Racquel Sanderson

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

June 4, 2025 · 5:30 am

Closest Approach (Shortest Distance) e-activity (Casio Classpad)

At 1pm, object $H$ travelling with constant velocity $\begin{pmatrix}200\\10\end{pmatrix}$ km/h is sighted at the point with position vector $\begin{pmatrix}-90\\-100\end{pmatrix}$ km. At 2pm object $J$ travelling with constant velocity $\begin{pmatrix}100\\-100\end{pmatrix}$ km/h is sighted at the point with position vector $\begin{pmatrix}20\\-120\end{pmatrix}$ km. Determine the minimum distance between $H$ and $J$ and when this occurs.

OT Lee Mathematics Specialist Year 11 Unit 1 and 2 Exercise 10.1 Question 6.

(1) $\begin{equation*}\mathbf{r_H}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}\end{equation*}$

(2) $\begin{equation*}\mathbf{r_J}=\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix}\end{equation*}$

$\begin{pmatrix}-80\\-20\end{pmatrix}$ is the position vector of $J$ at 1pm.

Find the relative displacement of $H$ to $J$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\mathbf{r_H}-\mathbf{r_J}\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-90\\-100\end{pmatrix}+t\begin{pmatrix}200\\10\end{pmatrix}-(\begin{pmatrix}-80\\-20\end{pmatrix}+t\begin{pmatrix}100\\-100\end{pmatrix})\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

Find the relative velocity of $H$ to $J$

$\begin{equation*}\mathbf{_H}\mathbf{v_J}=\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

The relative displacement is perpendicular to the relative velocity at the closest approach.

That is

(3) $\begin{equation*}\mathbf{_H}\mathbf{r_J}\cdot\mathbf{_H}\mathbf{v_J}=0\end{equation*}$

$\begin{equation*}(\begin{pmatrix}-10\\-80\end{pmatrix}+t\begin{pmatrix}100\\110\end{pmatrix})\cdot(\begin{pmatrix}100\\110\end{pmatrix})=0\end{equation}$

$\begin{equation*}(-10+100t)(100)+(-80+110t)(110)=0\end{equation}$

$\begin{equation*}-1000+10 000t-8800+12100t=0\end{equation}$

$\begin{equation*}22100t=9800\end{equation}$

$\begin{equation*}t=\frac{98}{221}\end{equation}$

Substitute $t=\frac{98}{221}$ into the relative displacement and find the magnitude.

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}-10\\-80\end{pmatrix}+\frac{98}{221}\begin{pmatrix}100\\110\end{pmatrix}\end{equation}$

$\begin{equation*}\mathbf{_H}\mathbf{r_J}=\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\end{equation}$

$\begin{equation*}\|\begin{pmatrix}34\frac{76}{221}\\-31\frac{49}{221}\end{pmatrix}\|=46.4\end{equation}$

The closest objects $H$ and $J$ get to each other is $46.4$ km at 1:27pm.

I have made an e-activity for this.

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Filed under Classpad Skills, Closest Approach, Vectors, Year 11 Mathematical Methods

Tagged as casio classpad, eactivity, vectors closest approach

May 28, 2025 · 1:33 pm

A Counting Question

How many three digit numbers can you form from the digits 1, 2, 3, 4 and 5 if

(a) the digits must occur in increasing order?

(b) adjacent digits differ by 2?

Cambridge Year 11 Specialist Mathematics Skill Sheet 1A

(a) There are $5\times 4\times 3=60$ permutations of three digits from the five digits, but how many of those are in the right order?

Each set of 3 digits has 6 arrangements ( $3\times 2\times 1=6$ ).

For example, if the set is ${1, 2, 3}$ , then the possible arrangements are:

123, 132, 213, 231, 312, and 321.

Only one of those arrangements is in numerical order.

Hence what we want is $\begin{pmatrix}5\\3\end{pmatrix}=10$

(b) I think this one is more about creating a list. I shall start with 1

135, 531, 131, 242, 353, 424, 535, 315

There are 8 possibilities.

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Filed under Counting Techniques, Year 11 Specialist Mathematics

Tagged as counting techniques, Year 11 Specialist Mathematics

May 24, 2025 · 10:02 am

Circle Geometry Problem

In the diagram, points $A, B, C, D$ and $Q$ lie on a circle centre $O$ , radius $6$ cm and diameter $BQ, \angle{ABQ}=50^\circ, AB$ is parallel to $DO$ and point $P$ lies on diameter $BQ$ such that $OP=DP=4$ cm.

(a) Find $\angle{BCD}$

(b) Determine the length of $PC$ .

$\angle{BOD}=180^\circ-50^\circ=130^\circ$ (Co-interior angles in parallel lines are supplementary.)

$\angle{BCD}=\frac{1}{2}\times 130=65^\circ$ (Angles subtended by the same arc. The angle at the centre is twice the angle at the circumference.)

$\angle{BCD}=65^\circ.$

Let $PC=x$

From the intersecting chord theorem

$\begin{equation*}4\times x=2\times 10\end{equation}$

$\begin{equation*}4x=20\end{equation}$

$\begin{equation*}x=5\end{equation}$

$PC=5cm$

A chord $AB$ of a circle $O$ is extended to $C$ . The straight line bisecting $\angle{OAB}$ meets the circle at $E$ . Let $\angle{BAE}=x$ . Prove that $EB$ bisects $\angle{OBC}$ .

$\angle{BAO}=2x$ ( $AE$ bisects $\angle {BAO}$ )

$\Delta AOB$ is isosceles ( $AO=B0$ radii of the circle)

$\angle{ABO}=2x$ (Equal angles in isosceles triangle)

Therefore $\angle {AOB}=180^\circ-4x$ (angle sum of a triangle)

$\angle {BEA}=90^\circ-2x$ (angle at the circumference is half angle at the centre)

$\angle{ABE}=180^\circ-x-(90^\circ-2x)=90^\circ+x$ (angle sum of a triangle)

$\angle{CBE}=180^\circ-(90^\circ+x)=90^\circ-x$ (angles on a straight line)

$\angle{OBE}=90^\circ+x-2x=90^\circ-x$

$\angle{OBE}=\angle{CBE}$

Hence, $BE$ bisects $\angle{OBC}$

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Filed under Circle Theorems, Geometry, Uncategorized, Year 11 Specialist Mathematics

Tagged as circle geometry, circle theorems

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 6, 2025 · 7:27 am

Combinations Proof

Prove $\begin{pmatrix}n+1\\r\end{pmatrix}=\begin{pmatrix}n\\r-1\end{pmatrix}+\begin{pmatrix}n\\r\end{pmatrix}$

Let’s start with the right hand side.

$RHS=\begin{pmatrix}n\\r-1\end{pmatrix}+\begin{pmatrix}n\\r\end{pmatrix}$

$RHS=\frac{n!}{(n-(r-1))!(r-1)!}+\frac{n!}{(n-r)!r!}$

Simplify

$RHS=\frac{n!}{(n+1-r)!}+\frac{n!}{(n-r)!r!}$

There is a common denominator of $(n+1-r)!r!$

$RHS=\frac{n!}{(n+1-r)!}\times \frac{r}{r}+\frac{n!}{(n-r)!r!}\times \frac{n+1-r}{n+1-r}$

$RHS=\frac{n!r+n!(n+1-r)}{(n+1-r)r!}$

$RHS=\frac{n!r+(n+1)n!-n!r}{(n+1-r)!r!}$

$RHS=\frac{(n+1)!}{(n+1-r)!r!}$

$RHS=\begin{pmatrix}n+1\\r\end{pmatrix}$

$RHS=LHS$

We know this intuitively from Pascal’s triangle

Where each entry is the sum of the two entries above it – for example,

$6+4=10$

Remember, Pascal’s triangle can be written as combinations,

so $\begin{pmatrix}5\\3\end{pmatrix}=\begin{pmatrix}4\\2\end{pmatrix}+\begin{pmatrix}4\\3\end{pmatrix}$

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Filed under Algebra, Counting Techniques, Year 11 Mathematical Methods

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 30, 2025 · 6:08 am

Factorising Non-Monic Quadratics

The general equation of a quadratic is $ax^2+bx+c$

Let’s explore different methods of factorising a non-monic quadratic (the $a$ term is not $1$ )

Factorise $6x^2+x-12$

We need to find two numbers that add to $1$ and multiply to $-72$ (i.e. add to $b$ and multiply to $a\times c)$

The two numbers are $9$ and $-8$

Method 1 – Splitting the middle term

This is the method I teach the most often

$\begin{equation*}6x^2+x-12\end{equation}$

Split the middle term (the $b$ term) into the two numbers

$\begin{equation*}6x^2-8x+9x-12\end{equation}$

The order doesn’t matter.

Find a common factor for the first term terms, and then for the last two terms.

$\begin{equation*}2x(3x-4)+3(3x-4)\end{equation}$

There is a common factor of $(3x-4)$ , factor it out.

$\begin{equation*}(3x-4)(2x+3)\end{equation}$

Method two – Fraction

$\begin{equation*}6x^2+x-12\end{equation}$

Put $6x$ into both factors and divide by $6$

$\begin{equation*}\frac{(6x-8)(6x+9)}{6}\end{equation}$

Factorise

$\begin{equation*}\frac{2(3x-4)3(2x+3)}{6}\end{equation}$

$\begin{equation*}\frac{6(3x-4)(2x+3)}{6}\end{equation}$

$\begin{equation*}(3x-4)(2x+3)\end{equation}$

Method 3 – Monic to non-monic

$\begin{equation*}y=6x^2+x-12\end{equation}$

Multiply both sides of the equation by $a$

$\begin{equation*}6y=6(6x^2+x-12)\end{equation}$

$\begin{equation*}6y=6^2x^2+6x-72\end{equation}$

$\begin{equation*}6y=(6x)^2+6x-72\end{equation}$

Let $A=6x$

$\begin{equation*}6y=A^2+A-72\end{equation}$

Factorise

$\begin{equation*}6y=(A+9)(A-8)\end{equation}$

Replace the $A$ with $6x$

$\begin{equation*}6y=(6x+9)(6x-8)\end{equation}$

$\begin{equation*}6y=3(2x+3)2(3x-4)\end{equation}$

$\begin{equation*}6y=6(2x+3)(3x-4)\end{equation}$

$\begin{equation*}y=(2x+3)(3x-4)\end{equation}$

Method 4 – Cross Method

$\begin{equation*}6x^2+x-12\end{equation}$

Place the two numbers in the cross

Place the two numbers that add to $1$ and multiply to $-72$ in the other parts of the cross.

Divide these two numbers by $6$ (i.e $a$ )

Simplify

Hence,

$\begin{equation*}(x-\frac{4}{3})(x+\frac{3}{2})\end{equation}$

Which is

$\begin{equation*}(3x-4)(2x+3)\end{equation}$

Method 5 – By Inspection

This is my least favourite method – although students get better with practice

$\begin{equation*}6x^2+x-12\end{equation}$

The factors of $a$ are $1, 2, 3,$ and $6$ and the factors of $12$ are $1, 2, 3, 4, 6, 12$

We know one number is positive and one number negative.

Which give us all of these possibilities

$\begin{equation*}(2x+3)(3x-4)\end{equation}$

With a bit of practice you don’t need to check all of the possibilties, but I find students struggle with this method.

Method 6 – Grid

$\begin{equation*}6x^2+x-12\end{equation}$

Create a grid like the one below


	$6x^2$
		$-12$

Find the two numbers that multiply to $-72$ and add to $1$ and place them in the other grid spots (see below)


	$6x^2$	$-8x$
	$9x$	$-12$

Find the HCF (highest common factor) of each row and put in the first column.

Row $1$ HCF= $2x$ , Row $2$ HCF= $3$


$2x$	$6x^2$	$-8x$
$3$	$9x$	$-12$

For the columns, calculate what is required to multiple the HCF to get the table entry.

For example, what do you need to multiple $2x$ and $3$ by to get $6x^2$ and $9x$ ? In this case it is $3x$ . It’s always going to be the same thing, so just use one value to calculate it,

	$3x$	$-4$
$2x$	$6x^2$	$-8x$
$3$	$9x$	$-12$

The factors are column $1$ and row $1$
$(2x+3)(3x-4)$

The two methods I use the most are splitting the middle term, and the cross method, but I can see value in the grid method.

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Filed under Algebra, Factorising, Factorising, Polynomials, Quadratic, Quadratics

Tagged as Factorising, factorising non-monic trinomials, non-monic quadratics

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

April 21, 2025 · 11:08 am

Area/Geometry Problem

This problem is from The Geometry Forum Problem of the Week June 1996

In triangle ABC, AC=18 and D is the point on AC for which AD=5. Perpendiculars drawn from D to AB and CB have lengths of 4 and 5 respectively. What is the area of triangle ABC?

I put together a diagram (in Geogebra)

Add points P and Q

Triangle APD and triangle DQC are right angled. Using pythagoras, $AP=3$ and $QC=12$

$BQDP$ is a cyclic quadrilateral and $BD$ is the diameter. I am not sure if this is useful, but it is good to notice.

$\begin{equation*}sin(A+B+C)=sin(180)=0\end{equation}$

$\begin{equation*}sin((A+C)+B)=sin(A+C)cosB+sinBcos(A+C)=0\end{equation}$

$\begin{equation*}cosB(sinAcosC+sinCcosA)+sinB(cosAcosC-sinAsinC)=0\end{equation}$

$\begin{equation*}cosB(\frac{4}{5}\times\frac{12}{13}+\frac{5}{13}\times\frac{3}{5})+sinB(\frac{3}{5}\times\frac{12}{13}-\frac{4}{5}\times\frac{5}{13})=0\end{equation}$

$\begin{equation*}cosB(\frac{48}{65}+\frac{15}{65})+sinB(\frac{36}{65}-\frac{20}{65})=0\end{equation}$

$\begin{equation*}\frac{63}{65}cosB+\frac{16}{65}sinB=0\end{equation}$

$\begin{equation*}63cosB+16sinB=0\end{equation}$

$\begin{equation*}63+16tanB=0\end{equation}$

$\begin{equation*}tanB=\frac{-63}{16}\end{equation}$

If $tanB=\frac{-63}{16}$ then $sinB=\frac{63}{65}$

Now,

$\begin{equation*}\frac{y+12}{sinA}=\frac{18}{sinB}\end{equation}$

$\begin{equation*}y+12=\frac{4}{5}(18)\frac{65}{63}\end{equation}$

$\begin{equation*}y+12=\frac{104}{7}\end{equation}$

Hence the Area is

$\begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})sinC\end{equation}$

$\begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})\frac{5}{13}\end{equation}$

$\begin{equation*}A=\frac{360}{7}=51.43\end{equation}$

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Filed under Area, Finding an area, Geometry, Identities, Non-Right Trigonometry, Pythagoras, Trigonometry

Tagged as area, area of triangles, problem solving

Possible factorisations	$b$ term of expansion
$(x-1)(6x+12)$	$12x-6x=6x$	No
$(x-2)(6x+6)$	$6x-12x=-6x$	No
$(x-3)(6x+4)$	$4x-18x-14x$	No
$(2x-1)(3x+12)$	$24x-3x=21x$	No
$(2x-2)(3x+6)$	$12x-6x=6x$	No
$(2x-3)(3x+4)$	$8x-9x=-1x$	Almost, switch the signs
$(2x+3)((3x-4)$	$-8x+9x=1x$	Yes

Square Root Puzzle

Closest Approach (Shortest Distance) e-activity (Casio Classpad)

A Counting Question

Circle Geometry Problem

Perfect Squares

Combinations Proof

Geometry Puzzle

Factorising Non-Monic Quadratics

Method 1 – Splitting the middle term

Method two – Fraction

Method 3 – Monic to non-monic

Method 4 – Cross Method

Method 5 – By Inspection

Method 6 – Grid

Problem Solving

Pentagon

Hexagon

Septagon

Octagon

Area/Geometry Problem

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