Cambridge Prelim MX1 Textbook Marathon/Q&A (5 Viewers)

leehuan · Jul 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

AfroNerd said:
Okay tyvm I understand now. I was originally looking in the year 11 3u Cambridge book for permutations and combinations but couldn't find it.

It's in the Yr 12 Cambridge textbook

HeroicPandas · Jul 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Linh_N said:
I keep getting 5/26 for part (C). Can anyone explain how to work out the question?

Did all for fun... but still a bit rusty so it most likely has at least one error
____
We are only concerned on placement of Es so the sample space is $\binom{13}{3}$ .

a) Group 3 Es as a blob. There are $11$ ways you can place this blob so the probability is $\frac{11}{\binom{13}{3}} = \frac{1}{26}$ .

b) Group 2 Es as a blob. Suppose this blob (*) is at one end: * _ _ _ _ _ _ _ _ _ _ _. Then there are $10$ ways you can place the last E: * _ + + + + + + + + + +, where the + represents a possible location for the remaining E. Similarly, there are $10$ ways you can place the last E given that the blob is on the other end (symmetry).

Total # of ways if blob is at either end: $2 * 10$ .

Suppose the blob is not at either ends. For example, _ * _ _ _ _ _ _ _ _ _ _. Then there are $9$ ways you can place the last
E: _ * _ + + + + + + + + +.
It is clear that you can place this blob in $10$ ways given that it's not at either ends.

Total # of ways if blob is not at either end: $10 * 9$ .

Probability is $\frac{2*10 + 10 * 9}{\binom{13}{3}} = \frac{5}{13}$ .

c) P(all Es apart) = 1 - P(all Es together) = 1 - [P(3 Es together) + P(2 Es together and one apart)] = $1- \left( \frac{1}{26} + \frac{5}{13}\right) = \frac{15}{26}$ .

d) We have the following situation: E _ _ _ _ _ _ _ _ _ _ _ E. This initial position occurs in $\binom{3}{2}$ combinations. Then the remaining E can be placed in $11$ ways.

Probability is $\frac{C(3,2)* 11}{\binom{13}{3}} = \frac{3}{26}$ .

leehuan · Jul 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

HeroicPandas said:
Did all for fun... but still a bit rusty so it most likely has at least one error
____
We are only concerned on placement of Es so the sample space is $\binom{13}{3}$ .

a) Group 3 Es as a blob. There are $11$ ways you can place this blob so the probability is $\frac{11}{\binom{13}{3}}$ .

b) Group 2 Es as a blob. Suppose this blob (*) is at one end: * _ _ _ _ _ _ _ _ _ _ _. Then there are $10$ ways you can place the last E: * _ + + + + + + + + + +, where the + represents a possible location for the remaining E. Similarly, there are $10$ ways you can place the last E given that the blob is on the other end (symmetry).

Total # of ways if blob is at either end: $2 * 10$ .

Suppose the blob is not at either ends. For example, _ * _ _ _ _ _ _ _ _ _ _. Then there are $9$ ways you can place the last
E: _ * _ + + + + + + + + +.
It is clear that you can place this blob in $10$ ways given that it's not at either ends.

Total # of ways if blob is not at either end: $10 * 9$ .

Probability is $\frac{2*10 + 10 * 9}{\binom{13}{3}}$ .

c) P(all Es apart) = 1 - P(all Es together) = 1 - [P(3 Es together) + P(2 Es together and one apart)] = $1- \frac{11}{\binom{13}{3}} - \frac{2*10 + 10*9}{\binom{13}{3}}$ .

d) We have the following situation: E _ _ _ _ _ _ _ _ _ _ _ E. This initial position occurs in $\binom{3}{2}$ combinations. Then the remaining E can be placed in $11$ ways.

Probability is $\frac{C(3,2)* 11}{\binom{13}{3}}$

Did you cater for the fact the T's are not distinct? We have to assume the T's are the same.

I'm finding a sample space of 13!/(3!3!) \neq 13!(3!10!)

HeroicPandas · Jul 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
Did you cater for the fact the T's are not distinct? We have to assume the T's are the same.

I'm finding a sample space of 13!/(3!3!) \neq 13!(3!10!)

I'm just looking at the possible ways you can place the Es (or combinations) such that the conditions are satisfied. The placement of the rest of the letters can be ignored since it's all the same for each part of the question

leehuan · Jul 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

HeroicPandas said:
I'm just looking at the possible ways you can place the Es (or combinations) such that the conditions are satisfied. The placement of the rest of the letters can be ignored since it's all the same for each part of the question

I guess normally in Ext 1 they just use Pr(Event) = Favourable Outcomes/Total Outcomes to make it easier, because the shortcuts can cause confusion.

si2136 · Jul 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

How frequent is Newton's Method of Approx. in the HSC? Like once? Thanks.

leehuan · Jul 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

si2136 said:
How frequent is Newton's Method of Approx. in the HSC? Like once? Thanks.

Yes.

They can do the very weird thing of asking you for two applications for it but there will never be more than one question on it.....it's just applying a formula.

porcupinetree · Jul 10, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

si2136 said:
How frequent is Newton's Method of Approx. in the HSC? Like once? Thanks.

leehuan said:
Yes.

They can do the very weird thing of asking you for two applications for it but there will never be more than one question on it.....it's just applying a formula.

Yeah, usually they don't like to waste too many marks on these sort of questions.

dlau · Jul 18, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Help with Question 14 b and c please.

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InteGrand · Jul 18, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

dlau said:
Help with Question 14 b and c please.

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$\noindent (b) Without loss of generality, let the circles have radius $2$. Note that the edges $AO,AP,BO,BP,PO$ all have length $2$ as each is a radius of $\mathcal{C}$ or $\mathcal{D}$. So $\triangle AOP$ is equilateral. So $\angle AOM = 60^\circ$ (angle in equilateral triangle). From a previous question, we know that $\triangle AOM$ is right-angled with $\angle M$ a right angle. So $\triangle AOM$ is a ``30-60-90'' triangle, so $AM=\sqrt{3}$ as $AM = AO \sin 60^\circ$ and $AO=2$ (radius). So $AB = 2AM = 2\sqrt{3}$, as $M$ bisects $AB$, from a previous question. Since $OP=2$ (radius), we have $\frac{AB}{OP}=\frac{2\sqrt{3}}{2}=\sqrt{3}$.$

$\noindent (c) Consider the \emph{segment} $AOB$, belonging to circle $\mathcal{D}$. Let its area be $x$. Since $\angle APB = 120^\circ = \frac{2\pi}{3}$ (as $\angle APM = 60^\circ = \angle BPM$, and angle sum of adjacent angles; in fact proved in part (a) I realise now), we have (using the formula for the segment area) $x = \frac{1}{2}\times 2^2 \left( \frac{2\pi}{3} - \sin \frac{2\pi}{3}\right)= 2 \left( \frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)$. Thus the total overlapping area is $2x$. The area of the circle $\mathcal{C}$ is $\pi \times 2^2 = 4\pi$. So the desired ratio is$

$\begin{align*}\frac{2x}{4\pi} &= \frac{x}{2\pi} \\ &= \frac{2 \left( \frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)}{2\pi} \\ &= \frac{ \frac{2\pi}{3} - \frac{\sqrt{3}}{2}}{\pi} \end{align*}$

dlau · Jul 23, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Help with q5 d please.

leehuan · Jul 23, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$\begin{align*}\sqrt{x+5}+\sqrt{x-2}&=\sqrt{5x-6}\\ (x+5)+2\sqrt{(x+5)(x-2)}+(x-2)&=5x-6\\ 2\sqrt{x^2+3x-10}&=3x-9\\ 4(x^2+3x-10)&=3(x^2-6x+9)\end{align*}$
That's the hard part. I'll let you continue from there.

dlau · Jul 23, 2016

Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

leehuan said:
$\begin{align*}\sqrt{x+5}+\sqrt{x-2}&=\sqrt{5x-6}\\ (x+5)+2\sqrt{(x+5)(x-2)}+(x-2)&=5x-6\\ 2\sqrt{x^2+3x-10}&=3x-9\\ 4(x^2+3x-10)&=3(x^2-6x+9)\end{align*}$
That's the hard part. I'll let you continue from there.

When I try solving for X I don't get a rational number ( it should be 11)

InteGrand · Jul 24, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

dlau said:
When I try solving for X I don't get a rational number ( it should be 11)

There was a typo in the last line of leehuan's work. He meant the RHS there to have a 9 outside the bracket instead of 3 (because he squared 3(x – 3) from the previous line).

Paradoxica · Jul 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

dlau said:
Assistance with question 28 please. From exercise 8F

Consider the circle on the Cartesian plane

x² + y² = r²

In the negative y quadrants, construct a chord parallel to the x-axis using the points P(rcosθ,-rsinθ) and P'(-rcosθ,-rsinθ), which has length 2rcosθ (Take θ to be in the first quadrant for the signs to make sense). This eliminates one degree of freedom, from which there are three

The point A(rcosα,rsinα) and the point A'(-rcosα,rsinα) generate generate identically equivalent triangles when joined with PP'.

The area of said triangle is 1⁄2(2rcosθ)|rsinα+rsinθ| = r²cosθ|sinα+sinθ|

For the time being, consider θ to be fixed. The area of the triangle then varies from r²cosθ(1-sinθ) to 0 when A is below the chord and from 0 to r²cosθ(1+sinθ), which is when A is above the chord.

The longer the distance from A to PP', the bigger the area of the triangle.
So, clearly, the maximum area is when A is the point (0,r), which gives the area as r²cosθ(1+sinθ)

Now that two degrees of freedom have been eliminated, the problem is now a single variable maximisation problem.

The chord (and it's angle) is the final variable to analyse.

This is impossible (I think) to transform into a quadratic function, so the only way to maximise this is to consider it as a trigonometric function, which should be fairly trivial for you to accomplish.

dlau · Jul 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Paradoxica said:
Consider the circle on the Cartesian plane

x² + y² = r²

In the negative y quadrants, construct a chord parallel to the x-axis using the points P(rcosθ,-rsinθ) and P'(-rcosθ,-rsinθ), which has length 2rcosθ (Take θ to be in the first quadrant for the signs to make sense). This eliminates one degree of freedom, from which there are

The point A(rcosα,rsinα) and the point A'(-rcosα,rsinα) generate generate identically equivalent triangles when joined with PP'.

The area of said triangle is 1⁄2(2rcosθ)|rsinα+rsinθ| = r²cosθ|sinα+sinθ|

For the time being, consider θ to be fixed. The area of the triangle then varies from r²cosθ(1-sinθ) to 0 when A is below the chord and from 0 to r²cosθ(1+sinθ), which is when A is above the chord.

The longer the distance from A to PP', the bigger the area of the triangle.
So, clearly, the maximum area is when A is the point (0,r), which gives the area as r²cosθ(1+sinθ)

Now that two degrees of freedom have been eliminated, the problem is now a single variable maximisation problem.

The chord (and it's angle) is the final variable to analyse.

This is impossible (I think) to transform into a quadratic function, so the only way to maximise this is to consider it as a trigonometric function, which should be fairly trivial for you to accomplish.

Thanks for this! Hmm haven't actually learnt trigonometric functions yet, but will have a go.

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Paradoxica · Jul 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

dlau said:
Thanks for this! Hmm haven't actually learnt trigonometric functions yet, but will have a go.

Oh. Hm. Well in that case, I only have two other proofs which bypass calculus completely. Whatever you want, I can try.

dlau · Jul 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Paradoxica said:
Oh. Hm. Well in that case, I only have two other proofs which bypass calculus completely. Whatever you want, I can try.

What are the two other options which bypass calculus? I have only learnt to do first derivative…

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Paradoxica · Jul 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

dlau said:
What are the two other options which bypass calculus? I have only learnt to do first derivative…

They employ clever tricks (Both use the Arithmetic Mean Geometric Mean Inequality, and one uses a relatively unknown fact about circumscribed triangles)

I am willing to pass these on in private message, if you want.

Anyway, continuing from the expression for the locally optimal area:

It suffices to maximise the trigonometric expression:

A = cosθ(1+sinθ) ≡ cosθ + cosθsinθ

Take the derivative with respect to θ

dA/dθ = -sinθ - sin²θ + cos²θ = -(2sin²θ + sinθ - 1) = (1-2sinθ)(1+sinθ)

Since θ was taken to be in the first quadrant, the only zero of the function derivative is θ = π/6.

The derivative decreases from 1 to 0 at π/6 and then enters the negative for the remaining values of θ.

So the function increases and maximises at π/6, then decreases for the rest of θ.

So this value is indeed the maximum.

Then from this, the length of the chord is r√3, and the other two side lengths are easily shown to be the same.

The conclusion follows.

InteGrand · Jul 31, 2016

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Given any inscribed triangle T in the circle that has at least one vertex V not "diametrically" opposite its opposite side (which is a chord of the circle), the area of T can be increased by moving the vertex V to this point on the circle that is "diametrically" opposite this chord. This is because given a side S of the triangle, the area is proportional to the perpendicular distance of the opposite vertex V from this side (using A = (1/2)bh), and from the geometry of a circle this perpendicular distance h is clearly maximal when V is "diametrically" opposite this side S.

Thus for the maximal-area inscribed triangle T, it must be one where each vertex is "diametrically" opposite its opposite side. In other words, all the altitudes are also the sides' medians, which implies this is an equilateral triangle. (*)

---------

By a vertex V being "diametrically" opposite its opposite side S, I mean the line connecting the midpoint M of S and the vertex V passes through the circle's centre. By HSC circle geometry theorems, this is equivalent to saying MV is the perpendicular bisector of S (use the theorem "line through centre to midpoint of chord is its perpendicular bisector" and the converse of this). So MV is both a median and an altitude.

(*) To show a triangle where the medians are also the altitudes is equilateral, you only really need to show the following result: "Let T be a triangle where one of its medians is also the altitude. Then T is isosceles." Then apply this result twice to show (*).

To show this isosceles result, you can do it by proving certain sub-triangles are congruent, using SAS (draw a diagram and it should be clear).

Cambridge Prelim MX1 Textbook Marathon/Q&A (5 Viewers)

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