Category Archives: Year 11 Specialist Mathematics

March 18, 2025 · 6:04 am

Intersecting Secant Theorem

$CD$ is a tangent to the circle.

Prove $c^2=a(a+b)$

I am going to add two chords to the circle

$\angle{BDC}=\angle{CAD}$ (angles in alternate segments are congruent)

$\angle{BCD}=\angle{DCA}$ (shared angle)

$\therefore \Delta BDC\cong \Delta{DAC}$ (AA)

Hence

$\frac{DC}{AC}=\frac{BC}{DC}$ (Corresponding sides in similar triangles)

$\frac{c}{a+b}=\frac{a}{c}$

$\therefore c^2=a(a+b)$

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Filed under Circle Theorems, Geometry, Year 11 Specialist Mathematics

Tagged as circle geometry, Intersecting Secants Theorem

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

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

Tagged as intersecting chord theorem, intersecting secant theorem, pythagoras

March 9, 2025 · 11:26 am

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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Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

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

ABCD	BACD	CABD	DABC
ACBD	BADC	CADB	DACB
ACDB	BCAD	CBAD	DBAC
ABDC	BCDA	CBDA	DBCA
ADBC	BDAC	CDAB	DCAB
ADCB	BDCA	CDBA	DCBA

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, $D_3=4!-\begin{pmatrix}4\\1\end{pmatrix} \times 3!+\begin{pmatrix}4\\2\end{pmatrix} \times 2!-\begin{pmatrix}4\\3\end{pmatrix} \times 1!+\begin{pmatrix}4\\4\end{pmatrix} \times 0!=9$

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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Filed under Counting Techniques, Interesting Mathematics, Year 11 Specialist Mathematics

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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Filed under Counting Techniques, Tree Diagram, Year 11 Specialist Mathematics

Tagged as counting techniques, Year 11 Specialist Mathematics

November 14, 2024 · 4:19 am

Vectors – Closest Approach (2D)

At 10am, object $M$ travelling with constant velocity $(20, 10)$ km/h is sighted at the point with position vector $(-100, 100)$ km. At 11am object $N$ travelling with constant velocity $v=(x,y)$ km/h is sighted at the point with position vector $(-20,-50)$ km respectively. Use a scalar product method to determine $v$ given that the two objects were closest together at a distance of $40$ km at 4pm.

OT Lee Mathematics Specialist Year 11

At 4pm $M$ is at the point with position vector $(-100+6\times 20, -100+6\times 10)=(20, -40)$

and $N$ is at the point with position vector $(-20+5\times x, -50+5\times y)=(-20+5x, -50+5y)$

We know the distance between $M$ and $N$ at 4pm is $40$ km.

Hence,

$\begin{equation*}40^2=(20-(-20+5x))^2+(-40-(-50+5y))^2\end{equation}$

$\begin{equation*}40^2=(40-5x)^2+(10-5y)^2\end{equation}$

$\begin{equation*}1600=25x^2-400x+1600+25y^2-100y+100\end{equation}$

(1) $\begin{equation*}0=x^2+y^2-16x-4y+4\end{equation*}$

In the diagram below, I have found the position vector of $N$ relative to $M$ $(_Nr_M)$ and the velocity of $N$ relative to $M$ $(N_v_M)$

We know that when $_Nr_M\cdot_Nv_M=0$ $M$ and $N$ are the closest distance apart.

$\begin{equation*}(-40+5x,-10+5y)\cdot(x-20,y-10)=0\end{equation}$

$\begin{equation*}(-40+5x)(x-20)+(-10+5y)(y-10)\end{equation}$

$\begin{equation*}5x^2+5y^2-140x-60y+900=0\end{equation}$

(2) $\begin{equation*}x^2+y^2-28x-12y+180=0\end{equation*}$

Two equations and two unknowns which we can solve simultaneously. Both equations are circles.

Equation $(1)$ becomes

(3) $\begin{equation*}(x-8)^2+(y-2)^2=64\end{equation*}$

and equation $(2)$ becomes

(4) $\begin{equation*}(x-14)^2+(y-6)^2=52\end{equation*}$

From equation $(3)$

$\begin{equation*}x=\sqrt{64-(y-2)^2}+8\end{equation}$

We will worry about the negative version later.

Substitute for $x$ into equation $(4)$

$\begin{equation*}(\sqrt{64-(y-2)^2}+8-14)^2+(y-6)^2=52\end{equation}$

$\begin{equation*}(\sqrt{64-(y-2)^2}-6)^2+(y-6)^2=52\end{equation}$

$\begin{equation*}64-(y-2)^2-12\sqrt{64-(y-2)^2}+36+y^2-12y+36=52\end{equation}$

$\begin{equation*}136-y^2+4y-4+y^2-12y-12\sqrt{64-(y-2)^2}=52\end{equation}$

$\begin{equation*}-12\sqrt{64-(y-2)^2}=-80+8y\end{equation}$

$\begin{equation*}\sqrt{64-(y-2)^2}=\frac{20}{3}-\frac{2}{3}y\end{equation}$

Square both sides of the equation

$\begin{equation*}64-(y-2)^2=\frac{400}{9}-\frac{80}{9}y+\frac{4}{9}y^2\end{equation}$

$\begin{equation*}60-y^2+4y=\frac{400}{9}-\frac{80}{9}y+\frac{4}{9}y^2\end{equation}$

$\begin{equation*}0=\frac{13}{9}y^2-\frac{116}{9}y-\frac{140}{9}\end{equation}$

$\begin{equation*}0=13y^2-116y-140\end{equation}$

$\begin{equation*}0=13y^2-130y+14y-140\end{equation}$

$\begin{equation*}0=13y(y-10)+14(y-10)\end{equation}$

$\begin{equation*}0=(y-10)(13y+14)\end{equation}$

(5) $\begin{equation*}\therefore y=10, y=-\frac{14}{13}\end{equation*}$

Substitute $y=10$ into $x=\sqrt{64-(y-2)^2}+8$

$\begin{equation*}x=\sqrt{64-(y-2)^2}+8\end{equation}$

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

Substitute $y=-\frac{14}{13}$ into $x$

$\begin{equation*}x=\sqrt{64-(-\frac{14}{13}-2)^2}+8\end{equation}$

(7) $\begin{equation*}x=\frac{200}{13}\end{equation*}$

Now we need to consider the negative version of $x$ . If you work through (like I did above) you end with the same equation for $y$ .

Hence our two values for $v$ are $v=(8,10)$ or $v=(\frac{200}{13},-\frac{14}{13})$ .

Would someone be expected to do this in an exam? I hope not, but I think its worth doing.

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Filed under Closest Approach, Vectors, Year 11 Specialist Mathematics

Category Archives: Year 11 Specialist Mathematics

Intersecting Secant Theorem

Circle Geometry Question 2

Circle Geometry Question

Derangements

Counting Techniques

Vectors – Closest Approach (2D)

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