\end{equation}$

$\begin{equation*}t=\frac{48}{9}=5\frac{1}{3}\end{equation}$

Hence the time is 5:20pm

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Filed under Algebra, Puzzles, Simplifying fractions, Solving Equations

Tagged as algebra, solving linear equations, time

December 10, 2025 · 7:54 am

Geometry Problem

Problem from *50 Maths Problems with Solutions* – Ajmal KP

The blue shaded area is the area of triangles $APO$ and $AQO$ subtract the sector $POQ$ .

We can use Heron’s law to find the area of the triangle $\Delta{ABC}$

$\begin{equation*}A=\sqrt{s(s-a)(s-b)(s-c)}\end{equation}$

where $s=\frac{a+b+c}{2}$

$\begin{equation*}A=\sqrt{20(20-16)(20-10)(20-14)}=40\sqrt{3}\end{equation}$

We also know the area of triangle $\Delta{ABC}=sr$ where $r$ is the radius of the inscribed circle.

Hence, $40\sqrt{3}=20r$ and $r=2\sqrt{3}$

We know $AP=AQ, CQ=CR$ , and $BP=BR$ – tangents to a circle are congruent.

$\begin{equation*}14-x=6+x\end{equation}$

(1) $\begin{equation*}8=2x\end{equation*}$

(2) $\begin{equation*}x=4\end{equation*}$

Area $\Delta{AQO}=\frac{1}{2}10\times 2\sqrt{3}=10\sqrt{3}$

Area $\Delta{APO}=$ Area $\Delta{AQO}$

$\begin{equation*}tan(\theta)=\frac{10}{2\sqrt{3}}\end{equation}$

$\begin{equation*}\theta=70.9^{\circ}\end{equation}$

Area of sector $OPQ=\frac{2\times70.9}{360}\pi (2\sqrt{3})^2=14.8$

Blue area = $20\sqrt{3}-14.8=19.8cm^2$

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Filed under Algebra, Area, Finding an angle, Finding an area, Geometry, Heron's Law, Interesting Mathematics, Puzzles, Radius and Semi-Perimeter, Right Trigonometry, Solving Equations, Trigonometry

Tagged as geometry puzzle

December 4, 2025 · 12:52 pm

Area of a triangle from the semi-perimeter and the radius of the incircle.

$\begin{equation*}A=sr\end{equation}$

Where $s$ is the semi-perimeter, $s=\frac{a+b+c}{2}$ and $r$ is the radius of the incircle.

$AB, BC$ and $AC$ are tangents to the circle. And the radii are perpendicular to the tangents.

Add line segments $AO, CO$ and $BO$ .

$\Delta{ABC}$ is split into three triangles, $\Delta{AOB}, \Delta{AOC}$ and $\Delta{BOC}$ .

Hence Area $\Delta{ABC}=\Delta{AOB}+\Delta{AOC}+\Delta{BOC}$

$\Delta{ABC}=\frac{1}{2}cr+\frac{1}{2}br+\frac{1}{2}ar$

$\Delta{ABC}=\frac{1}{2}r(a+b+c)$

Remember $s=\frac{1}{2}(a+b+c)$

$\Delta{ABC}=sr$

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Filed under Area, Finding an area, Geometry, Interesting Mathematics, Radius and Semi-Perimeter

November 26, 2025 · 10:57 am

Trigonometric Equation

Solve $cos(4\theta)+cos(2\theta)+cos(\theta)=0$ for $0\le \theta \le\pi$

Remember the identity

(1) $\begin{equation*}cos(A)+cos(B)=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})\end{equation*}$

Hence

$\begin{equation*}cos(4\theta)+cos(2\theta)=2cos(3\theta)cos(\theta)\end{equation}$

Now I have

$\begin{equation*}2cos(3\theta)cos(\theta)+cos(\theta)=0\end{equation}$

$\begin{equation*}cos(\theta)(2cos(3\theta)+1)=0\end{equation}$

$cos(\theta)=0$ or $cos(3\theta)=\frac{-1}{2}$

$\theta=\frac{\pi}{2}$

$cos(3\theta)=-\frac{1}{2}$ for $0 \le \theta \le 3\pi$

$3\theta=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}$

$\theta=\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}$

Hence $\theta =\frac{\pi}{2},\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}$

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Filed under Identities, Quadratic, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

Tagged as solving trig equations, sum to product trig identity

November 22, 2025 · 6:10 am

Polynomial Long Division

I usually choose to use synthetic division when factorising polynomials, but I know some teachers are unhappy when their students do this. So for completeness, here is my PDF for Polynomial Long Division.

Polynomial Long Division Download

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Filed under Algebra, Cubics, Factorisation, Factorising, Factorising, Polynomials, Quadratics, Solving, Solving, Solving Equations, Year 11 Mathematical Methods

Tagged as polynomial long division, solving cubic equations

November 16, 2025 · 5:54 am

Geometry Circle Question

In the diagram below, $A, B, C$ and $D$ lie on the circle with centre $O$ . If $\angle{DBC} = 41^{\circ}$ and $\angle{ACD} = 53^{\circ}$ , determine with reasoning $\angle{BAC}$ and $\angle{AOB}$