HSC 2013-14 MX1 Marathon (archive) (3 Viewers)

Stygian · Aug 11, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Yes well done

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$\\ $A number is divisible by 3 if the sum of its digits are divisible by 3$ \\ \\ $Find how many 4 digit numbers are even and divisible by 3$$

is it 1500?

(tried to do it in my head lol)

Sy123 · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Stygian said:
is it 1500?

(tried to do it in my head lol)

Yea it is, my initial fact was useless

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$\\ \lim_{c \to \infty} \int_{-c}^c \frac{1}{e^{x} + e^{-x}} \ dx$

Stygian · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Yea it is, my initial fact was useless

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$\\ \lim_{c \to \infty} \int_{-c}^c \frac{1}{e^{x} + e^{-x}} \ dx$

Yep i was thinking it was a red herring, all you had to do was determine the number of multiples of 6 between 1000 and 9999

randomly_generated · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Yea it is, my initial fact was useless

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$\\ \lim_{c \to \infty} \int_{-c}^c \frac{1}{e^{x} + e^{-x}} \ dx$

dafaq? that's not 3U. I can't use latex but my answer is Pi/2 I rationalsed and integrated.

my question:

find all solutions for:

3cos(2a) - sin(a) +2 = 0

dunjaaa · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Screen shot 2014-08-13 at 12.33.57 PM.png

randomly_generated · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

dunjaaa said:
View attachment 30760

I did it first

mreditor16 · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

from a past 3u trial paper

post images

aDimitri · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

mreditor16 said:
from a past 3u trial paper

post images

does this mean only 1 replacement occurs, and from the second draw on they stay out of the bag? terribly ambiguous question. i'm going to assume you mean you replace them all.

i) (1/3)^2 * (2/3)^2 * 4C2 = 24/81
ii) 4 are drawn, for the sum to be greater than 9, need 2 or more white counters.
P = 24/81 + 8/81 + 1/81 = 33/81

Kurosaki · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

aDimitri said:
does this mean only 1 replacement occurs, and from the second draw on they stay out of the bag? terribly ambiguous question. i'm going to assume you mean you replace them all.

i) (1/3)^2 * (2/3)^2 = 4/81
ii) 4 are drawn, for the sum to be greater than 9, need 2 or more white counters.
P = 4/81 + 2/81 + 1/81 = 7/81

Think you need to multiply by $\binom{4}{2}$ for the first bit to account for ordering.
Same for second bit (haven't done the problem myself though, just glancing through so I might be wrong).

mreditor16 · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Kurosaki! done with trials?

aDimitri · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
Think you need to multiply by $\binom{4}{2}$ for the first bit to account for ordering.
Same for second bit (haven't done the problem myself though, just glancing through so I might be wrong).

oh true that. didn't account for ordering hold up let me fix

Stygian · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

aDimitri said:
does this mean only 1 replacement occurs, and from the second draw on they stay out of the bag? terribly ambiguous question. i'm going to assume you mean you replace them all.

i) (1/3)^2 * (2/3)^2 = 4/81
ii) 4 are drawn, for the sum to be greater than 9, need 2 or more white counters.
P = 4/81 + 2/81 + 1/81 = 7/81

this is the more logical assumption so you would go with it imo

would you say it would be erroneous to consider for part ii (in a situation where you weren't given part i) to work with 1-(probability of 3 or more black counters)? Should this logically give the same result?

Stygian · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
Think you need to multiply by $\binom{4}{2}$ for the first bit to account for ordering.
Same for second bit (haven't done the problem myself though, just glancing through so I might be wrong).

^this should fix that

Kurosaki · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

mreditor16 said:
Kurosaki! done with trials?

Almost, I have 2 next week and then I'm done, but I have the rest of this week off haha

.

Pretty nice question in my 3U trial that I'd like to share:

PQRS is a cyclic quadrilateral. Angle QPS is 90 degrees, angle PSR is $\theta$ , PQ=a, PS=b, QR=c.
Show $(a^2+b^2) \sin ^2 \theta = a^2 + c^2 +2ac \cos \theta$

aDimitri · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
Almost, I have 2 next week and then I'm done, but I have the rest of this week off haha .

Pretty nice question in my 3U trial that I'd like to share:

PQRS is a cyclic quadrilateral. Angle QPS is 90 degrees, angle PSR is $\theta$ , PQ=a, PS=b, QR=c.
Show $(a^2+b^2) \sin ^2 \theta = a^2 + c^2 +2ac \cos \theta$

spent 10 minutes trying to figure this out then noticed the question said CYCLIC.
fml

mreditor16 · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
Almost, I have 2 next week and then I'm done, but I have the rest of this week off haha .

Pretty nice question in my 3U trial that I'd like to share:

PQRS is a cyclic quadrilateral. Angle QPS is 90 degrees, angle PSR is $\theta$ , PQ=a, PS=b, QR=c.
Show $(a^2+b^2) \sin ^2 \theta = a^2 + c^2 +2ac \cos \theta$

ah nice, good luck with the remaining ones!

aDimitri · Aug 13, 2014

Re: HSC 2013 3U Marathon Thread

Kurosaki said:
Almost, I have 2 next week and then I'm done, but I have the rest of this week off haha .

Pretty nice question in my 3U trial that I'd like to share:

PQRS is a cyclic quadrilateral. Angle QPS is 90 degrees, angle PSR is $\theta$ , PQ=a, PS=b, QR=c.
Show $(a^2+b^2) \sin ^2 \theta = a^2 + c^2 +2ac \cos \theta$

$\\ \text{Since PQRS is cyclic, } \angle{PQR} = 180-\theta \\ \text{By use of the cosine rule in triangle PQR, } \\ (PR)^{2} = a^{2} + c^{2} - 2accos(180-\theta) = a^{2} + c^{2} + 2accos\theta \\ \text{By use of the sine rule in triangle PRS, } \frac{PR}{sin\theta} = \frac{b}{sin(\angle{PRS})} \qquad (1) \\ \text{Now by the angles in the same segment theorem, } \angle{PRS} = \angle{PQS} \\ \text{By squaring and rearranging (1), } (PR)^{2} = \frac{b^{2}sin^{2}\theta}{sin^{2}(\angle{PQS})} \qquad (2) \\ \text{Now, by elementary trig in triangle PQS, } sin(\angle{PQS}) = \frac{b}{SQ} \\ sin^{2}(\angle{PQS}) = \frac{b^{2}}{b^{2} + a^{2}} \\ \text{Substituting back into (2), } (PR)^{2} = (a^2+b^2)sin^{2}\theta \\ \text{Hence, } (a^2+b^2)sin^{2}\theta = a^{2} + c^{2} + 2accos\theta$

Sy123 · Aug 14, 2014

Re: HSC 2013 3U Marathon Thread

$\\ $Prove by mathematical induction that the sum of the interior angles of an$ \ n \ $sided polygon is$ \ \pi(n-2)$

aDimitri · Aug 14, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\\ $Prove by mathematical induction that the sum of the interior angles of an$ \ n \ $sided polygon is$ \ \pi(n-2)$

are we allowed to make the assumption that the angle sum of a triangle is pi?

aDimitri · Aug 14, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\\ $Prove by mathematical induction that the sum of the interior angles of an$ \ n \ $sided polygon is$ \ \pi(n-2)$

$\text{Testing for n=3, we get a triangle, and we know that the angle sum of a triangle is } \pi \\ \text{Therefore, statement is true for n=3} \\ \text{Assuming to be true for n=k} \\ \text{i.e. Angle sum of k sided polygon is given by } \pi(k-2) \\ \text{Construct a k sided polygon...} \\ \text{In order to transform it into a k+1 sided polygon, add a new vertex next to the kth side} \\ \text{It is quite clear that the only change that has been made is the addition of a new triangle adjacent to the pre-existing side} \\ \text{Hence, angle sum of k+1 sided polygon } = \pi(k-2) + \pi \\ = \pi((k+1) - 2) \\ \text{Therefore, blah blah blah true by mathematical induction}$

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