# HSC 2018-2019 MX2 Marathon (1 Viewer)

#### TheOnePheeph

##### Active Member
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

##### Member
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..

#### HeroWise

##### Active Member
Do you remember what it was? @DrEuler

#### stupid_girl

##### Active Member
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

##### Member
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.

##### -insert title here-
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:

#### HeroWise

##### Active Member
Whey use NumPy when u can make it urself :dab2:

#### stupid_girl

##### Active Member
Geometry question could be a killer 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

##### Member
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.

#### sharky564

##### Member
Geometry question could be a killer 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 $\bg_white N$, $\bg_white P$, $\bg_white Q$ of $\bg_white BC$, $\bg_white AC$, $\bg_white BD$ respectively. Also, let $\bg_white AB \cap CD = R$, $\bg_white AB \cap NQ = R'$ and $\bg_white AC \cap BD = S$. By Midpoint Theorem, we have $\bg_white MP \parallel CD \parallel NQ$ and $\bg_white PN \parallel AB \parallel MQ$ so $\bg_white MPNQ$ is a parallelogram with $\bg_white MP = NQ = \frac{1}{2} CD$ and $\bg_white PN = MQ = \frac{1}{2} AB$. Since $\bg_white AB = CD$, we have $\bg_white MPNQ$ is a rhombus.
Furthermore, since $\bg_white \angle BMC = 90^{\circ}$, $\bg_white N$ is the centre of the circle passing through $\bg_white BMC$ by semicircle theorem. Thus, $\bg_white NM = \frac{1}{2} BC = NB = NQ$ as $\bg_white AB = BC = CD$. Therefore, $\bg_white QM = MN = NQ$ and $\bg_white PM = MN = NP$ so $\bg_white PMN$ and $\bg_white QMN$ are equilateral triangles. In particular, $\bg_white \angle PNQ = 120^{\circ}$.
As a result of these parallel lines, we have $\bg_white \angle BRD = \angle BR'Q = \angle PNQ = 120^{\circ}$, so $\bg_white \angle DSC = \angle DBC + \angle DCB = \frac{1}{2} \angle BCR + \frac{1}{2} \angle CBR = \frac{1}{2} (180^{\circ} - \angle BRC) = 30^{\circ}$.
Thus, the angle between the diagonals is $\bg_white 30^{\circ}$, so the area of $\bg_white ABCD$ is $\bg_white \frac{1}{2} AC \cdot BD \sin(30^{\circ}) = \frac{1}{4} AC \cdot BD$.

Only if direction:
Note that the area condition is equivalent to $\bg_white \angle DSC = 30^{\circ}$. We proceed in a similar manner to get $\bg_white MPNQ$ is a rhombus. By reversing our final angle chase, we get $\bg_white \angle BRC = 120^{\circ}$, so $\bg_white \angle PNQ = 120^{\circ}$. Since $\bg_white MPNQ$ is a rhombus, we get $\bg_white PMN$ and $\bg_white QMN$ are equilateral triangles. This implies $\bg_white NM = NP = \frac{1}{2} AB = \frac{1}{2} BC = NB = NC$, so $\bg_white N$ is equidistant from $\bg_white M, B, C$. This yields $\bg_white BMC$ is a semicircle with centre $\bg_white N$, so $\bg_white \angle BMC = 90^{\circ}$.

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#### DrEuler

##### Member
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 $\bg_white N$, $\bg_white P$, $\bg_white Q$ of $\bg_white BC$, $\bg_white AC$, $\bg_white BD$ respectively. Also, let $\bg_white AB \cap CD = R$, $\bg_white AB \cap NQ = R'$ and $\bg_white AC \cap BD = S$. By Midpoint Theorem, we have $\bg_white MP \parallel CD \parallel NQ$ and $\bg_white PN \parallel AB \parallel MQ$ so $\bg_white MPNQ$ is a parallelogram with $\bg_white MP = NQ = \frac{1}{2} CD$ and $\bg_white PN = MQ = \frac{1}{2} AB$. Since $\bg_white AB = CD$, we have $\bg_white MPNQ$ is a rhombus.
Furthermore, since $\bg_white \angle BMC = 90^{\circ}$, $\bg_white N$ is the centre of the circle passing through $\bg_white BMC$ by semicircle theorem. Thus, $\bg_white NM = \frac{1}{2} BC = NB = NQ$ as $\bg_white AB = BC = CD$. Therefore, $\bg_white QM = MN = NQ$ and $\bg_white PM = MN = NP$ so $\bg_white PMN$ and $\bg_white QMN$ are equilateral triangles. In particular, $\bg_white \angle PNQ = 120^{\circ}$.
As a result of these parallel lines, we have $\bg_white \angle BRD = \angle BR'Q = \angle PNQ = 120^{\circ}$, so $\bg_white \angle DSC = \angle DBC + \angle DCB = \frac{1}{2} \angle BCR + \frac{1}{2} \angle CBR = \frac{1}{2} (180^{\circ} - \angle BRC) = 30^{\circ}$.
Thus, the angle between the diagonals is $\bg_white 30^{\circ}$, so the area of $\bg_white ABCD$ is $\bg_white \frac{1}{2} AC \cdot BD \sin(30^{\circ}) = \frac{1}{4} AC \cdot BD$.

yeah he did

#### stupid_girl

##### Active Member
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

##### Member
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

##### Active Member
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.

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

##### Well-Known Member
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

##### Active Member
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

##### Well-Known Member
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

##### Member
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.

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 $\bg_white O=0$ is the origin of the Argand Plane. Also, let $\bg_white A=a, B=b, C=c, D=d$ such that $\bg_white |a|=|b|=1$. Then, we can assume $\bg_white c=ka$ and $\bg_white d=lb$, since they lie on $\bg_white OA$ and $\bg_white OB$ (where $\bg_white k$ and $\bg_white l$ are non-zero real numbers). Note that we are trying to find constraints on $\bg_white k$ and $\bg_white l$ so they satisfy for all $\bg_white a$ and $\bg_white b$.

Since $\bg_white \angle ADC = -\angle BCD$, $\bg_white \arg \left ( \frac{d - a}{d - c} \right ) = \arg \left ( \frac{c - d}{c - b} \right )$. Thus, $\bg_white \frac{d - a}{d - c} \div \frac{c - d}{c - b} = r$, where $\bg_white r$ is a real number. Substituting in the values from above, we get $\bg_white \frac{(lb - a)(ka - b)}{-(lb - ka)^2} = r$. Since $\bg_white r = \overline{r}$, we have $\bg_white \frac{(l\overline{b} - \overline{a})(k\overline{a} - \overline{b})}{-(l\overline{b} - k\overline{a})^2} = \frac{(lb - a)(ka - b)}{-(lb - ka)^2}$.

Now, since $\bg_white |a|=1$, $\bg_white a\overline{a} = |a|^2 = 1$, so $\bg_white \overline{a}=\frac{1}{a}$. Similar holds for $\bg_white b$. Substituting this above yields

\bg_white
\begin{aligned}
\frac{(lb - a)(ka - b)}{(lb - ka)^2} &= \frac{\left ( \frac{l}{b} - \frac{1}{a} \right ) \left (\frac{k}{a} - \frac{1}{b} \right )}{\left ( \frac{l}{b} - \frac{k}{a} \right )^2}\\
&= \frac{a^2b^2 \left ( \frac{l}{b} - \frac{1}{a} \right ) \left (\frac{k}{a} - \frac{1}{b} \right )}{a^2 b^2 \left ( \frac{l}{b} - \frac{k}{a} \right )^2}\\
&= \frac{ \left ( la - b \right ) \left ( kb - a \right )}{\left (la - kb \right )^2}\\
(lb - a)(ka - b)(la - kb)^2 &= (la - b)(kb - a)(lb - ka)^2\\
[\text{Expanding and collecting like }&\text{terms yields}]\\
(a^4 - b^4)kl(k - l) + (a^3 b - ab^3)(k - 1)(2kl - k^2l - kl^2 - k - l) &= 0\\
(k - l)(a - b)(a + b)((a^2 + b^2)kl + ab(2kl - k^2l - kl^2 - k - l)) &= 0\\
\end{aligned}

Thus, we must have $\bg_white k = l$ or $\bg_white (a^2 + b^2)kl + ab(2kl - k^2l - kl^2 - k - l) = 0 \implies kl(a + b)^2 = (kl + 1)(k + l)ab$. Substituting $\bg_white a=1, b=-1$ yields $\bg_white (kl+1)(k+l) = 0$, so $\bg_white kl(a+b)^2 = 0$, which means $\bg_white kl=0$, which violates $\bg_white k$ and $\bg_white l$ being non-zero reals. Therefore, we must have $\bg_white k = l$, which implies $\bg_white \frac{OA}{OC} = \frac{OB}{OD}$, so $\bg_white AB \parallel CD$.