HSC 2016 MX1 Marathon (archive) (1 Viewer)

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Speed6

Retired '16
Hi there 2016'ers (including myself)!

Welcome to the 2016 HSC 3U Marathon,

I invite and encourage all former and most importantly current students in particular to participate in this marathon.

Some simple housekeeping rules:

1. Only post questions within the level and scope of 3U HSC Mathematics

2. Provide neat working out when possible

3. Don't flood the thread with multiple unanswered questions

4. Since HSC 3U Mathematics can be tested up to 30% of the maths you did in preliminary, feel free to post the often odd and challenging questions

5. And remember to have fun and enjoy the beauty of mathematics (I don't really know if this is a rule lol)

To fire up, the first question is by calamebe:

$\bg_white By Mathematical Induction show that 9^{n}-8n-1 is divisible by 64 for all integers \textit{n}\geq2.$

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leehuan

Well-Known Member
Re: HSC 2016 3U Marathon

Keep in mind not every school is up to induction yet. Series is the one topic I've found to be taught in all sorts of parts of the year (as early as prelim, late first term HSC or mid third term HSC).

I'll answer it some days later if nobody does.

Blitz_N7

Member
Re: HSC 2016 3U Marathon

$\bg_white A rectangle is inscribed in an isosceles triangle with one of the sides of the rectangle on the base of the triangle. Prove that the rectangle of greatest area occupies half the area of the triangle.$

calamebe

Active Member
Re: HSC 2016 3U Marathon

Keep in mind not every school is up to induction yet. Series is the one topic I've found to be taught in all sorts of parts of the year (as early as prelim, late first term HSC or mid third term HSC).

I'll answer it some days later if nobody does.
Oh really? My school is doing it soon, so I thought it would have been largely the same for most schools. Well that's a bit unlucky.

Speed6

Retired '16
Re: HSC 2016 3U Marathon

I had the opportunity to do MI in Year 11 MX1. Generally, its taught in the HSC year.

davidgoes4wce

Well-Known Member
Re: HSC 2016 3U Marathon

My question is in the solution Samir El Hosri says this

"Angle in the alternate segment with chord AC and tangent AD"

Would either of the responses below suffice?

"Angle between the tangent AD and chord AC equal the angle in the alternate segment"
or
"Angle between the tangent AD and circle C2 and the chord equals the angle in the alternate segment"

InteGrand

Well-Known Member
Re: HSC 2016 3U Marathon

My question is in the solution Samir El Hosri says this

"Angle in the alternate segment with chord AC and tangent AD"

Would either of the responses below suffice?

"Angle between the tangent AD and chord AC equal the angle in the alternate segment"
or
"Angle between the tangent AD and circle C2 and the chord equals the angle in the alternate segment"
You can just say ''alternate segment theorem''.

davidgoes4wce

Well-Known Member
Re: HSC 2016 3U Marathon

OK that seems alot easier then

davidgoes4wce

Well-Known Member
Re: HSC 2016 3U Marathon

But still curious to know if my responses for either of those 2 in relation to that question would still be right?

InteGrand

Well-Known Member
Re: HSC 2016 3U Marathon

But still curious to know if my responses for either of those 2 in relation to that question would still be right?
Out of

'"Angle between the tangent AD and chord AC equal the angle in the alternate segment"
or
"Angle between the tangent AD and circle C2 and the chord equals the angle in the alternate segment"
'

the first one is OK to use, but I would avoid the second one, because angle between a tangent and a circle is ambiguous.

leehuan

Well-Known Member
Re: HSC 2016 3U Marathon

A rectangle is inscribed in an isosceles triangle with one of the sides of the rectangle on the base of the triangle. Prove that the rectangle of greatest area occupies half the area of the triangle.

^Blitz's question done without LaTeX. It seems to be messing up on my computer again.

davidgoes4wce

Well-Known Member
Re: HSC 2016 3U Marathon

Its Monday morning and I can't figure out this. I had a big weekend is my excuse.

Carrotsticks

Retired
Re: HSC 2016 3U Marathon

Its Monday morning and I can't figure out this. I had a big weekend is my excuse.
Connect O to B and observe that triangle OAB is right angled.

Sent from my SM-G928I using Tapatalk

davidgoes4wce

Well-Known Member
Re: HSC 2016 3U Marathon

OK after looking at this question again

Im assuming both tangents have the same angle
so Angle DAB= alpha (isosceles triangle), ANGLE ABO=90 degrees (tangent to a circle the radius is perpendicular to the centre of the circle)

ANGLE BOA= 180-(90+alpha)=90-alpha (sum of the angles of a triangle)

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leehuan

Well-Known Member
Re: HSC 2016 3U Marathon

OK after looking at this question again

Im assuming both tangents have the same angle
so Angle DAB= alpha (isosceles triangle), ANGLE ABO=90 degrees (tangent to a circle the radius is perpendicular to the centre of the circle)

ANGLE BOA= 180-(90+alpha)=90-alpha (sum of the angles of a triangle)
It is quite easy to prove that triangles BOA and COA are congruent using either SSS, SAS or RHS.

davidgoes4wce

Well-Known Member
Re: HSC 2016 3U Marathon

OK I finally managed to get it, angle BOA is twice the angle of BDO. (Angle as the centre twice the angle at the circumference). It took me a while to get there but I finally got it.

Arucerious

New Member
Re: HSC 2016 3U Marathon

I don't understand how to do this.
A form has ten questions in order, each of which requires the answer 'Yes' or 'No'. Find the number of ways the form can be filled in:
(g) if exactly three answers are 'Yes', and they occur together,

integral95

Well-Known Member
Re: HSC 2016 3U Marathon

OK I finally managed to get it, angle BOA is twice the angle of BDO. (Angle as the centre twice the angle at the circumference). It took me a while to get there but I finally got it.
The theorem doesn't work as they are not subtended from a chord (or arc) in the circle

InteGrand

Well-Known Member
Re: HSC 2016 3U Marathon

The theorem doesn't work as they are not subtended from a chord (or arc) in the circle
It does work; to see this, we can call the angles BOE and BDE instead, and then the angles are subtended by arc BE (and are essentially the same angles).

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