Author Topic: Help with 2015 Independent Math Trial Paper (again) (Read 4935 times) Tweet Share

joey9911 · « **on:** July 23, 2016, 04:13:15 pm »

I need help with part ii. I've already done part i. Been stuck on this one for a while now

RuiAce · « **Reply #1 on:** July 23, 2016, 04:46:27 pm »

Quote from: joey9911 on July 23, 2016, 04:13:15 pm

I need help with part ii. I've already done part i. Been stuck on this one for a while now

Hahaha I remember this question. All I'm worried about is the final answer being wrong because I hiccuped.
$\text{Needed: Opp. sides on a rectangle are equal, quadratics}$
$\text{Taking proportional sides on the similar triangles (any pair, just has to have DE):}\\ \begin{align*}\frac{AD}{EC}&=\frac{ED}{BC}\\ \frac{3}{EC}&=\frac{ED}{3}\\ ED&=\frac{9}{DC-ED}\\ ED&=\frac{9}{8-ED}\\ 8DE-DE^2&=9 \\ DE^2-8DE+9&=0 \end{align*}$
$\text{Use the quadratic formula or completing the square etc. to solve:}\\ DE=4\pm \sqrt{7}$
$\text{Logically taking the plus case, }DE=4+\sqrt{7}\\ \text{(Note: I reckon }4-\sqrt{7}\text{ is just the length of }EC)$
$\text{Edit: They also said }DE > CE$

Ruqiaab1997 · « **Reply #2 on:** July 23, 2016, 08:36:23 pm »

I really need help with ancient history

joey9911 · « **Reply #3 on:** July 23, 2016, 10:43:36 pm »

Thank you so much again RuiAce!

One last question!! (This is the last question in the paper!)

I need help with the whole thing

Refer to attachment.

Thanks again!!!

jamonwindeyer · « **Reply #4 on:** July 24, 2016, 12:26:27 am »

Quote from: Ruqiaab1997 on July 23, 2016, 08:36:23 pm

I really need help with ancient history

Hey there, welcome to the forums!! You can go get some help in our Ancient History Threads

or check out some free downloadable notes!!

jamonwindeyer · « **Reply #5 on:** July 24, 2016, 01:49:08 am »

Quote from: RuiAce on July 24, 2016, 01:25:19 am

This one was really really dodgy. I promise if nobody gets an answer in then I will delete this post and replace it with an answer tomorrow.

In the interests of honesty I won't have time to sit down and tackle this properly until Sunday evening

As some guidance (not a proper written solution) for the poster, the first question is using the area of a triangle formula:

$A = \frac{1}{2}ab\sin{C}$

Both a and b are your radii, and notice this as well (by double angle formula for sine, this is an extension paper right?):

$\frac{1}{2}\sin{C}=\sin{\frac{C}{2}}\cos{\frac{C}{2}}$

So, to match the result in the question, you just need to prove that the angle in the middle is:

$C = \frac{2\pi}{n}$

This comes directly from the properties of a regular polygon. Think about the triangular slices of a hexagon, for example. There are six of them, so each angle at the centre is 360/6=60 degrees. For n sides, it is 360/n degrees at the centre. Apply this principle to radians and your answer comes out!

To find the similar expression, you'll use a similar idea, but there will be an extra trick to it. Can't spot it right this second (2am might be a bit late to be doing Math

) but Rui may be able to enlighten. The last part will simply be dividing the two answers and maybe (probably) using a Pythagorean identity or something similar (must be something to it for 2 marks)

This might get you started!!

joey9911 · « **Reply #6 on:** July 24, 2016, 09:23:21 am »

Thanks Jamon for the help! I still don't know how to do it, so I might wait till RuiAce comes up with an answer (If he is willing to!)

And this was the last question in the 2unit 2015 Mathematics Independent Trial Paper, not Extension 1. All of my friends said this last question was fit for Extension 1 students only, (they need help with it too), so I don't know why a question this hard is in a 2u paper :\

RuiAce · « **Reply #7 on:** July 24, 2016, 09:52:41 am »

No no no this can definitely be done by 2U methods only.
$\text{We know that the lengths }OA_1=OA_2=r\\ \text{The reason why }\angle A_1ON=\angle A_2ON=\frac{\pi}{n}\text{ is as Jamon said.}$
$A_1,A_2,\dots,A_n\text{ describe a REGULAR polygon with }n\text{ sides.}\\ \text{For a REGULAR polygon, each side subtends angles that are EQUAL, at the CENTRE of the polygon.}\\ \text{This means }\angle A_1OA_2 = \angle A_2OA_3 = \dots = \angle A_nOA_1 = \frac{2\pi}{n}$
$\triangle A_1OA_2\text{ defines an isosceles triangle with }A_1=A_2\\ \text{Hence, the altitude }ON\text{ bisects }\angle A_1OA_2\\ \therefore A_1ON=\angle A_2ON = \frac{\pi}{n}$
$\triangle A_1ON\text{ defines a right-angled triangle with }ON \perp A_1N\\ \text{So we can use }A=\frac{1}{2}bh\text{ to find the area of this triangle.}\\ \text{What is the base and height?}$
$\text{Using trigonometry:}\\ \sin \frac{\pi}{n}=\frac{A_1n}{r} \iff A_1n=r\sin \frac{\pi}{n}\\ \cos \frac{\pi}{n}=\frac{ON}{r} \iff ON = r\cos \frac{\pi}{n}$
$\text{So the area of }A_1ON\text{ is }\frac{1}{2}\left(r\sin \frac{\pi}{n}\right)\left(r\cos \frac{\pi}{n}\right)=\frac{1}{2}r^2\sin \frac{\pi}{n}\cos \frac{\pi}{n}$
$\textbf{However, }\text{it can be shown that the area of }A_2ON\text{ is ALSO }\frac{1}{2}r^2\sin \frac{\pi}{n}\cos \frac{\pi}{n}\\ \text{Adding the two up, the area of }\triangle A_1OA_2\text{ is }\\A_{\triangle A_1OA_2}=r^2\sin \frac{\pi}{n}\cos \frac{\pi}{n}$
$\text{For }\triangle B_1OM\text{, we now have }OM=r\text{ and }\angle B_1OM=\frac{\pi}{n}\\ \text{We still need the length of }BM\text{. Using trigonometry:}\\ \tan \frac{\pi}{n}=\frac{BM}{r} \iff BM=r\tan\frac{\pi}{n}=\frac{\sin\frac{\pi}{n}}{\cos \frac{\pi}{n}}\\ \text{Note - I used the ratio identity tan = sin/cos}$
$\text{So the area of }\triangle B_1OM\text{ is therefore }A=\frac{1}{2}r^2\left(\frac{\sin\frac{\pi}{n}}{\cos \frac{\pi}{n}}\right)\\ \text{Using a similar proof to find }\triangle B_2OM\text{ has the same area we get our desired area:}\\ A_{\triangle B_1OB_2}=\frac{r^2\sin\frac{\pi}{n}}{\cos \frac{\pi}{n}}$
_____________________________________________
$\text{The second part is surprisingly easier than the first.}\\ \text{Now, because we have a regular polygon, the ENTIRE polygon bounded by }A_1,A_2,\dots,A_n\text{ will have }n\textbf{ identical copies }\text{of this triangle.}\\ \therefore A_{A_i}=nr^2\sin \frac{\pi}{n}\cos \frac{\pi}{n}\\ \text{The polygon bounded by }B_1,B_2,\dots,B_n\text{ will also have }n\text{ identical copies of the OTHER triangle.}\\ \therefore A_{B_i}=\frac{nr^2\sin\frac{\pi}{n}}{\cos \frac{\pi}{n}}$
$\text{So now we can just compare the ratios.}\\ \begin{align*}\frac{\textit{Area of outside polygon}}{\textit{Area of inside polygon}}&=\frac{\frac{nr^2\sin\frac{\pi}{n}}{\cos \frac{\pi}{n}}}{nr^2\sin \frac{\pi}{n}\cos \frac{2\pi}{n}}\\ &=\frac{1}{\cos^2 \frac{\pi}{n}}\\&=\sec^2 \frac{\pi}{n}\end{align*}$

joey9911 · « **Reply #8 on:** July 24, 2016, 11:06:36 am »

Wow! Thank you so much!

I just have a one question from your amazing solution.

For the area of the second triangle, A2-O-N, why are the angles 2pi/n? If it's the same area, same angle, shouldn't it be pi/n?

Thanks again!

RuiAce · « **Reply #9 on:** July 24, 2016, 11:13:56 am »

Quote from: joey9911 on July 24, 2016, 11:06:36 am

Wow! Thank you so much!

I just have a one question from your amazing solution.

For the area of the second triangle, A2-O-N, why are the angles 2pi/n? If it's the same area, same angle, shouldn't it be pi/n?

Thanks again!

After I finished my solution, I realised that I made heaps of typos. Tried getting rid of all of them but it seems I missed that one! I'll edit it now

joey9911 · « **Reply #10 on:** July 24, 2016, 11:17:08 am »

All good! Thank you so much RuiAce & Jamon! Going to recommend ATARNotes and its forums to all my friends! You guys have helped me out heaps!

Thanks again!!!

jamonwindeyer · « **Reply #11 on:** July 24, 2016, 11:45:48 am »

Quote from: joey9911 on July 24, 2016, 11:17:08 am

All good! Thank you so much RuiAce & Jamon! Going to recommend ATARNotes and its forums to all my friends! You guys have helped me out heaps!

Thanks again!!!

Thanks Joey, you are awesome!! Haha sorry I jumped to conclusions, this would definitely be pushing the 2U students, very tough without Extension methods!

Rui as usual smashes out the solution anyway, but yep, a very challenging question

olivercutbill · « **Reply #12 on:** July 24, 2016, 09:51:33 pm »

Quote from: joey9911 on July 23, 2016, 04:13:15 pm

I need help with part ii. I've already done part i. Been stuck on this one for a while now

How was part i done here?

RuiAce · « **Reply #13 on:** July 24, 2016, 10:09:19 pm »

Quote from: olivercutbill on July 24, 2016, 09:51:33 pm

How was part i done here?

$\text{By working through a ton of angles}$
$\begin{align*}\text{Let }\angle AED &= x\\ \text{Then }\angle DAE &= 90^\circ - x \text{ (angle sum of }\triangle AED)\\ \text{Then }\angle BAE&=x\, \, (\angle BAD\text{ is a right angle as it lies in a rectangle)}\\ \text{Then }\angle ABE &= 90^\circ -x \text{ (angle sum of }\triangle ABE)\\ \text{Then }\angle EBC&=x\, \, (\angle ABC\text{is a right angle as it lies in a rectangle)}\\ \text{Then }\angle BEC&=90^\circ - x\text{ (angle sum of }\triangle CEB)\end{align*}$
$\begin{align*}\text{In }\triangle AED&\text{ and} \triangle EBC\\ \angle AED&=\angle EBC=x\quad \text{(from above)}\\ \angle DAE&=\angle CEB = 90^\circ - x\, \text{(from above)}\\ \therefore \triangle AED & ||| \triangle EBC \quad \text{ (equiangular)}\end{align*}$

olivercutbill · « **Reply #14 on:** July 24, 2016, 10:11:34 pm »

Quote from: RuiAce on July 24, 2016, 10:09:19 pm

$\text{By working through a ton of angles}$
$\begin{align*}\text{Let }\angle AED &= x\\ \text{Then }\angle DAE &= 90^\circ - x \text{ (angle sum of }\triangle AED)\\ \text{Then }\angle BAE&=x\, \, (\angle BAD\text{ is a right angle as it lies in a rectangle)}\\ \text{Then }\angle ABE &= 90^\circ -x \text{ (angle sum of }\triangle ABE)\\ \text{Then }\angle EBC&=x\, \, (\angle ABC\text{is a right angle as it lies in a rectangle)}\\ \text{Then }\angle BEC&=90^\circ - x\text{ (angle sum of }\triangle CEB)\end{align*}$
$\begin{align*}\text{In }\triangle AED&\text{ and} \triangle EBC\\ \angle AED&=\angle EBC=x\quad \text{(from above)}\\ \angle DAE&=\angle CEB = 90^\circ - x\, \text{(from above)}\\ \therefore \triangle AED & ||| \triangle EBC \quad \text{ (equiangular)}\end{align*}$

Ahhhh I had thought this was probably the case.

Your a legend Rui!

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Author Topic: Help with 2015 Independent Math Trial Paper (again) (Read 4935 times) Tweet Share

joey9911

Help with 2015 Independent Math Trial Paper (again)

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Ruqiaab1997

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