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HSC 2018-2019 MX2 Marathon (1 Viewer)

TheOnePheeph

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Love seeing debates about Mathematics lol, reminds me of when my schools math department and me had a huge debate in which I ended up proving a question wrong in our trial.
Lol I feel like such a dick arguing about maths, especially on a public forum haha. Fun nonetheless, and definitely worth it if you end up proving a question wrong.
 

DrEuler

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Lol I feel like such a dick arguing about maths, especially on a public forum haha. Fun nonetheless, and definitely worth it if you end up proving a question wrong.
No it's such a good thing. Debating about it really opens up your mind to others viewpoints and it can be quite beneficial for your understanding..
 

stupid_girl

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Love seeing debates about Mathematics lol, reminds me of when my schools math department and me had a huge debate in which I ended up proving a question wrong in our trial.
Interesting. Do you still remember that question?
 

DrEuler

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Interesting. Do you still remember that question?
It was a probability question(as expected).
Unfortunately what I meant when I said "it was in our trial" is that it was in my schools trial, not the one I sat but rather the year before. I was able to see the paper and well I was puzzled with their answer.

It was rather hard to explain and convince the teachers because with any other topic you can present a perfectly logical and intuitive proof but with probability you have to get the teacher to understand your intuition which is really difficult.

However it helped a lot that my school had heaps of smart people in 4u, so enough of us noticed it.
 

Paradoxica

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It was rather hard to explain and convince the teachers because with any other topic you can present a perfectly logical and intuitive proof but with probability you have to get the teacher to understand your intuition which is really difficult.
Just do an enormous computational bash in NumPy :dab:
 

stupid_girl

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Geometry question could be a killer :devil: but I think it will be gone from MX2 soon.

ABCD is a quadrilateral with three equal sides AB,BC and CD. Show that the mid-point of AD lies on a circle with diameter BC if and only if the area of ABCD is a quarter of the product of its diagonals.
 

sharky564

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But if they flip the same number of heads, then Ben will flip more tails, since he flips n+1 coins and Amy only flips n, therefore putting it in the 2nd case of Ben flipping more tails than Amy.
yeah my bad, good job
 

sharky564

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Geometry question could be a killer :devil: but I think it will be gone from MX2 soon.

ABCD is a quadrilateral with three equal sides AB, BC and CD. Show that the mid-point of AD lies on a circle with diameter BC if and only if the area of ABCD is a quarter of the product of its diagonals.
Pretty sure this is an old Tournament of Towns problem but wasn't too difficult (maybe IMOSL G1?)... It is a bit of a pain to convert to directed angles so left as an exercise to the reader.

If direction:
We construct midpoints , , of , , respectively. Also, let , and . By Midpoint Theorem, we have and so is a parallelogram with and . Since , we have is a rhombus.
Furthermore, since , is the centre of the circle passing through by semicircle theorem. Thus, as . Therefore, and so and are equilateral triangles. In particular, .
As a result of these parallel lines, we have , so .
Thus, the angle between the diagonals is , so the area of is .

Only if direction:
Note that the area condition is equivalent to . We proceed in a similar manner to get is a rhombus. By reversing our final angle chase, we get , so . Since is a rhombus, we get and are equilateral triangles. This implies , so is equidistant from . This yields is a semicircle with centre , so .
 
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DrEuler

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Pretty sure this is an old Tournament of Towns problem but wasn't too difficult (maybe IMOSL G1?)... It is a bit of a pain to convert to directed angles so left as an exercise to the reader.
We construct midpoints , , of , , respectively. Also, let , and . By Midpoint Theorem, we have and so is a parallelogram with and . Since , we have is a rhombus.
Furthermore, since , is the centre of the circle passing through by semicircle theorem. Thus, as . Therefore, and so and are equilateral triangles. In particular, .
As a result of these parallel lines, we have , so .
Thus, the angle between the diagonals is , so the area of is .
Have you had olympiad training by any chance?
 

stupid_girl

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Maybe it wasn't too difficult for mathematical olympiad...but should be enough of a headache for MX2.
If it appears in MX2 paper (final chance in 2019!!!), the performance will likely be worse than the geometry question last year.
 

sharky564

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Maybe it wasn't too difficult for mathematical olympiad...but should be enough of a headache for MX2.
If it appears in MX2 paper (final chance in 2019!!!), the performance will likely be worse than the geometry question last year.
Yeah, I never thought that the geo q in last year's paper was actually difficult but that's probs becos I had the intuition to solve those probs.
 

stupid_girl

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Let's continue to have fun with geometry. If a similar question appears in MX2 paper, I guess half of the candidates may give up.:p

A and B are two points on a circle with centre O. Extend OA to C and OB to D such that ∠ADC=∠BCD. Prove that AB||CD.
Hint: Consider two cases - AB is the diameter and AB is not the diameter.
 

blyatman

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This is a good question that utilizes Ext 1 mathematics (and a bit of thinking outside the box):

What is the average chord length in a unit circle? I.e. if you randomly drew an infinite amount of chords within the unit circle, what would be the average length of those chords?

A few years back, I was offered a job as a financial trader. In the initial stages of the selection process, we had to pass a math test consisting of 5 questions. I don't remember the others, but this one stuck with me since it was super interesting.
 
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StudyOnly

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This is a good question that utilizes Ext 1 mathematics (and a bit of thinking outside the box):

What is the average chordlength in a unit circle? I.e. if you randomly drew an infinite amount of chords within the unit circle, what would be the average length of those chords?

A few years back, I was offered a job as a financial trader. In the initial stages of the selection process, we had to pass a math test consisting of 5 questions. I don't remember the others, but this one stuck with me since it was super interesting.
Did you get the job? 🤨
 

blyatman

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Did you get the job? 🤨
Yeh I was offered the job (I got 4/5 of the questions and did well in the interview). I decided not to take it though, since I wanted to be an engineer for the early part of my career (a lot of engineers go into trading later in life, but it would've been hard to go back to engineering if I had started my career in trading). Definitely miss the trading salary though...

Anyway, the question is pretty interesting as you need to apply a few things from 2u/3u math.

Hints:
1. Find a way to calculate the chord length for some given parameter (i.e. parameterize the chord length in terms of some variable).
2. The chord length will be a function based on some parameter. Think about what it means to find the average of some function. In particular, if you were given the velocity equation/curve of a particle, how would you determine it's average velocity? Using this concept, you can determine the average of the chord length function.
 
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sharky564

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Let's continue to have fun with geometry. If a similar question appears in MX2 paper, I guess half of the candidates may give up.:p

A and B are two points on a circle with centre O. Extend OA to C and OB to D such that ∠ADC=∠BCD. Prove that AB||CD.
Hint: Consider two cases - AB is the diameter and AB is not the diameter.
Let's make it more fun by using complex numbers lol (and here's a crash course of doing geometry with complex numbers).

We can assume the circle is the unit circle, and is the origin of the Argand Plane. Also, let such that . Then, we can assume and , since they lie on and (where and are non-zero real numbers). Note that we are trying to find constraints on and so they satisfy for all and .

Since , . Thus, , where is a real number. Substituting in the values from above, we get . Since , we have .

Now, since , , so . Similar holds for . Substituting this above yields



Thus, we must have or . Substituting yields , so , which means , which violates and being non-zero reals. Therefore, we must have , which implies , so .
 

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