HSC 2012 MX1 Marathon #1 (archive) (4 Viewers)

[Trebla]

Re: 2012 HSC MX1 Marathon

Suppose that there are two angles subtended from a common interval AB (say angles ACB and ADB). If angle ACB is twice the size of angle ADB and AC = BC show that the points A, B and D form a circle with C as the centre.

Woops. Corrected it.

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

Suppose that there are two angles subtended from a common interval AB (say angles ACB and ADB). If angle ACB is twice the size of angle ADB and AC = BC show that the points A, B and D form a circle with C as the centre.

Not entirely happy with the way I've worded this, but it's late and I'm tired.

EDIT: Just realised something - That's not always the case:


Feel free to correct me if I'm wrong, which I might well be.

 

[mnmaa]

Re: 2012 HSC MX1 Marathon

good heavens....i got 95 in the ext1 hsc - and now 3months of not touching a single page of maths - i've forgotten over 80% of the course.

G fucking G

Your not alone...

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ \textrm{Construct a circle passing through the points A, B and D, and let the center of the circle be O.}\\ \angle AOB=2\angle ADB~\textrm{(Angle at centre is twice the angle at the arc)}\\ ~\\ \textrm{But,}~ \angle ACB=2\angle ADB\\ \therefore \angle ACB=\angle AOB\\ ~\\ \textrm{Now,}~AC=BC~\textrm{(given) and}~AO=BO~\textrm{(radius of circle ABD)}\\ ~\\ \textrm{Hence, O and C are the same point - i.e. C is the centre of circle ABD}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ \textrm{Construct a circle passing through the points A, B and D, and let the center of the circle be O.}\\ \angle AOB=2\angle ADB~\textrm{(Angle at centre is twice the angle at the arc)}\\ ~\\ \textrm{But,}~ \angle ACB=2\angle ADB\\ \therefore \angle ACB=\angle AOB\\ ~\\ \textrm{Now,}~AC=BC~\textrm{(given) and}~AO=BO~\textrm{(radius of circle ABD)}\\ ~\\ \textrm{Hence, O and C are the same point - i.e. C is the centre of circle ABD}" title="\\ \textrm{Construct a circle passing through the points A, B and D, and let the center of the circle be O.}\\ \angle AOB=2\angle ADB~\textrm{(Angle at centre is twice the angle at the arc)}\\ ~\\ \textrm{But,}~ \angle ACB=2\angle ADB\\ \therefore \angle ACB=\angle AOB\\ ~\\ \textrm{Now,}~AC=BC~\textrm{(given) and}~AO=BO~\textrm{(radius of circle ABD)}\\ ~\\ \textrm{Hence, O and C are the same point - i.e. C is the centre of circle ABD}" /></a>

Not entirely happy with the way I've worded this, but it's late and I'm tired.

It seems like you used the proof from the question to prove itself.

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

It seems like you used the proof from the question to prove itself.

I can't see how... can you please elaborate?

EDIT: I'm going to bed, I'll check back here when I get home tomorrow.

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

I can't see how... can you please elaborate?

An immediate result from the original question is that the angle subtended from the centre is twice the angle at the arc.

However, you used this theorem to answer the question.

 

[Trebla]

Re: 2012 HSC MX1 Marathon

Not entirely happy with the way I've worded this, but it's late and I'm tired.

EDIT: Just realised something - That's not always the case:
View attachment 24267

Feel free to correct me if I'm wrong, which I might well be.

In the case where C and D are on opposite sides of AB, the result is true when you compare the reflex angle ACB to ADB.

What the question is essentially asking is to prove the 'converse' of the well known circle geometry theorem that if C is the centre of the circle passing through by A, B and D then angle ACB is twice of angle ADB (the proof of which is far simpler).

i.e. Prove that if angle ACB is twice of angle ADB and AC = BC (which is a necessary condition) then C is the centre of the circle passing through A, B and D. Assume that C and D are on the same side of AB.

 

[largarithmic]

Re: 2012 HSC MX1 Marathon

It seems like you used the proof from the question to prove itself.

You can do this sort of thing and make it be perfectly valid - its called "reverse reconstruction". You need to be a bit careful though about the uniqueness of points and such (nooblet94's proof isnt actually right because of things lying on different sides of lines).

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

An immediate result from the original question is that the angle subtended from the centre is twice the angle at the arc.

However, you used this theorem to answer the question.

But that theorem is easily proved without assuming the result of the question is true. I would've proved it in my answer but (as far as I'm aware) that result is assumed knowledge for the HSC Course.

In the case where C and D are on opposite sides of AB, the result is true when you compare the reflex angle ACB to ADB.

What the question is essentially asking is to prove the 'converse' of the well known circle geometry theorem that if C is the centre of the circle passing through by A, B and D then angle ACB is twice of angle ADB (the proof of which is far simpler).

i.e. Prove that if angle ACB is twice of angle ADB and AC = BC (which is a necessary condition) then C is the centre of the circle passing through A, B and D.

Not entirely sure what you mean by that. As far as I can see, the point C in my diagram satisfies the conditions of the question -angle ACB = 2angle ADB and is equidistant from A and B - but isn't the centre of the circle, so the result isn't always true.

 

[Trebla]

Re: 2012 HSC MX1 Marathon

The counterexample seems to be a special case of where it is a right angle where one can relabel the external point as C. I don't think that scenario would work otherwise.

What I meant to say was if C and D are on opposite sides of AB then take the reflex angle ACB as twice of angle ADB for the result.

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Here's a bit of an interesting problem.

Consider a sphere of radius R placed inside a cube such that the sphere is tangential to all sides of the cube. However in the gaps at the corners, there are to be smaller spheres of radius r, such that each of the smaller spheres remain in contact with 3 sides of the cube, and the larger sphere.

Show that the largest possible corner-sphere has radius:

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

Here's a bit of an interesting problem.

Consider a sphere of radius R placed inside a cube such that the sphere is tangential to all sides of the cube. However in the gaps at the corners, there are to be smaller spheres of radius r, such that each of the smaller spheres remain in contact with 3 sides of the cube, and the larger sphere.

Show that the largest possible corner-sphere has radius:

I've got it. Give me about an hour and a half to write it up all nice and neat (I've got an ortho appointment in about 10 minutes and my current working out is hardly legible). Thoroughly enjoyed that problem

EDIT: My scanner has decided it doesn't want to work and I can't be bothered typing up the whole thing in, so here's a brief summary:


Basically, you find two expressions for the length between the point of contact of the two spheres and the vertex of the cube. The first expression involves finding the length of the diagonal of the cube, subtracting the diameter and dividing that in two. You end up with a value of .

The second expression involves constructing a cube with vertices at the vertex of the larger cube, the 3 points of contact and the centre of the smaller sphere and finding the length of its diagonal. The distance from the centre of the smaller sphere to the point of contact of the two spheres is r. Adding the two together you get a value of .

Upon equating the two expressions you end up with

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

A slightly different approach to what I did, but very nice. I quite like these problems myself.

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

Can you describe your approach? I'm curious

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Just finished that X-Men Origins movie. Pretty good.

Anyway here you go.

 

[deswa1]

Re: 2012 HSC MX1 Marathon

I'm not sure if this is strictly 3U but here goes:

It is known that if f(0)=0 and f'(0) is greater than or equal to zero for x>0, then f(x) is greater than or equal to zero for x>0.
Use this to show that sinx-1+x^3/6 is greater than or equal to 0 for x>0

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

I'm not sure if this is strictly 3U but here goes:

It is known that if f(0)=0 and f'(0) is greater than or equal to zero for x>0, then f(x) is greater than or equal to zero for x>0.
Use this to show that sinx-1+x^3/6 is greater than or equal to 0 for x>0

I think you've written that wrong.

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

I think you've written that wrong.

I think he meant:

 

[deswa1]

Re: 2012 HSC MX1 Marathon

Woops, it was sinx -x + x^3/6. It's a past 4U HSC question I think (94) but some 3U students might be able to do it

 

[Nooblet94]

Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ f(x)=\sin x -x@plus; \frac{x^3}{6}\\ f'(x)=\cos x -1@plus;\frac{x^2}{2}\\ f''(x)=-\sin x @plus;x\\ f'''(x)=-\cos x @plus;1\\ ~\\ f(0)=\sin 0 -0@plus; \frac{0^3}{6}=0\\ f'(0)=\cos 0 -1@plus;\frac{0^2}{2}=0\\ f''(0)=-\sin 0 @plus;0=0\\ ~\\ \textrm{Now,}~f'''(x)\geq 0~\textrm{for}~x>0~\textrm{(Since the maximum value of}~\cos x~\textrm{is 1)}\\ ~\\ \textrm{Hence, by the given theorem}\\ f''(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f'(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f(x)\geq 0~\textrm{for}~x>0\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ f(x)=\sin x -x+ \frac{x^3}{6}\\ f'(x)=\cos x -1+\frac{x^2}{2}\\ f''(x)=-\sin x +x\\ f'''(x)=-\cos x +1\\ ~\\ f(0)=\sin 0 -0+ \frac{0^3}{6}=0\\ f'(0)=\cos 0 -1+\frac{0^2}{2}=0\\ f''(0)=-\sin 0 +0=0\\ ~\\ \textrm{Now,}~f'''(x)\geq 0~\textrm{for}~x>0~\textrm{(Since the maximum value of}~\cos x~\textrm{is 1)}\\ ~\\ \textrm{Hence, by the given theorem}\\ f''(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f'(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f(x)\geq 0~\textrm{for}~x>0\\" title="\\ f(x)=\sin x -x+ \frac{x^3}{6}\\ f'(x)=\cos x -1+\frac{x^2}{2}\\ f''(x)=-\sin x +x\\ f'''(x)=-\cos x +1\\ ~\\ f(0)=\sin 0 -0+ \frac{0^3}{6}=0\\ f'(0)=\cos 0 -1+\frac{0^2}{2}=0\\ f''(0)=-\sin 0 +0=0\\ ~\\ \textrm{Now,}~f'''(x)\geq 0~\textrm{for}~x>0~\textrm{(Since the maximum value of}~\cos x~\textrm{is 1)}\\ ~\\ \textrm{Hence, by the given theorem}\\ f''(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f'(x)\geq 0~\textrm{for}~x>0\\ \Rightarrow f(x)\geq 0~\textrm{for}~x>0\\" /></a>


Here's one from my MX1 assessment task last year that hardly anyone got:
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ \textrm{The constant term in each of the binomials} \left(x@plus;\frac{4}{x}\right)^{2n}~\textrm{and}~\left(\frac{x}{a}@plus;\frac{b}{x} \right )^{2n}~\textrm{are equal.}\\ ~\\ \textrm{If}~a@plus;b=-5~\textrm{find the values of}~a~\textrm{and}~b" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ \textrm{The constant term in each of the binomials} \left(x+\frac{4}{x}\right)^{2n}~\textrm{and}~\left(\frac{x}{a}+\frac{b}{x} \right )^{2n}~\textrm{are equal.}\\ ~\\ \textrm{If}~a+b=-5~\textrm{find the values of}~a~\textrm{and}~b" title="\\ \textrm{The constant term in each of the binomials} \left(x+\frac{4}{x}\right)^{2n}~\textrm{and}~\left(\frac{x}{a}+\frac{b}{x} \right )^{2n}~\textrm{are equal.}\\ ~\\ \textrm{If}~a+b=-5~\textrm{find the values of}~a~\textrm{and}~b" /></a>