HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)

largarithmic · Jan 7, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
Geometry problems are fun to do, but they are the absolute worst when it comes to typing them down.

I dont think anyone would complain if you just drew a diagram and drew in an angle chase

seanieg89 · Jan 7, 2012

Re: 2012 HSC MX1 Marathon

Will leave the first geometry question for someone else to do, it's fairly straightforward. Refer to my diagram for the second, it is not too difficult to show that there is no diagram dependence in the following working:

The blue lines in the diagram represent our four initial lines.

A,B,C,D,E,F are the points of intersection of these lines.

The circles ABC and ADF meet at a new point X in addition to A. It suffices to show that CEFX and BDFX are cyclic quads. Angle chasing is enough to establish this. (Not going to write down all of the reasons.)

$\angle XCF=\pi-\angle BCX=\angle BAX=\angle DAX=\pi-\angle DEX=\angle XEF\\$ Hence $CEFX$ is cyclic.$\\ \\$Furthermore:\\$\angle BXC=\angle BAC = \angle DAE = \angle DXE\\ \Rightarrow \angle BXD= \angle CXE -\angle DXE + \angle BXC=\angle CXE=\angle CFE=\angle BFD\\$ Hence $BDFX$ is cyclic. $$

largarithmic · Jan 7, 2012

Re: 2012 HSC MX1 Marathon

seanieg89 said:
View attachment 24071

Will leave the first geometry question for someone else to do, it's fairly straightforward. Refer to my diagram for the second, it is not too difficult to show that there is no diagram dependence in the following working:

The blue lines in the diagram represent our four initial lines.

A,B,C,D,E,F are the points of intersection of these lines.

The circles ABC and ADF meet at a new point X in addition to A. It suffices to show that CEFX and BDFX are cyclic quads. Angle chasing is enough to establish this. (Not going to write down all of the reasons.)

$\angle XCF=\pi-\angle BCX=\angle BAX=\angle DAX=\pi-\angle DEX=\angle XEF\\$ Hence $CEFX$ is cyclic.$\\ \\$Furthermore:\\$\angle BXC=\angle BAC = \angle DAE = \angle DXE\\ \Rightarrow \angle BXD= \angle CXE -\angle DXE + \angle BXC=\angle CXE=\angle CFE=\angle BFD\\$ Hence $BDFX$ is cyclic. $$

^^

now try this question, and explain why it directly implies that problem (B) (i.e. show how to solve problem (B) without angle chasing:

Given any points A,B,C,D on the plane with no three collinear, explain how to find the spiral centre that sends AB to CD. I.e. construct the point X such that ABX is similar to CDX with point A corresponding to C, B corresponding to D, X to X, and in that orientation.

king chopper · Jan 9, 2012

Re: 2012 HSC MX1 Marathon

The angles of a certain pentagon are in an arithmetic progression. The largest angle is double the smallest, find in radians, the size of each angle

deswa1 · Jan 10, 2012

Re: 2012 HSC MX1 Marathon

Angle sum of pentagon is 3pi. Let the smallest angle be a. Therefore the angles are:

a,a+d,a+2d,a+3d,a+4d where d is the common difference.
5a+10d=3pi (1)

But a+4d=2a (largest angle twice smallest)
Therefore 4d=a
Sub into (1): 30d=3pi
d=pi/10 a=2pi/5

Therefore the angles are 2pi/5,pi/2,3pi/5,7pi/10,4pi/5

Sanical · Jan 10, 2012

Re: 2012 HSC MX1 Marathon

:O No one did the question

Is it because you can't do it or because it's a pain to write down the solution? If you have done it, at least give some hints lol, I don't know how to do it.

deswa1 · Jan 10, 2012

Re: 2012 HSC MX1 Marathon

To be honest I've got no idea what the question means. What is a horizontal triangle and a vertical hexagon? Is the question asking to prove that angle CPA is tan root3/5 or the other angle (CP to the left of P -> there's no letter there.) If you can explain those, I'll have a go and let you know

Sanical · Jan 10, 2012

Re: 2012 HSC MX1 Marathon

deswa1 said:
To be honest I've got no idea what the question means. What is a horizontal triangle and a vertical hexagon? Is the question asking to prove that angle CPA is tan root3/5 or the other angle (CP to the left of P -> there's no letter there.) If you can explain those, I'll have a go and let you know

No it should be 90 - angle CPA = tan root3/5.

I've been staring at it so much I started to think the left-most triangle is 3D :O.

king chopper · Jan 10, 2012

Re: 2012 HSC MX1 Marathon

OP and OQ are radii length r cm of a circle centred at O. Arc PQ of the circle subtends an angle of theta radians at O and perimeter of sector OPQ is 12cm.
Show that Area A of sector is given by A = 72theta/ (2+theta)^2
Hence find the maximum area of the sector

deterministic · Jan 10, 2012

Re: 2012 HSC MX1 Marathon

Sanical said:
:O No one did the question

Is it because you can't do it or because it's a pain to write down the solution? If you have done it, at least give some hints lol, I don't know how to do it.

This is a 3D trig and PC is above the horizontal plane. Define G as the point the perpendicular from C meets the extended line of AB. The aim of the question is to show that CG/PG=sqrt(3)/5

Let x=AB so AP=2x
Also BC=x and angle CBG=60.
Then use simple trig to find GC and GB in terms of x

To find PG in terms of x you use Pythagoras on AG(=AB+GB) and PA. Rest should follow.

Carrotsticks · Jan 10, 2012

Re: 2012 HSC MX1 Marathon

king chopper said:
OP and OQ are radii length r cm of a circle centred at O. Arc PQ of the circle subtends an angle of pheta radians at O and perimeter of sector OPQ is 12cm.
Show that Area A of sector is given by A = 72pheta / (2+pheta)^2
Hence find the maximum area of the sector

haha did you mean 'theta'?

king chopper · Jan 10, 2012

Re: 2012 HSC MX1 Marathon

Carrotsticks said:
haha did you mean 'theta'?

haha yeh my bad ....

OP and OQ are radii length r cm of a circle centred at O. Arc PQ of the circle subtends an angle of theta radians at O and perimeter of sector OPQ is 12cm.
Show that Area A of sector is given by A = 72theta/ (2+theta)^2
Hence find the maximum area of the sector

nightweaver066 · Jan 16, 2012

Re: 2012 HSC MX1 Marathon

king chopper said:
haha yeh my bad ....

OP and OQ are radii length r cm of a circle centred at O. Arc PQ of the circle subtends an angle of theta radians at O and perimeter of sector OPQ is 12cm.
Show that Area A of sector is given by A = 72theta/ (2+theta)^2
Hence find the maximum area of the sector

$A = \frac{r^2\theta}{2}$

$P = r\theta$

$\therefore 12 - 2r = r\theta$

$r = \frac{12}{\theta + 2}$

$\therefore A = \frac{144\theta}{2(\theta + 2)^2}$

$= \frac{72\theta}{(\theta + 2)^2}$

$\frac{dA}{d\theta} = \frac{72(\theta + 2)^2 - 144\theta(\theta + 2)}{(\theta + 2)^4}$

$$For minimum/maximum area, Let $ \frac{dA}{dx} = 0$

$\therefore 72(\theta + 2)^2 - 144\theta(\theta + 2) = 0$

$(\theta + 2)(\theta + 2 - 2\theta) = 0$

$(\theta + 2)(-\theta + 2) = 0$

$\therefore \theta = \pm 2$

Test theta = 2, it will be a maximum, $$Max Area $ = \frac{144}{16} = 9cm^2$

New question..

such_such · Jan 20, 2012

Re: 2012 HSC MX1 Marathon

Given <a href="http://www.codecogs.com/eqnedit.php?latex=sin18=\frac{1}{4}(\sqrt{5}-1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?sin18=\frac{1}{4}(\sqrt{5}-1)" title="sin18=\frac{1}{4}(\sqrt{5}-1)" /></a>, find cos36 in surd form

nightweaver066 · Jan 20, 2012

Re: 2012 HSC MX1 Marathon

such_such said:
Given <a href="http://www.codecogs.com/eqnedit.php?latex=sin18=\frac{1}{4}(\sqrt{5}-1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?sin18=\frac{1}{4}(\sqrt{5}-1)" title="sin18=\frac{1}{4}(\sqrt{5}-1)" /></a>, find cos36 in surd form

$cos36 = 1 - 2sin^218$

$= 1 - 2(\frac{6 - 2\sqrt{5}}{16})$

$= \frac{1 + \sqrt{5}}{4}$

SpiralFlex · Jan 20, 2012

Re: 2012 HSC MX1 Marathon

nightweaver066 said:
$A = \frac{r^2\theta}{2}$

$P = r\theta$

$\therefore 12 - 2r = r\theta$

$r = \frac{12}{\theta + 2}$

$\therefore A = \frac{144\theta}{2(\theta + 2)^2}$

$= \frac{72\theta}{(\theta + 2)^2}$

$\frac{dA}{d\theta} = \frac{72(\theta + 2)^2 - 144\theta(\theta + 2)}{(\theta + 2)^4}$

$$For minimum/maximum area, Let $ \frac{dA}{dx} = 0$

$\therefore 72(\theta + 2)^2 - 144\theta(\theta + 2) = 0$

$(\theta + 2)(\theta + 2 - 2\theta) = 0$

$(\theta + 2)(-\theta + 2) = 0$

$\therefore \theta = \pm 2$

Test theta = 2, it will be a maximum, $$Max Area $ = \frac{144}{16} = 9cm^2$

New question..

$2\cos x=\frac{1}{2\cos x}$

$(2\cos x-1)(2\cos x+1)=0$

$x=\frac{\pi}{3}, -\frac{\pi}{3}$

Their oordinates are at 1.

$V=2\pi \int_{0}^{\frac{\pi}{3}}(4\cos^2 x - \frac{1}{4}\sec^2 x)\; dx$

$=2\pi [2x+\sin 2x-\frac{1}{4}\tan x]$ (With the limits.)

$\therefore V=....$

-Questions coming up-

zeebobDD · Jan 20, 2012

Re: 2012 HSC MX1 Marathon

Sketch the graph y^2= x[(1-x)^2]

i) Find the area contained within the loop between x=0 and x=1

ii) Find the volume of the solid formed when this region is rotated about the x-axis

yes i know were i got the question from , you dnt have to tell me. not too hard

Carrotsticks · Jan 20, 2012

Re: 2012 HSC MX1 Marathon

nightweaver066 said:
$cos36 = 1 - 2sin^218$

$= 1 - 2(\frac{6 - 2\sqrt{5}}{16})$

$= \frac{1 + \sqrt{5}}{4}$

$\cos 36^{\circ} = \frac{\varphi}{2} $ where $ \varphi = $ limiting ratio of successive terms in the Fibbonaci Sequence $ \lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n} $ where $ F_n $ denotes the $ n^{th} $ term of the series.$$

SpiralFlex · Jan 20, 2012

Re: 2012 HSC MX1 Marathon

zeebobDD said:
Sketch the graph y^2= x[(1-x)^2]

i) Find the area contained within the loop between x=0 and x=1

ii) Find the volume of the solid formed when this region is rotated about the x-axis

yes i know were i got the question from , you dnt have to tell me. not too hard

*Cough* *Cough* Terry Lee. Also this is a 3U thread haha.

Carrotsticks · Jan 20, 2012

Re: 2012 HSC MX1 Marathon

SpiralFlex said:
*Cough* *Cough* Terry Lee. Also this is a 3U thread haha.

The function $y= \sqrt {x} (1-x)$ is integrable using methods taught in Extension 1.

Perhaps the sketching part is inappropriate for MX1. Everything else seems okay.

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