Solving An Oxford Elliptic Curve Problem

0:00 Intro
1:00 (i) find a & b
2:02 (ii) find δ
3:35 (iiI) symmetric about y=δ
6:06 (iv) find a cubic eq
7:03 (v) find α+β+γ
9:40 Check out Brilliant
10:36 (vi) circle intersects the elliptic curve

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Thank you,
blackpenredpen

[helo]

Fermat's Last Theorem is proved with Elliptical Curves so I am excited.

[toby.barnett]

This was the one question I couldn’t solve hahaha I saw the elliptic curve and thought oh no 🤣

[thetheoreticalnerd]

My take on part 2 (using calculus): We can differintiate both sides as such: dy/dx (2y - 1) = 3x^2 - 1. Dividing 2y - 1 both sides, we have dy/dx = (3x^2 - 1)/(2y - 1). Since y value on the graph has an undefined slope (vertical), we just set the denominator of dy/dx equal to 0: 2y - 1 = 0. Thus, we have y = 1/2.

[korbleu]

I sat this exam last year for Oxford/Imperial and I absolutely loved this question! Very exciting seeing you cover it !! :)

[josephwong]

At the two points (gamma, delta) or (alpha, delta), the slope of the tangent lines satisfy the property dx/dy=0, so I just implicit differentiate the equation to find dx/dy=(2y-1)/(3x^2-1), and set y=1/2.

[davidbizzozero]

I believe that Greek letter at 16:32 is a lower case squigma.

[vladimirkhazinski]

For part iii i think a better method would be to shift the function downward one half, then plug in -y for y and see if it yeilds the same equation.

[saturten]

I was not familiar with Vieta's formulas so the explanation for the last part of section vi left me very confused.
the missing piece was (-a sub n-1 / a sub n) = -1 / 1 = the sum of roots, where a sub i is the coefficient of the ith term in the cubic equation

[kaishang]

set y = 0. x intecpt values∈{ -1, 0, 1} . due to the restrictions, the values are 1, -1.

set x = 0. y intercept values∈{ 0, 1}
y is a parabola, since parabola's vortex is midway between the y-intercepts, the vortex of the graph is symmetric on the line y=0.5

since parabolas are symmetric on two sides of the vortex and the other side is independent of y, the graph is symmetric on the line y=0.5. (there are some circular reasoning that I'm not energized enough to resolve)

since the three numbers already exist on the same horizontal line y=0.5, replace y with 0.5 in the original equation and move everything to one side gives x^3-x+0.25=0. replace the 0 with a new y gives a cubic equation that has the correct roots.

the equation of the circle can be written as y=sqrt( (b-a)^2/4+(x-(b+a)/2)^2 )+0.5 which can be simplified to sqrt( - (x-a) (x-b) )+0.5
sub it into the "y" of the original problem can probably yield the correct answer. however, the amount of calculation already feels like this is the wrong path.
however, by continuing the next step, the roots actually cancel out nicely to 0 = then I got stuck since I don't know the cubic roots formula. so maybe this is indeed the wrong path.
try again and define the circle as a function of y(right half) might work, but i really need to be doing my homework by now.

[vladimir]

for part 3 you can say that if (a, b) is on the curve then (a, 1-b) is on the curve and the distance to the line y=1/2 from each is abs(b-1/2) without writing any equations. Once I wrote it it is probably not faster than what you did.

[diffusegd]

I did this question in the actual exam and it was definitely the one I had most doubts about

[ilickcatnip]

Enjoyed the Question and ofcourse the video too. You should also have a look at the JEE advanced paper. It has some nice Questions in calculus and other math in general, could be good inspiration for your channel's audience!

[babajani]

Hi love ur vids, I had a request. Could you plz make some videos on the STEP Exam. It is the Cambridge entrance exam for High School students and the questions in it are absolutely brutal. There are 3 papers in total and they increase in difficulty from STEP 1 being the easiest and STEP 3 is the hardest. I think you will really enjoy some of the questions since in my opinion, having looked at both the JEE advance maths questions and STEP 3, I would say that a lot of the STEP 3 are actually even harder than Jee so plz give it a go. But if you want the really hard ones then do STEP 3. STEP 2 AND 1 are still hard but not as hard. Also, if you want the hardest ones, even from STEP 3, then there is a mark scheme which has all the answers of the questions from the past years and towards the end of that, you can go to STEP 3 and they tell you which questions were done well or poorly. Hence which were hardest and which were easiest.

[dylandejonge]

When I did this a few months ago I wasn’t able to do parts v and vi, the others ones I got correct but the last 2 where hard. I could’ve known v when I saw how it was done but vi was too hard for me

[alejrandom]

another way to solve (ii) is to find dx/dy and set it equal to 0. you get (2y-1)/(3x^3-1)=0 which means y=1/2

[maxryder]

For some reason it is still slightly stressful to see the MAT paper even though I am literally already in Oxford student

[Jkauppa]

try this: elliptic integral of the 2nd kind, integrate sqrt(1+c*sin^2(z) ] dz = (2/3)*csc^2(z)*(c*sin^2(z)+1)^(3/2), the c-value is negative, csc has a zero-point issue

[physicsmania]

Hey blackpenredpen! Can u solve the following integral: ∫(x³ - 4x² + 3x + 6)/√(x² + 3x + 5) dx evaluated from 0 to 4.
I challenge you to solve this without using any of the hyperbolic functions or their inverses... use the logarithm instead... By the way, great video as always!! I am a huge fan of yours!

[JPiMaths]

The MAT brings back some interesting memories - excited to watch this one!!

[AlephThree]

I did the Oxford maths entrance exam in 1993, and scored 98% on the pure paper. The funny thing was that loads of other people got very similar marks. They’d just made the paper a little too easy / similar to previous years. In contrast very strong candidates were getting in with very low scores (like 40%) the year before - the 1992 paper was just a bit too hard. The margins can be quite small at this level for a bunch of pretty well prepared candidates.