MX2 Marathon (3 Viewers)

stupid_girl · Sep 19, 2019

DrEuler said:
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 · Sep 22, 2019

stupid_girl said:
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 · Sep 24, 2019

DrEuler said:
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 · Sep 25, 2019

Whey use NumPy when u can make it urself :dab2:

stupid_girl · Sep 28, 2019

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 · Sep 29, 2019

TheOnePheeph said:
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 · Sep 29, 2019

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

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

DrEuler · Sep 29, 2019

sharky564 said:
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 $N$ , $P$ , $Q$ of $BC$ , $AC$ , $BD$ respectively. Also, let $AB \cap CD = R$ , $AB \cap NQ = R'$ and $AC \cap BD = S$ . By Midpoint Theorem, we have $MP \parallel CD \parallel NQ$ and $PN \parallel AB \parallel MQ$ so $MPNQ$ is a parallelogram with $MP = NQ = \frac{1}{2} CD$ and $PN = MQ = \frac{1}{2} AB$ . Since $AB = CD$ , we have $MPNQ$ is a rhombus.
Furthermore, since $\angle BMC = 90^{\circ}$ , $N$ is the centre of the circle passing through $BMC$ by semicircle theorem. Thus, $NM = \frac{1}{2} BC = NB = NQ$ as $AB = BC = CD$ . Therefore, $QM = MN = NQ$ and $PM = MN = NP$ so $PMN$ and $QMN$ are equilateral triangles. In particular, $\angle PNQ = 120^{\circ}$ .
As a result of these parallel lines, we have $\angle BRD = \angle BR'Q = \angle PNQ = 120^{\circ}$ , so $\angle DSC = \angle DBC + \angle DCB = \frac{1}{2} \angle BCR + \frac{1}{2} \angle CBR = \frac{1}{2} (180^{\circ} - \angle BRC) = 30^{\circ}$ .
Thus, the angle between the diagonals is $30^{\circ}$ , so the area of $ABCD$ is $\frac{1}{2} AC \cdot BD \sin(30^{\circ}) = \frac{1}{4} AC \cdot BD$ .

Have you had olympiad training by any chance?

HeroWise · Sep 29, 2019

yeah he did

sharky564 · Sep 29, 2019

DrEuler said:
Have you had olympiad training by any chance?

indubitably

stupid_girl · Sep 29, 2019

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 · Sep 29, 2019

stupid_girl said:
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 · Oct 3, 2019

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.

StudyOnly · Oct 3, 2019

blyatman said:
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?

sharky564 · Oct 4, 2019

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

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

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

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

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

sharky564 · Oct 4, 2019

blyatman said:
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.

I believe its $\frac{4}{\pi}$ ?

DrEuler · Oct 4, 2019

sharky564 said:
I believe its $\frac{4}{\pi}$ ?

You sure it's constant for circles of any radius?

sharky564 · Oct 4, 2019

DrEuler said:
You sure it's constant for circles of any radius?

He says its a unit circle. Of course, for a circle of radius $R$ , the answer is just $\frac{4R}{\pi}$ .

DrEuler · Oct 4, 2019

sharky564 said:
He says its a unit circle. Of course, for a circle of radius $R$ , the answer is just $\frac{4R}{\pi}$ .

That's me not reading questions properly again lol.

TheOnePheeph · Oct 4, 2019

sharky564 said:
I believe its $\frac{4}{\pi}$ ?

You get 4/pi if you use the angle as the parameter, but if you use the perpendicular distance from the center to the chord you get pi/2, which I find kind of weird. Two different definitions of average or something I guess?

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