HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

asianese · Jul 10, 2013

Re: HSC 2013 3U Marathon Thread

Since P is the midpoint of AB, then AP/AB = 1/2

Also since Q is the midpoint of AC, then AQ/AC = 1/2

Angle PAQ = Angle CAB (common) and so the triangles APQ and ABC are similar.

Since sides of similar triangles are in the same ratio, PQ/BC = 1/2 as well and thus PQ is half the length of BC.

Since all angles of similar triangles are equal, Angle APQ = Angle ABC. Thus PQ is parallel to BC since corresponding angles in these lines are equal.

asianese · Jul 10, 2013

Re: HSC 2013 3U Marathon Thread

$Using the expansion of $ (1+x)^n, $ show that$

$\dfrac{2^{n+2}-3-n}{(n+1)(n+2)} = \dfrac{\binom{n}{0}}{1\cdot2}+\dfrac{\binom{n}{1}}{2\cdot3}+\dfrac{\binom{n}{2}}{3\cdot4}+\cdots+\dfrac{\binom{n}{n}}{(n+1)(n+2)}$

Sy123 · Jul 13, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
$Using the expansion of $ (1+x)^n, $ show that$

$\dfrac{2^{n+2}-3-n}{(n+1)(n+2)} = \dfrac{\binom{n}{0}}{1\cdot2}+\dfrac{\binom{n}{1}}{2\cdot3}+\dfrac{\binom{n}{2}}{3\cdot4}+\cdots+\dfrac{\binom{n}{n}}{(n+1)(n+2)}$

For those who haven't done it yet, think integration.

=====

$$Prove that$ \ \ \tan^{-1}(x) + \tan^{-1} \left(\frac{1}{x} \right ) = \frac{\pi}{2}$

RealiseNothing · Jul 13, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$Prove that$ \ \ \tan^{-1}(x) + \tan^{-1} \left(\frac{1}{x} \right ) = \frac{\pi}{2}$

Differentiate LHS:

$\frac{1}{1+x^2} + \frac{\frac{-1}{x^2}}{1+\frac{1}{x^2}}$

$\frac{1}{1+x^2} + \frac{\frac{-1}{x^2}}{\frac{1+x^2}{x^2}}$

$\frac{1}{1+x^2} - \frac{1}{1+x^2}$

$0$

Thus the LHS is a constant. Now let $x=\sqrt{3}$ and the LHS becomes:

$\tan^{-1}(\sqrt{3}) + \tan^{-1}(\frac{1}{\sqrt{3}})$

$\frac{\pi}{3} + \frac{\pi}{6}$

$\frac{\pi}{2}$

Carrotsticks · Jul 13, 2013

Re: HSC 2013 3U Marathon Thread

RealiseNothing said:
Now let $x=\sqrt{3}$

Probably easier to let x=1.

Makematics · Jul 13, 2013

Re: HSC 2013 3U Marathon Thread

why cant you construct a right angled triangle and let alpha=arctanx and beta=arctan(1/x)? you add them and there is your pi/2

Makematics · Jul 13, 2013

Re: HSC 2013 3U Marathon Thread

Carrotsticks said:
Probably easier to let x=pi/4.

by pi/4 you mean 1 right?

Sy123 · Jul 13, 2013

Re: HSC 2013 3U Marathon Thread

$Simplify$

$\binom{n}{0} \binom{n}{2} + \binom{n}{2} \binom{n}{4} + \binom{n}{4}\binom{n}{6} + \dots + \binom{n}{n-2} \binom{n}{n}$

Carrotsticks · Jul 13, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
by pi/4 you mean 1 right?

Haha yep! Sorry about that!

Carrotsticks · Jul 13, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
why cant you construct a right angled triangle and let alpha=arctanx and beta=arctan(1/x)? you add them and there is your pi/2

I personally would have done this.

Sy123 · Jul 13, 2013

Re: HSC 2013 3U Marathon Thread

$A parabola is a plane curve, whereby it is a locus of the points that are equidistant from a fixed line, the directrix, and a fixed focus$

$By considering these elements of a parabola and some variable point, rotate the parabola$ \ \ x^2=4ay \ \ $about the origin,$ \ \ 45^{\circ} \ \ $anti-clockwise$

Makematics · Jul 14, 2013

Re: HSC 2013 3U Marathon Thread

Carrotsticks said:
I personally would have done this.

hmm so you're allowed to assume it's right angled?

or are you proving that it is right angled IFF the statement is true?

Sy123 · Jul 14, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
hmm so you're allowed to assume it's right angled?

or are you proving that it is right angled IFF the statement is true?

Contructing a right angled triangle with short sides: x and 1 can always be done. (x =/= 0 )
One angle will be atan(x), and the other one will be atan(1/x), by definition of what inverse tan means

Makematics · Jul 14, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Contructing a right angled triangle with short sides: x and 1 can always be done. (x =/= 0 )
One angle will be atan(x), and the other one will be atan(1/x), by definition of what inverse tan means

alright then.
i was just wondering, if you use the inverse tan rule for that question -->
arctan(a)+arctan(b)=arctan( (a+b)/(1-ab) ) and it is undefined, is that enough to say that the angle is pi/2, or is that just bullshitting it?

Sy123 · Jul 14, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
alright then.
i was just wondering, if you use the inverse tan rule for that question -->
arctan(a)+arctan(b)=arctan( (a+b)/(1-ab) ) and it is undefined, is that enough to say that the angle is pi/2, or is that just bullshitting it?

It may open the door to rigour issues, I'm not sure whether it is allowed or not to do so, and then make the denominator of the RHS zero (i.e. ab=1)

An alternative is to add $\tan^{-1}(1)$ to both sides, so on the right hand side you have pi/2 + pi/4 = 3pi/4
On the RHS you may be able to add it using an identity similar to the one you jsut wrote but with 3 tans instead

Makematics · Jul 14, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
It may open the door to rigour issues, I'm not sure whether it is allowed or not to do so, and then make the denominator of the RHS zero (i.e. ab=1)

An alternative is to add $\tan^{-1}(1)$ to both sides, so on the right hand side you have pi/2 + pi/4 = 3pi/4
On the RHS you may be able to add it using an identity similar to the one you jsut wrote but with 3 tans instead

yeah that's true. well ill stick to the right angled triangle.

Sy123 · Jul 14, 2013

Re: HSC 2013 3U Marathon Thread

4U people please refrain from answering this one unless considerable time has passed.

$$Using the substitution$ \ \ u = \frac{\pi}{2} -x$

$Evaluate$

$\int_0^{\pi /2}x \sin^2 x \ dx$

RealiseNothing · Jul 14, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
why cant you construct a right angled triangle and let alpha=arctanx and beta=arctan(1/x)? you add them and there is your pi/2

I didn't even think of that, that's so much better than what I did.

HeroicPandas · Jul 14, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Simplify$

$\binom{n}{0} \binom{n}{2} + \binom{n}{2} \binom{n}{4} + \binom{n}{4}\binom{n}{6} + \dots + \binom{n}{n-2} \binom{n}{n}$

Apply this to the second thing of each term
$\binom{n}{r} = \binom{n}{n-r}$

$\therefore \binom{n}{0}\binom{n}{n-2} + \binom{n}{2}\binom{n}{n-4} +\dots+ \binom{n}{n-2}\binom{n}{0}$

Consider:

$(1+x)^n (1+x)^n \equiv (1+x)^{2n}$

Expand LHS and RHS and equate the coefficient of x^(n-2)

$\therefore \binom{n}{0}\binom{n}{n-2} + \binom{n}{2}\binom{n}{n-4} +\dots \binom{n}{n-2}\binom{n}{0} = \binom{2n}{n-2}$

HeroicPandas · Jul 15, 2013

Re: HSC 2013 3U Marathon Thread

$$Consider $(1+x)^n$

By differentiating and integrating it prove that:

$\sum_{r=1}^{n} \binom{n}{r}\frac{r}{r+1} = 2^n \frac{n-1}{n+1}$

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