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Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

dlau

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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. (*)

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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).
Thanks so much for the help. I will try this.


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eyeseeyou

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

Help

 

InteGrand

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

not very many....
Well quite a few at any rate, e.g. discriminant, complete square, ratio of difference of cubes to difference, single variable calculus, multivariable calculus, use of positive definite matrix etc.
 

appleibeats

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

Screen Shot 2016-08-11 at 5.20.50 PM.png

I got c , but the answers says b. Not sure why?

Any one have any clue??
 

integral95

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

For 12a

The equation of the parabola is in the form

The tangent has the equation

Sub that into the parabola to get a quadratic in terms of x and a, then use the discriminate and let that equal to 0 (since it's a tangent, so there's only one solution) to solve for a. Then sub that value back into the first equation.
 

Memphis Depay

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

Make sure the tangent is effectively portrayed
 

davidgoes4wce

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

For 12a

The equation of the parabola is in the form

The tangent has the equation

Sub that into the parabola to get a quadratic in terms of x and a, then use the discriminate and let that equal to 0 (since it's a tangent, so there's only one solution) to solve for a. Then sub that value back into the first equation.




















I haven't checked the answer but would this be right?
 

davidgoes4wce

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

Not sure if directrix on the diagonal is covered in Maths Extension 1. I did draw a graph of the points and am aware that the formula from the Focal Point to the straight line is:



Im not sure if the parabola is parallel to the x-axis or parallel to the y-axis.

Here was the diagram I tried to work from :

 

jathu123

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

13a


there might be a shorter way
 
Last edited:

davidgoes4wce

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

13a


there might be a shorter way
Think its the first time I have done a question that involved a parabola with a diagonal directrix. (Don't think I have seen too many of those type in past exam papers)

Think I also have to memorise the distance of a point on a parabola to the focus is equivalent to point on a parabola to the directrix. (PS=PM)

When I looked at your final equation, without using a calculator my initial thoughts were it a circle or hyperbola.

But drawing it on desmos it looked like that (not sure how would you describe that graph).

 

davidgoes4wce

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

Q 7 (b) EX 9A

I just wondering has there been a mistake in this answer of the 3U Year 11 textbook?









ii) ==> PA² + PB² + PC² = 3x² + 3y² - 14x - 6y + 78 = 77

==> 3x² + 3y² - 14x - 6y = -1

==> x² + y² - 14x/3 - 2y = -1/3

Grouping, (x² - 14x/3) + (y² - 2y) = -1/3

==> (x² - 14x/3 + 49/9) + (y² - 2y + 1) = -1/3 + 49/9 + 1

==> (x - 7/3)² + (y - 1)² = 55/9








The back of the book has the answer as (2,1) for centre and radius of
 

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