HSC 2013 MX2 Marathon (archive) (2 Viewers)

GoldyOrNugget · Apr 26, 2013

Re: HSC 2013 4U Marathon

study1234 said:
Nup. It's a circle, and also I think you have to use cases.

The answer is 105/7 = 15. You don't have to use cases -- you can prove that for any of these bead strings, none of its rotations will be identical to itself, so there's no risk of double counting: 7!/(4!*2!*1!) will count every rotation precisely 7 times.

For the record: ['ggrggyr', 'gggrgyr', 'ggrggry', 'ggggyrr', 'ggrgygr', 'gggrygr', 'gggrrgy', 'ggggrry', 'ggrrggy', 'ggggryr', 'gggrgry', 'ggrgrgy', 'gggyrgr', 'gggygrr', 'ggygrgr']

RealiseNothing · Apr 26, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
The answer is 105/7 = 15. You don't have to use cases -- you can prove that for any of these bead strings, none of its rotations will be identical to itself, so there's no risk of double counting: 7!/(4!*2!*1!) will count every rotation precisely 7 times.

For the record: ['ggrggyr', 'gggrgyr', 'ggrggry', 'ggggyrr', 'ggrgygr', 'gggrygr', 'gggrrgy', 'ggggrry', 'ggrrggy', 'ggggryr', 'gggrgry', 'ggrgrgy', 'gggyrgr', 'gggygrr', 'ggygrgr']

Couldn't you just treat it as a circle:

$\frac{6!}{4!2!} = 15$

anomalousdecay · Apr 26, 2013

Re: HSC 2013 4U Marathon

study1234 said:
Nup. It's a circle, and also I think you have to use cases.

WOW, where did I get the 3! from lol.
And, realise nothing beat me to it.

Sy123 · Apr 26, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
$\\ $Suppose $ \alpha + \beta + \gamma = \frac{\pi}{2} \\\\ $Prove that $ \sin 2 \alpha + \sin 2 \beta + \sin 2 \gamma = 4 \cos \alpha \cos \beta \cos \gamma$

$\alpha + \beta = \frac{\pi}{2} - \gamma$

$\sin(\alpha + \beta) = \cos(\gamma)$

$\therefore \ \ \sin(\alpha + \beta) \sin(\alpha + \gamma) \sin(\beta + \gamma) = \cos \alpha \cos \beta \cos \gamma=RHS$

$RHS =\sin(\beta + \gamma) (\frac{1}{2}(\cos(\alpha + \beta -\alpha - \gamma) - \cos(\alpha + \beta + \alpha + \gamma)))$

$2RHS = \cos(\beta - \gamma) \sin(\beta + \gamma) - \cos(2\alpha + \beta + \gamma) \sin(\beta + \gamma)$

$\sin 2\alpha + \sin 2\beta + \sin 2\gamma = 4\cos \alpha \cos \beta \cos \gamma$
$since$ $\cos B \sin A = 0.5(\sin(A+B)+\sin(A-B))$

GoldyOrNugget · Apr 26, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Couldn't you just treat it as a circle:

$\frac{6!}{4!2!} = 15$

Only once you've made the observation that all rotations are different to each other. With your method, if the beads were RGRG (in which case, you can rotate it by two beads and get RGRG again), you'd get 3!/(2!2!) = 6/4 (uh oh...)

bleakarcher · Apr 26, 2013

Re: HSC 2013 4U Marathon

Carrotsticks said:
Each cross section, taken perpendicular to the base, is a rectangle.

By symmetry, it's just pi*r^2*H/4

Sy123 · Apr 26, 2013

Re: HSC 2013 4U Marathon

$$If$ \ \ y=\cos(n \cdot \cos^{-1}x)$

$i) Prove that$

$y= \frac{1}{2} ((x+\sqrt{1-x^2})^n + (x-\sqrt{1-x^2})^n)$

$ii) Prove that if$ \ \ n \ \ $is an odd number, it can be expressed as a polynomial$

$y = a_k x^{2k+1} + a_{k-1}x^{2k-1} + \dots + a_1 x^3 + a_0 x$

$iii) Find$

$\sum_{r=0}^{k} a_r$

study1234 · Apr 26, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
The answer is 105/7 = 15. You don't have to use cases -- you can prove that for any of these bead strings, none of its rotations will be identical to itself, so there's no risk of double counting: 7!/(4!*2!*1!) will count every rotation precisely 7 times.

For the record: ['ggrggyr', 'gggrgyr', 'ggrggry', 'ggggyrr', 'ggrgygr', 'gggrygr', 'gggrrgy', 'ggggrry', 'ggrrggy', 'ggggryr', 'gggrgry', 'ggrgrgy', 'gggyrgr', 'gggygrr', 'ggygrgr']

The answer is actually 9. You have to consider when the bracelet if flipped (i.e. anticlockwise is the same as clockwise). This makes combinations such as ggrggry the same as ggrggyr. These are the combinations: (ggrggyr, gggrgyr, ggggyrr, ggrgygr, gggrrgy, ggrrggy, ggggryr, gggrgry, ggrgrgy).

However, does anyone know to get the answer as 9 by using permutations and combinations, rather than trial and error?

Sy123 · Apr 27, 2013

Re: HSC 2013 4U Marathon

$$If$ \ \ I_n = \int_0^1 x^n \sqrt{x(1-x)} \ dx$

$Prove that$

$I_n = \frac{4n+2}{4n+5} I_{n-1}$

HeroicPandas · Apr 27, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$If$ \ \ I_n = \int_0^1 x^n \sqrt{x(1-x)} \ dx$

$Prove that$

$I_n = \frac{4n+2}{4n+5} I_{n-1}$

$I_n = \int_0^1 x^n \sqrt{x(1-x)} \ dx = \int_{0}^{1}x^{n+\frac{1}{2}}\sqrt{(1-x)}dx \\ \\ $Let u = x^{n+\frac{1}{2}}, du = (n+ \frac{1}{2})x^{n-\frac{1}{2}} \\ dv = \sqrt{(1-x)}.dx, v = \frac{2(1-x)^{\frac{3}{2}}}{-3} \\ \\ \therefore I_n = \left ( x^{n+\frac{1}{2}. \frac{2(1-x)^{\frac{3}{2}}}{-3}} \right )_0 ^1 + \frac{2}{3}(n+\frac{1}{2}) $\int_{0}^{1}x^{n-\frac{1}{2}}\sqrt{(1-x)}(1-x)dx \\ I_n = \frac{2}{3}(n+\frac{1}{2}) \int_{0}^{1}x^{n-\frac{1}{2}}\sqrt{(1-x)}dx-x^{n+\frac{1}{2}}\sqrt{(1-x)}dx \\ $ if I_n = $\int_{0}^{1}x^{n+\frac{1}{2}} \sqrt{(1-x)}, $ $ then I_{n-1} = \int_{0}^{1}x^{n-\frac{1}{2}}\sqrt{(1-x)} \\ \therefore I_n = \frac{2}{3}(n+\frac{1}{2})I_{n-1} - \frac{2}{3}(n+\frac{1}{2})I_n \\ \therefore I_n = \frac{2n+1}{4+2n}I_{n-1}$

I got different!

Sy123 · Apr 27, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$I_n = \int_0^1 x^n \sqrt{x(1-x)} \ dx = \int_{0}^{1}x^{n+\frac{1}{2}}\sqrt{(1-x)}dx \\ \\ $Let u = x^{n+\frac{1}{2}}, du = (n+ \frac{1}{2})x^{n-\frac{1}{2}} \\ dv = \sqrt{(1-x)}.dx, v = \frac{2(1-x)^{\frac{3}{2}}}{-3} \\ \\ \therefore I_n = \left ( x^{n+\frac{1}{2}. \frac{2(1-x)^{\frac{3}{2}}}{-3}} \right )_0 ^1 + \frac{2}{3}(n+\frac{1}{2}) $\int_{0}^{1}x^{n-\frac{1}{2}}\sqrt{(1-x)}(1-x)dx \\ I_n = \frac{2}{3}(n+\frac{1}{2}) \int_{0}^{1}x^{n-\frac{1}{2}}\sqrt{(1-x)}dx-x^{n+\frac{1}{2}}\sqrt{(1-x)}dx \\ $ if I_n = $\int_{0}^{1}x^{n+\frac{1}{2}} \sqrt{(1-x)}, $ $ then I_{n-1} = \int_{0}^{1}x^{n-\frac{1}{2}}\sqrt{(1-x)} \\ \therefore I_n = \frac{2}{3}(n+\frac{1}{2})I_{n-1} - \frac{2}{3}(n+\frac{1}{2})I_n \\ \therefore I_n = \frac{2n+1}{4+2n}I_{n-1}$

I got different!

You may be correct, I did a the substitution x=sin^2 u which got me to a trig recurrence. So maybe I made a mistake.
But I'm not sure I don't have the answer with me.

Sy123 · Apr 27, 2013

Re: HSC 2013 4U Marathon

$$A cyclic quadrilateral has the side lengths$ \ \ a, b, c, d$

$Prove that the area of the cyclic quadrilateral is given by$

$A = \sqrt{(s-a)(s-b)(s-c)(s-d)}$

$$Where$ \ \ s=\frac{1}{2} (a+b+c+d)$

hayabusaboston · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$A cyclic quadrilateral has the side lengths$ \ \ a, b, c, d$

$Prove that the area of the cyclic quadrilateral is given by$

$A = \sqrt{(s-a)(s-b)(s-c)(s-d)}$

$$Where$ \ \ s=\frac{1}{2} (a+b+c+d)$

Was about to ask if you come up with all these questions yourself Sy, but then I think I saw this question in cambridge somewhere. Hm.
Anyway, im here to request your participation in the "Accurate IQ test" thread started by Hypem. That is all.

hayabusaboston · Apr 28, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$I_n = \int_0^1 x^n \sqrt{x(1-x)} \ dx = \int_{0}^{1}x^{n+\frac{1}{2}}\sqrt{(1-x)}dx \\ \\ $Let u = x^{n+\frac{1}{2}}, du = (n+ \frac{1}{2})x^{n-\frac{1}{2}} \\ dv = \sqrt{(1-x)}.dx, v = \frac{2(1-x)^{\frac{3}{2}}}{-3} \\ \\ \therefore I_n = \left ( x^{n+\frac{1}{2}. \frac{2(1-x)^{\frac{3}{2}}}{-3}} \right )_0 ^1 + \frac{2}{3}(n+\frac{1}{2}) $\int_{0}^{1}x^{n-\frac{1}{2}}\sqrt{(1-x)}(1-x)dx \\ I_n = \frac{2}{3}(n+\frac{1}{2}) \int_{0}^{1}x^{n-\frac{1}{2}}\sqrt{(1-x)}dx-x^{n+\frac{1}{2}}\sqrt{(1-x)}dx \\ $ if I_n = $\int_{0}^{1}x^{n+\frac{1}{2}} \sqrt{(1-x)}, $ $ then I_{n-1} = \int_{0}^{1}x^{n-\frac{1}{2}}\sqrt{(1-x)} \\ \therefore I_n = \frac{2}{3}(n+\frac{1}{2})I_{n-1} - \frac{2}{3}(n+\frac{1}{2})I_n \\ \therefore I_n = \frac{2n+1}{4+2n}I_{n-1}$

I got different!

Do you even know how much u confused me with ur polynomials question, where u made it a^2 not a, that had me going for hours and hours trying to figure out how u got a^2 and it turns out it was a all along lol. Btw If U have a cool conics question can u plz give to me.

RealiseNothing · Apr 28, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
$\therefore I_n = \frac{2n+1}{4+2n}I_{n-1}$

I got different!

This is correct, Sy must have made a silly mistake some where.

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

$$Find the smallest natural number$ \ \ n$

$which satisfies the following properties$

$1) It has 6 as its last digit$

$2) If the last digit 6 is erased and placed in front of the remaining digits$

$$ the resulting number is four times as large as the original number$ \ \ n$

Sy123 · Apr 28, 2013

Re: HSC 2013 4U Marathon

Trebla said:
totally not giving away the answer here

Hmm is there perhaps a really elegant solution here? It is an early IMO question and I was able to solve it with a bit of algebra and substitution.

RealiseNothing · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Find the smallest natural number$ \ \ n$

$which satisfies the following properties$

$1) It has 6 as its last digit$

$2) If the last digit 6 is erased and placed in front of the remaining digits$

$$ the resulting number is four times as large as the original number$ \ \ n$

153846

hayabusaboston · Apr 28, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
153846

Did u calculate that in ur head?

RealiseNothing · Apr 28, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Hmm is there perhaps a really elegant solution here? It is an early IMO question and I was able to solve it with a bit of algebra and substitution.

It's just backwards multiplication of two numbers. The original number ends in 6, and when multiplied by 4 gets the same number but with a 6 out the front instead of the end:

$6 \times 4 = 24$

Multiply by the 4 and carry the 2 (since you multiply by the units and carry the tens)

$4 \times 4 + 2 = 18$

Multiply by the 8 and carry the 1:

$8 \times 4 + 1 = 33$

Multiply by the 3 and carry the 3:

$3 \times 4 + 3 = 15$

Multiply by the 5 and carry the 1:

$5 \times 4 + 1 = 21$

Multiply by the 1 and carry the 2:

$1 \times 4 + 2 = 6$

Since we have just 6, we stop here, and so our number is all the numbers we multiplied by (if you read the 1st number of each step going from top to bottom) - 648351 - but we reverse it as we did the multiplcation in the reverse order to get 153846.

HSC 2013 MX2 Marathon (archive) (2 Viewers)

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