Racquel Sanderson

September 7, 2024 · 9:31 am

Geometry Problem

This problem is from Geometry Snacks by Ed Southall and Vincent Pantaloni – it’s a great book.

Two squares are constructed such that three vertices are collinear as shown. Find the value of the marked angle.

I started by marking in the right angles. And I added the diagonal of the larger square (pink line).

Because there are right angles at $O$ and $P$ , we know there is a circle, which has the diagonal of the square as its diameter (see second image below).

$\angle{RSP}$ is $45^{\circ}$ (Angle between the diagonal and side of a square)

$PORS$ is a cyclic quadrilateral.

In cyclic quadrilaterals opposite angles are supplementary.

Hence, $\angle{ROP}=180^{\circ}-45^{\circ}=135^{\circ}$

As $\angle{ROS}=90^{\circ}$ , $\angle{SOP}$ must be $45^{\circ}$

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Filed under Finding an angle, Geometry

Tagged as circle geometry, finding an angle, geometry, squares

August 28, 2024 · 6:42 am

Using the Hungarian Algorithm for an Assignment Problem

The following question is from the 2019 ATAR Mathematics Applications Paper – Calculator Free Question 3.

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Filed under Networks, Year 12 Mathematics Applications

Tagged as assignment problem, hungarian algorithm, Mathematics Applications, year 12 Apps

August 23, 2024 · 7:41 am

Sketching in the Complex Plane Using a TI-nspire CX 11

Let $R$ be the region of the complex plane where the inequalities $|z-i|\le2$ and $|z-\bar{z}|\ge3$ hold simultaneously.

First find the Cartesian equations.

Second, sketch each of the functions.

The section that is shaded twice is our region.

Determine the minimum value of $Re(z)$ in $R$ .

We can find the point of intersection between the circle and the line.

$Re(z)=-1.94$

Or if you want exact values

Use the Solve Systems of Equations tool – Menu – Algebra – Solve Systems of Equations – Solve Systems of Equations.

$Re(z)=-\frac{\sqrt{15}}{2}$

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Filed under Complex Numbers, Sketching Complex Regions, TI nspire CX 11

August 8, 2024 · 5:42 am

Pythagoras Question

This was a question one of my year 9s had to tackle:

A hemisphere of radius length 5cm is partially filled with water. The top of the hemisphere is horizontal and the surface of the water is a circle of radius 4cm. Find the depth of the water.

ICE-EM Mathematics 9, page 70, question 2

Below is a cross section of the hemisphere

The depth of the water is $5-x$

We can find $x$ using the Pythagorean theorem

$x^2=5^2-4^2$

$x=3$ (it’s the classic 3-4-5 triangle)

Hence the depth of the water is 2cm.

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Filed under Pythagoras

Tagged as depth of water in hemispherical bowl, pythagoras

July 25, 2024 · 7:05 am

Function Composition and Domain and Range

My year 12 Specialist students have been working on function composition and the domain and range of the resulting composition. And they have been struggling a bit with why the composition doesn’t exist.

For example,

The functions $f$ and $g$ are defined by $f(x)=x^2-2$ and $g(x)=\sqrt{x}$

(a) Explain why $g(f(x))$ is not defined.

(b) By suitably restricting the domain of $f$ , obtain a function $f_1$ such that $g(f_1(x))$ is defined.

For the composite function to exist the range of the inner function (in this case $f(x)$ ) must be a subset of the domain of the outer function (in this case $g(x)$ ).

Start by finding the domain and range of each function.

We can see the range of $f(x)$ is not a subset of the domain of $g(x)$

i.e. $R_{f(x)}\nsubseteq D_{g(x)}$

We can restrict the range of $f(x)$ by restricting the domain.

$f(x)\geq0$

$x^2-2\geq0$

$x^2\geq2$

$x\leq -\sqrt{2}$ or $x\geq \sqrt{2}$

Therefore $f_1(x)=x^2-2, x\leq -\sqrt{2}$ or $x\geq \sqrt{2}$

and $g(f_1(x))=\sqrt{x^2-2}, x\leq -\sqrt{2}$ or $x\geq \sqrt{2}$

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Filed under Composition, Functions

July 16, 2024 · 7:35 am

Australian Mathematics Competition – Polynomial Question

I came across this question from the 2010 Senior Australian Mathematics Competition:

A polynomial $f$ is given. All we know about it is that all its coefficients are non-negative integers, $f(1)=6$ and $f(7)=3438$ . What is the value of $f(3)$
Australian Mathematics Competition 2006-2012

I thought ‘excellent, a somewhat hard polynomial question for my students’ and then I tried it. Now I know why only 1% of students got it correct.

As we don’t know the order of the polynomial, let

$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0$

We know all of the coefficients are greater than or equal to zero. We also know

$f(1)=a_n+a_{n-1}+...+a+a_0=6$

Which means that all of the coefficients are between zero and six

$0\le{a_n}\le{6}$

We have also been given $f(7)$

$f(7)=7^na_n+7^{n-1}a_{n-1}+ ... +7a+1=3438$

As all of the coefficients are between zero and six, this is $3438$ written in base 7.

Let’s calculate a few powers of 7

	Powers of 7
$7^0$	1
$7^1$	7
$7^2$	49
$7^3$	343
$7^4$	2401
$7^5$	16807

Hence $3438$ written in base $7$ is $13011$

Therefore $f(x)=x^4+3x^3+x+1$

$f(3)=3^4+3\times3^3+3+1$

$f(3)=81+81+4$

$f(3)=166$

I really like this question. I think it could work well as a class extension activity with a bit of scaffolding.

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Filed under Number Bases, Polynomials

Tagged as base 7, number bases, Polynomials

June 12, 2024 · 7:37 am

Year 12 Mathematics Applications Finance

Ming, a former high school student and now a successful business owner, wishes to set up a perpetuity of $6000 per year to be paid to a deserving student from her school. The perpetuity is to be paid at the start of the year in one single payment.

(a) A financial institution has agreed to maintain an account for the perpetuity paying a fixed rate of 5.9% p.a. compounded monthly. Show that an amount of $98 974, to the nearest dollar, is required to maintain this perpetuity.

(b) Ming allows herself five years to accumulate the required $98 974 by making regular quarterly payments into an account paying 5.4% p.a. compounded monthly. Determine the quarterly payment needed to reach the required amount after five years if Ming starts the account with an initial deposit of $1000.
SCSA 2017 CA 8

(a) For a perpetuity, we want the interest to equal the payment.

Remember the compound interest formula is

$A=P(1+\frac{r}{100n})^{nt}$

Where $P$ is the principal, $r$ is the interest rate (as a percent), $n$ is the number of compounding periods in a year, and $t$ is the time.

$\therefore I=A-P$

$I=P(1+\frac{r}{100n})^{nt}-P$

$6000=P(1.00492)^12-P$

$P=\frac{6000}{(1.00492^12-1)}$

$P=98 974.14$

Therefore an amount of $98 974 is required to maintain this perpetuity

For part (b) I will use the Finance Solver on a Classpad (Casio).

The quarterly payments are $4283.77.

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Filed under Classpad Skills, Finance, Finance, Year 12 Mathematics Applications

Tagged as casio classpad

· 7:02 am

The Logistic Equation

My year 12 Specialist students are working on logistic growth at the moment. An example might be helpful.

A new viral disease was found to spread according to the equation $\frac{dN}{dt}=kn(M-N)$ , where $M$ is the susceptible population, $N$ is the number of people infected at time $t$ months and $k=1.5\times 10^{-9}$ . In March 2010, it was thought only 100 people out of a population of 18 million were infected. Use the logistic model to find the number infected in:

(a) March 2011

(b) June 2012

(c) January 2017
Specialist 12 – Nelson Senior Maths

$\frac{dN}{dt}=kN(M-N)$

$\frac{dN}{dt}=(1.5\times10^{-9})N(18\times10^{6}-N)$

$\frac{dN}{N(18\times10^{6}-N)}=1.5\times10^{-9}dt$

Use partial fractions to separate the denominator $\frac{dN}{N(18\times10^{6}-N)}$

$\frac{1}{N(18\times10^{6}-N)}=\frac{A}{N}+\frac{B}{18\times10^{6}-N}$

$1=A(18\times10^{6}-N)+BN$

When $N=0$

$1=A(18\times10^{6})$

$A=\frac{1}{18\times10^{6}}$

When $N=18\times10^{6}$

$1=B(18\times10^{6})$

$B=\frac{1}{18\times10^{6}}$

$\frac{1}{18\times10^{6}}(\frac{1}{N}+\frac{1}{(18\times10^{6}-N)}dN)=1.5\times10^{-9}dt$

$\int\frac{1}{N}+\frac{1}{(18\times10^{6}-N)}dN=\int27\times10^{-3}dt$

$\ln|N|-\ln|18\times10^{6}-N|=(27\times10^{-3})t+c$

$\ln|\frac{N}{18\times10^{6}-N}|=(27\times10^{-3})t+c$

$\frac{N}{18\times10^{6}-N}=e^{(27\times10^{-3})t+c}$

Let $A=e^{c}$ and rearrange to make $N$ the subject.

$N=\frac{(18\times10^{6})Ae^{(27\times10^{-3})t}}{1+Ae^{(27\times10^{-3})t}}$

Divide by $Ae^{(27\times10^{-3})t}$

$N=\frac{(18\times10^{6})}{\frac{1}{A}e^{-(27\times10^{-3})t}+1}$

Initially 100 people were infected.

$100=\frac{(18\times10^{6})}{\frac{1}{A}+1}$

$A=\frac{1}{179999}$

$N=\frac{(18\times10^{6})}{179999e^{-(27\times10^{-3})t}+1}$

(a) $t=12, N=138.3$ , hence $138$

(b) $t=27, N=207.3$ , hence $207$

It is not necessary to solve the differential equation, you can use the formula

$\frac{dP}{dt}=rP(k-P)\leftrightarrowP=\frac{kP_0}{P_0+(k-P_0)e^{-rkt}}$

This formula is on the Year 12 Mathematics Specialist formula sheet for Western Australia.

For our question,

$\frac{dN}{dt}=(1.5\times10^{-9})N(18\times10^6-N)$

So, $P=\frac{18\times10^{6}\times100}{100+(18\times10^6-100)e^{-(1.5\times10^{-9})(18\times10^6)t}}$

And you can substitute values for $t$ .

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Filed under Differential Equations, Integration, Logistic Growth

Tagged as logistic growth, solving differential equations

June 1, 2024 · 7:43 am

Non-Right Trigonometry Problem

I worked on this question with one of my students (I don’t know where it is from).

Mike leaves the rose bush he was examining and walks 35m in the direction
S $20^{\circ}$ W towards a pond.
From there he walks 70m towards a rotunda. Mike is now 100m from the rose bush.
Find the bearing of the rotunda from the pond.

Let’s try to draw a diagram

Because we don’t the direction Mike walked from the pond, I have drawn a circle with radius 70m centred at the pond.

We know Mike is now 100m from the rose bush. As we don’t know the direction, I have drawn another circle with radius 100m centred at the rose bush. Where the two circles intersect are the possible locations of the rotunda.

First Position

Use the cosine rule to find the angle
$cos\theta=\frac{b^2+c^2-a^2}{2bc}$
$cos\theta=\frac{70^2+35^2-100^2}{2(70)(35)}$
$\theta=cos^{-1}(\frac{-3875}{5250})$
$\theta=137.6^{\circ}$

Using the fact that alternate angles in parallel lines are congruent, we can see
that the bearing from the pond to the rotunda is
$360-(137.6-20)=242.2^{\circ} T$

Second Position

It is the same triangle, so
$\theta=137.6^{\circ}.$

This time the bearing is
$20+137.6=157.6^{\circ}T$

Hence, the two possible bearings of the rotunda from the pond are $242.2^{\circ}T$ or $157.6^{\circ}T$ .

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Filed under Bearings, Non-Right Trigonometry, Trigonometry

Tagged as bearings, non-right trigonometry

May 28, 2024 · 6:52 am

Proof of the Sine Rule

As I have done a cosine rule proof, I thought I should also do a sine rule proof.

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}\textnormal{ or }\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

From the above diagram, we can find $h$ in two ways.

Set equation 1 equal to equation 2

$asinC=csinA$

$\frac{a}{sinA}=\frac{c}{sinC}$ or $\frac{sinC}{c}=\frac{sinA}{a}$

We could have put the altitude of the triangle from vertex A

Following the same process as above

Set equation 3 equal to equation 4.

$bsinC=csinB$

$\frac{b}{sinB}=\frac{c}{sinC}$

Now $\frac{c}{sinC}=\frac{a}{sinA}$ therefore

$\frac{b}{sinB}=\frac{c}{sinC}=\frac{a}{sinA}$ or $\frac{sinB}{b}=\frac{sinC}{c}=\frac{sinA}{a}$

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Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry

Tagged as proof of sine rule, sine rule, sine rule proof

As numbers	As Powers of 7
$3438=1\times2401+1037$	$3438=1\times7^4+1037$
$1037=3\times343+8$	$1037=3\times7^3+8$
$8=1\times7+1$	$8=1\times7^1+1$
$1=1\times1$	$1=1\times7^0$

Geometry Problem

Using the Hungarian Algorithm for an Assignment Problem

Sketching in the Complex Plane Using a TI-nspire CX 11

Pythagoras Question

Function Composition and Domain and Range

Australian Mathematics Competition – Polynomial Question

Year 12 Mathematics Applications Finance

The Logistic Equation

Non-Right Trigonometry Problem

First Position

Second Position

Proof of the Sine Rule

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