March 17, 2025 · 12:33 pm

Circle Geometry Question 2

One of my Year 11 Specialist students had this question

Triangle ABC touches the given circle at Points P, Q, R and S only. The secant BW touches the circle at V and W.

Diagram not drawn to scale

(a) Determine the lengths of the line segments marked x, y and z, leaving your answers as exact values.

(b) If the length of the line segment QW is 4 units, determine the exact radius of the circle.

(a) We are going to use the Intersecting Secant Theorem – the tangent version

c^2=a\times(a+b)

Hence, we have

    \begin{equation*}30^2=25(25+x+6)\end{equation}

    \begin{equation*}900=25(31+x)\end{equation}

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

Then we can use the intersecting chord theorem to find y.

    \begin{equation*}10\times y=6 \times x\end{equation}

    \begin{equation*}10y=30\end{equation}

    \begin{equation*}y=3\end{equation}

Back to the Intersecting Secant Theorem to find z

    \begin{equation*}z^2=4\times 17\end{equation}

    \begin{equation*}z=2\sqrt{17}\end{equation}

(b)


QW is part of a 3-4-5 triangle, therefore \angle{Q}=90^\circ

This is definitely the case of the diagram not being drawn to scale. If \angle{Q}=90^\circ, then the purple line must be the diameter.

We can use pythagoras to find the length of the diameter

    \begin{equation*}(2r)^2=13^2+4^2\end{equation}

    \begin{equation*}4r^2=185\end{equation}

    \begin{equation*}r=\frac{\sqrt{285}}{2}\end{equation}

The radius of the circle is \frac{\sqrt{285}}{2}

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March 9, 2025 · 11:26 am

Circle Geometry Question

In the above diagram O is the centre of the larger circle. A, B,D and E are points on the circumference of the larger circle. A, C, E and 0 are points on the circumference of the smaller circle. Show that \angle{CAB}=\angle{ABC}. AB, AC and BC are straight lines.

AO=OB (radii of the larger circle)

At a line from O to E (it is also a radius of the larger circle)

Let \angle{CAB}=\alpha.

ACEO is a cyclic quadrilateral.

Hence, \angle{CED}=180-\alpha (AECO is a cyclic quadrilateral)

As CB is a straight line \angle{OEB}=180-(180-\alpha)=\alpha.

\Delta OEB is an isosceles triangle.

Therefore, \angle{ABC}=\alpha

Therefore \angle{ABC}=\angle{CAB}

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March 5, 2025 · 2:00 am

Binomial Expansion Theorem

My Year 11 Mathematics Methods students are working on the Binomial Expansion Theorem.

But before we get onto that, remember Pascal’s triangle

First 8 rows of Pascal’s triangle

Now we can use combinations to find the numbers in each row. For example, 1 4 6 4 1 is \begin{pmatrix}4\\0\end{pmatrix}=1, \begin{pmatrix}4\\1\end{pmatrix}=4, \begin{pmatrix}4\\2\end{pmatrix}=6, \begin{pmatrix}4\\3\end{pmatrix}=4, \begin{pmatrix}4\\4\end{pmatrix}=1

ExpressionExpansionCo-efficients
(x+y)^2x^2+2xy+y^21, 2, 1
(x+y)^3x^3+3x^2y+3xy^2+y^31, 3, 3, 1
(x+y)^4x^4+4x^3y+6x^2y^2+4xy^3+y^41, 4, 6, 4, 1

As you can see, the coefficients are the row of pascal’s triangle corresponding to the power. So (x+y)^6 would have co-efficients from the sixth row of the table 1, 6, 15, 20, 15, 6, 1.

To generalise

(x+y)^n=\begin{pmatrix}n\\0\end{pmatrix}x^ny^0+\begin{pmatrix}n\\1\end{pmatrix}x^{n-1}y^1+\begin{pmatrix}n\\2\end{pmatrix}x^{n-2}y^2+ ...+\begin{pmatrix}n\\n-1\end{pmatrix}x^1{y^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}x^0y^n

Which we can condense to

(x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i

Worked Examples

(1) Expand (2x-3)^4

(2x-3)^4=\begin{pmatrix}4\\0\end{pmatrix}(2x)^4(-3)^0+\begin{pmatrix}4\\1\end{pmatrix}(2x)^3(-3)^1+\begin{pmatrix}4\\2\end{pmatrix}(2x)^2(-3)^2+\begin{pmatrix}4\\3\end{pmatrix}(2x)^1(-3)^3+\begin{pmatrix}4\\4\end{pmatrix}(2x)^0(-3)^4
(2x-3)^4=16x^4-96x^3+216x^2-216x+81

(2) Find the co-efficient of the x^3 term in the expansion of (2-5x)^5.

Remember (x+y)^n=\Sigma_{i=0}^n \begin{pmatrix}n\\i\end{pmatrix}x^{n-i}y^i, the x^3 is when i=3
\begin{pmatrix}5\\3\end{pmatrix}(2)^2(-5)^3=10\times 2\times -125=-5000

(3) Find the constant term in the expansion of (x^2+\frac{3}{x^4})^6

We need to find the term where the x‘s cancel out. Each term is \begin{pmatrix}6\\i\end{pmatrix}(x^2)^{6-i}(\frac{3}{x^4})^i.
\begin{pmatrix}6\\i\end{pmatrix}(x^{12-2i})(3^ix^{-4i}).
We need 12-2i-4i=0, hence i=2
Therefore, the co-efficient is \begin{pmatrix}6\\2\end{pmatrix}\times3^2=135

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March 3, 2025 · 2:51 am

Derangements

From Wikipedia

In combinatorial mathematics, a derangement is a permutation of the elements of a set in which no element appears in its original position.

For example, if we have the set {A, B, C}, there are 6 permutations

ABC, ACB, BAC, BCA, CAB, CBA

But only 2 of them are derangements – BCA and CAB

I did a practical question on this here. In that question I used a tree diagram, but there must be a way to determine the number of derangements given n elements.

We know 3 elements has 2 derangements, what about 4 elements? We know there are 4!=24 permutations

ABCDBACDCABDDABC
ACBDBADCCADBDACB
ACDBBCADCBADDBAC
ABDCBCDACBDADBCA
ADBCBDACCDABDCAB
ADCBBDCACDBADCBA

I have highlighted the derangements. when n=4 there are 9 derangements.

Let’s try to generalise.

  • In how many ways can one element be in its original position?
    \begin{pmatrix}4\\1\end{pmatrix} \times 3!=24
    Choose one element from the 4 to be in its original position, and then arrange the remaining three elements.

    So far, we have D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!=0

    Clearly we are counting some arrangements multiple times, for example ABDC has A and B in the correct position, so we need to add all of the arrangements with 2 elements in their original position.
  • In how many ways can two elements be in their original position?
    \begin{pmatrix}4\\2\end{pmatrix} \times 2!=12

    So now we have, D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!=12

    But once again we have counted some arrangements multiple times, so we need to subtract all of the arrangements with 3 elements in their original position.
  • In how many ways can three elements be in their original position?
    \begin{pmatrix}4\\3\end{pmatrix} \times 1!=4

    So now we have, D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!=8

    We now need to add all of the arrangements with 4 elements in their original position.
  • In how many ways can four elements be in their original position?
    \begin{pmatrix}4\\4\end{pmatrix} \times 0!=1

    So now we have, D_4=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!+\begin{pmatrix}4\\4\end{pmatrix} \times 0!=9

Hence,

D_n=\begin{pmatrix}n\\0\end{pmatrix}\times n!-\begin{pmatrix}n\\1\end{pmatrix}\times (n-1)!+\begin{pmatrix}n\\2\end{pmatrix}\times (n-2)!-... \mp \begin{pmatrix}n\\n\end{pmatrix}\times 0!

Which we can simplify to

D_n=\Sigma_{i=0}^{n}(-1)^n\begin{pmatrix}n\\i\end{pmatrix}\times(n-i)!

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February 25, 2025 · 9:28 am

Differentiating f(x)=e^x

We are going to differentiate f(x)=e^x from first principals.

Remember the definition of a derivative is

(1)   \begin{equation*}f'(x)=\lim_{\limits{h\to 0}}\frac{f(x+h)-f(x)}{h}\end{equation*}

If f(x)=e^x, then

    \begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^{x+h}-e^x}{h}\end{equation}

    \begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x\times e^h-e^x}{h}\end{equation}

    \begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x(e^h-1)}{h}\end{equation}

    \begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}

Let’s think about \lim_{\limits{h \to 0}}\frac{e^h-1}{h}

Remember e is defined as \lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n

(2)   \begin{equation*}\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation*}

Let y=e^h-1, as h \to 0, e^h-1 \to 1-1=0 hence y \to 0

If y=e^h-1, then h=ln(y+1)

(3)   \begin{equation*}\lim_{\limits{y \to 0}}\frac{y}{ln(y+1)}\end{equation*}

We are going to rewrite the equation 3 as

(4)   \begin{equation*}\lim_{\limits{y \to 0}}\frac{\frac{1}{ln(y+1)}}{y}\end{equation*}

And then we can write equation 4 as

(5)   \begin{equation*}{\lim_{\limits{y \to 0}}{\frac{1}{\frac{1}{y}ln(y+1)}\end{equation*}

Using log laws we can write equation 5 as

(6)   \begin{equation*}\lim_{\limits{y \to 0}}\frac{1}{ln(y+1)^{\frac{1}{y}}}\end{equation*}

Let y=\frac{1}{n}

As y \to 0, n \to \infty

equation 6 becomes

(7)   \begin{equation*}\lim_{\limits{n \to \infty}}\frac{1}{ln(\frac{1}{n}+1)^n}\end{equation*}

We can move the limit to inside the natural log

(8)   \begin{equation*}\frac{1}{ln(\lim_{\limits{n \to \infty}}(\frac{1}{n}+1)^n)}\end{equation*}

And we know from the definition of e that e=\lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n

Hence, equation 8 is

(9)   \begin{equation*}\frac{1}{ln(e)}=1\end{equation*}

Back to our derivative

    \begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}

We know that \lim_{\limits{h \to 0}}{\frac{e^h-1}{h}=1 hence

    \begin{equation*}f'(x)=e^x\end{equation}

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February 19, 2025 · 6:15 am

Finding a Recursive Rule for a Sequence (Year 12 Mathematics Applications)

How do we go about finding the rule for a first order linear recurrence relation?

Something like

    \begin{equation*}5, 7, 11, 19, ...\end{equation}

There isn’t a common difference (arithmetic sequence) or a common ratio (geometric sequence). Sometimes you can just see the rule, but an algorithm will be handy.

Let’s say our relationship is

(1)   \begin{equation*}T_{n+1}=bT_n+c, T_1=a\end{equation*}

Referring back to our sequence 5, 7, 11, 19, ..., we know

(2)   \begin{equation*}7=b\times5+c\end{equation*}

and

(3)   \begin{equation*}11=b\times7+c\end{equation*}

We can solve equation 2 and 3 simultaneously

equation 2- equation 3

    \begin{equation*}-4=-2b\end{equation}

Hence b=2

Substitute b=2 into equation 2

    \begin{equation*}7=2\times 5+c\end{equation}

    \begin{equation*}7=10+c\end{equation}

Hence c=-3

    \begin{equation*}T_{n+1}=2T_n-3, T_1=5\end{equation}

Let’s try to generalise

If T_{n+1}=bT_n+c, T_1=a, then

(4)   \begin{equation*}T_2=bT_1+c\end{equation*}

and

(5)   \begin{equation*}T_3=bT_2+c\end{equation*}

Equation 5 - equation 4

    \begin{equation*}T_3-T_2=(T_2-T_1)b\end{equation}

    \begin{equation*} b=\frac{T_3-T_2}{T_2-T_1}\end{equation}

Hence, b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}

Once you know b, substitute into either equation to find C.

Example

Find the recursive rule for the following

-8, -12, -20, -36, ...

    \begin{equation*}b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}=\frac{-20-(-12)}{-12-(-8)}=\frac{-8}{-4}=2\end{equation}

    \begin{equation*}-12=2\times-8+c\end{equation}

    \begin{equation*}-12=-16+c\end{equation}

Hence c=4 and T_{n+1}=2T_n+4, T_1=-8

It is also possible to find the rule using a Classpad (if it’s in the calculator section} by using an e-activity.

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February 15, 2025 · 8:35 am

Counting Techniques

Four teachers decide to swap desks at work. How many ways can this be done if no teacher sits at their previous desk?
Mathematics Specialist Units 1&2 Cambridge

I like this question as it seems easy until you start thinking about it. I think the best approach is a tree diagram.

If we think of the four teachers as A, B, C and D. Then A can no longer sit in A, so the options are B, C and D for the first desk.

For the second desk, If B is in the first desk, then A, C or D could be in the second. If C is in the first desk, then A or D could be in the second (B can’t be in the same desk). If D is in the first desk, then A or C can be in the second desk.

And so on, leaving 9 possibilities

BADC
BCDA
BDAC
CADB
CDAB
CDBA
DABC
DCAB
DCBA

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February 4, 2025 · 9:17 am

Differentiating the Tangent Function

Remember tan(x)=\frac{sin(x)}{cos(x)}.

I use the quotient rule to differentiate f(x)=tan(x).

(1)   \begin{equation*}\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\end{equation*}

If h(x)=tan(x)=\frac{sin(x)}{cos(x)} then from equation 1

(2)   \begin{equation*}h'(x)=\frac{cos(x)cos(x)-(-sin(x)sin(x))}{[cos(x)]^2}\end{equation*}

(3)   \begin{equation*}h'(x)=\frac{cos^2(x)+sin^2(x)}{cos^2(x)}\end{equation*}

Remember the Pythagorean identity

(4)   \begin{equation*}sin^2(x)+cos^2(x)=1\end{equation*}

Hence

    \begin{equation*}h'(x)=\frac{1}{cos^2(x)}=sec^2(x)\end{equation}

(5)   \begin{equation*}\frac{d}{dx}tan(x)=sec^2(x)\end{equation*}

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February 3, 2025 · 9:11 am

Differentiating Trigonometric Functions

In the last post we looked at two trig limits:

(1)   \begin{equation*}\lim_{x \to 0}\frac{sin(x)}{x}=1\end{equation*}

(2)   \begin{equation*}\lim_{x \to 0}\frac{1-cos(x)}{x}=0\end{equation*}

We are going to use these two limits to differentiate sine and cosine functions from first principals.

    \begin{equation*}f(x)=sin(x)\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x+h)-sin(x)}{h}\end{equation}

Use the trig identity

    \begin{equation*}sin(A+B)=sinAcosB+sinBcosA\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x)cos(h)+sin(h)cos(x)-sin(x)}{h}\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}(\frac{sin(x)(cos(h)-1)}{h}+\frac{sin(h)cos(x)}{h})\end{equation}

    \begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{(cos(h)-1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

    \begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{-(-cos(h)+1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

Evaluate the limits

    \begin{equation*}f'(x)=sin(x)\times 0+cos(x)\times (1)=cos(x)\end{equation}

Hence, \frac{d}{dx}sin(x)=cos(x).

Now we are going to do the same for f(x)=cos(x).

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x+h)-cos(x)}{h}\end{equation}

Use the trigonometric identity

    \begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}\end{equation}

    \begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)(cos(h)-1)-sin(x)sin(h)}{h}\end{equation}

    \begin{equation*}f'(x)=cos(x)\lim\limits_{h \to 0}\frac{-(1-cos(h))}{h}-sin(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}

Evaluate the limits

    \begin{equation*}f'(x)=cos(x)\times(0)-sin(x)\times (1)=-sin(x)\end{equation}

Hence \frac{d}{dx} cos(x)=-sin(x)

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February 2, 2025 · 7:46 am

Trigonometric Limits

\lim\limits_{x \to 0}\frac{sin(x)}{x}=?

Unit Circle

Remember cos(x)=\frac{OA}{OB}=\frac{OA}{1}, hence OA=cos(x) and the co-ordinate of A is (cos(x), 0).

sin(x)=\frac{AB}{OB}=\frac{AB}{1}, hence AB=sin(x) and the co-ordinate of B is (cos(x), sin(x))

And from the definition of tan(x) we know D is the point (1, tan(x))

Consider the areas of triangle OAB, sector OBC, and triangle OCD.

We know from inspection of the above diagram that

Area OAB< Area OCB<Area OCD

Which means,

\frac{1}{2}b_1 h_1<\frac{1}{2}r^2x<\frac{1}{2}b_2 h_2

We can ignore all of the halves.

cos(x)sin(x)<x<(1)tan(x)

Remember tan(x)=\frac{sin(x)}{cos(x)}

cos(x)sin(x)<x<\frac{sin(x)}{cos(x)}

Divide everything by sin(x) (as we are in the first quadrant we know sin(x)>0, so we don’t need to worry about the inequality)

cos(x)<\frac{x}{sin(x)}<\frac{1}{cos(x)}

Invert everything and change the direction of the inequalities)

\frac{1}{cos(x)}>\frac{sin(x)}{x}>cos(x)

I am going to rewrite it as follows

cos(x)<\frac{sin(x)}{x}<\frac{1}{cos(x)}

because I like to use less thans rather than greater thans.

Now what happens as x tends to 0?

cos(0)=1

1<\frac{sin(x)}{x}<\frac{1}{1}

Hence by the squeeze theorem \lim\limits_{x \to 0}\frac{sin(x)}{x}=1

Now we know this limit, we are going to use it to find \lim\limits_{x \to 0}\frac{1-cos(x)}{x}

Multiply by \frac{1+cos(x)}{1+cos(x)}

\lim\limits_{x \to 0}\frac{1-cos(x)}{x}\times \frac{1+cos(x)}{1+cos(x)}

\lim\limits_{x \to 0}\frac{1-cos^2(x)}{x(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin^2(x)}{x(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin(x)}{x}\times sin(x)(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin(x)}{x}\times \lim\limits_{x \to 0}sin(x)(cos(x)+1)

If we evaluate the limits,

(1)(sin(0)(cos(0)+1)=1\times 0 \times 2=0

Hence, \lim\limits_{x \to 0}\frac{1-cos(x)}{x}=0

In the next post we are going to use these limits to differentiate sine and cosine functions.

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