Check out the latest episode of mathmajorchat with Nate Mankovich: th-cam.com/video/WKcTyGMJiz4/w-d-xo.html We discus the importance of learning computer science for making a successful transition to graduate school for applied math.
I dont know wether you had some other requests for this but I only asked for it a week ago so to see it being incorporated is really good - keep them coming
I think step isn’t ever a trainwreck if you go through each possible route of approaching a problem in your head and understanding which one is most likely to get you to the right answer.
second day comenting this problem: find the function form Natural to natural that relationates the number of sides of a regular polygon to the numbers of points that are intersections between all the lines that connect each vertice of the polygon to all other vertices
7:08 what about when q=-r (of course it is a special case and depends on a, but still it needs to be mentioned and solved)? Q and R are symmetric points with regards to the X axis (and therefore P is the origin) QR is parallel to the Y axis and has no (real) slope. 8:10 vieta's formula is known. so x1+x2=-b/a (and x1*x2=c/a) therefore q+r=-p 10:42 also should consider if p=0 (the case I mentioned earlier)
I'm pretty sure we can ignore the q=-r case because the normal lines will intersect at the origin only when q=r=0, which breaks the uniqueness requirement, or when a=0, which breaks the requirement that a>0.
If this was a question at an interview I'd be giving a side eye to anybody not starting by drawing a diagram. And even updating the diagram after finding all those lines go through (0,-2a). Things like this that can make the difference between an offer and a rejection.
'Normal' means the lines are perpendicular to the curve (or more specifically, perpendicular to the tangent to the curve) at that point, hope that helps!
I wonder if there is some underlying algebraic geoemtry concepts behind this, especially after seeing p+q+r = 0 (like with how addition on elliptic curves is defined).
I only solved part one. My solution pretty much same as yours Stright line: x = my + c Parabola: x = nyy Equate: nyy - my - c = 0 Intersection: y = [ m +|- sqrt[mm + 4nc] ] / 2n Specifically... Parabola: x = yy/4a 1st Deriv: x' = y/2a Negative inverse: -1/x' = -2a/y Knowing point Q and the gradient, it's trivial to calculate the x-intercept Giving the normal line from point Q as: x = -y/q + a(qq+2) Using the formula for parabola intersection above y = [ m +|- sqrt[mm + 4nc] ] / 2n where m = -1/q n = 1/4a c = a(qq+2) ... gives the following roots y = (2a/q) [ -1 +|- (qq+1) ] The positive root is the y value at point Q: y = 2aq The negative root, therefore must be the y value at point P: y = (-2a/q)(qq+2) The same value is known in terms of p: y = 2ap Equate them: (-2a/q)(qq+2) = 2ap pq + qq + 2 = 0
Not really. All light beams that hit a satellite dish pass through the focus of the parabola. In this case, the "light beams" meet at another point on parabola
As someone who has actually sat STEP and been admitted to Cambridge to study maths I can tell you that is not true. Are there a lot of people from private schools? Sure. But would the department love to admit someone, regardless of background, if they are an exceptional mathematician? Absolutely.
@@SecondQuantisation there avg fees is waaaaaay too much for most students. Even if they could perform exceptionally, it is ridiculously hard to get in if ur not European or American.
Check out the latest episode of mathmajorchat with Nate Mankovich: th-cam.com/video/WKcTyGMJiz4/w-d-xo.html
We discus the importance of learning computer science for making a successful transition to graduate school for applied math.
As someone going to sit this exam in the summer for this exact reason I feel a little relieved I got it, step questions can be a real pain sometimes
I dont know wether you had some other requests for this but I only asked for it a week ago so to see it being incorporated is really good - keep them coming
Nice problem indeed. Great job, Michael.
This was a great problem.
Thank you, professor.
Nice! I recently walked through some of the multiple choice section on the Oxford MAT
Just did this problem today - the step grind is real 🔥🔥
The thumbnail is incorrect. Your question lists lines PQ and PR but your thumbnail shows lines PR and QR.
I did the same question while practicing conics for IITJEE exam
Please do some more interesting questions on conics
I have fond memories of studying for my STEP in summer 1991. Managed to get a 1 on it (top grade was an A), but hey, it still got me in...
Intresting, they must have changed it. Top grade is S now
Did this question the other week nice to see it here
Amazing, best exercice to go trhough after breaking the fast . Thank you .
lmao
im sitting S2 and S3 this summer, sometimes the questions seem doable and other times it's an absolute trainwreck
Yh this one was fairly easy for a step question. It was only slightly beyond further maths
I think step isn’t ever a trainwreck if you go through each possible route of approaching a problem in your head and understanding which one is most likely to get you to the right answer.
i remember doing this problem before taking the exam in 2018
you should do some from step 2 2020, they were nice
second day comenting this problem:
find the function form Natural to natural that relationates the number of sides of a regular polygon to the numbers of points that are intersections between all the lines that connect each vertice of the polygon to all other vertices
7:08 what about when q=-r (of course it is a special case and depends on a, but still it needs to be mentioned and solved)? Q and R are symmetric points with regards to the X axis (and therefore P is the origin) QR is parallel to the Y axis and has no (real) slope.
8:10 vieta's formula is known. so x1+x2=-b/a (and x1*x2=c/a) therefore q+r=-p
10:42 also should consider if p=0 (the case I mentioned earlier)
I'm pretty sure we can ignore the q=-r case because the normal lines will intersect at the origin only when q=r=0, which breaks the uniqueness requirement, or when a=0, which breaks the requirement that a>0.
I don't even hope to be one of those smart guys that get enrolled I'm just curious
To get into Cambridge no matter what your abilities just get a letter of recommendation from Mike Penn.
May we ask why didn’t you Mike Penn get into Cambridge or Oxford? Inquiring mathematicians want to know.
Probably for the same reason you didn’t get into Harvard or MIT
A diagram would have been nice.
If this was a question at an interview I'd be giving a side eye to anybody not starting by drawing a diagram. And even updating the diagram after finding all those lines go through (0,-2a). Things like this that can make the difference between an offer and a rejection.
I love these videos, but all I can hear is inhalation Dx
Does the word “normal” have any special meaning here because I don’t know any?
'Normal' means the lines are perpendicular to the curve (or more specifically, perpendicular to the tangent to the curve) at that point, hope that helps!
Thanks for the explanation, it really helps!
I wonder if there is some underlying algebraic geoemtry concepts behind this, especially after seeing p+q+r = 0 (like with how addition on elliptic curves is defined).
Ellipses and Parabolas are both conic sections, just brainstorming with you
I love this kind of problems! Keep up the good work!
I only solved part one.
My solution pretty much same as yours
Stright line:
x = my + c
Parabola:
x = nyy
Equate:
nyy - my - c = 0
Intersection:
y = [ m +|- sqrt[mm + 4nc] ] / 2n
Specifically...
Parabola:
x = yy/4a
1st Deriv:
x' = y/2a
Negative inverse:
-1/x' = -2a/y
Knowing point Q and the gradient, it's trivial to calculate the x-intercept
Giving the normal line from point Q as:
x = -y/q + a(qq+2)
Using the formula for parabola intersection above
y = [ m +|- sqrt[mm + 4nc] ] / 2n
where
m = -1/q
n = 1/4a
c = a(qq+2)
... gives the following roots
y = (2a/q) [ -1 +|- (qq+1) ]
The positive root is the y value at point Q:
y = 2aq
The negative root, therefore must be the y value at point P:
y = (-2a/q)(qq+2)
The same value is known in terms of p:
y = 2ap
Equate them:
(-2a/q)(qq+2) = 2ap
pq + qq + 2 = 0
So is this just showing that beams hitting a satellite dish will always bounce back to one location?
Not really. All light beams that hit a satellite dish pass through the focus of the parabola. In this case, the "light beams" meet at another point on parabola
15:00 Have a great day ☀️
Yo
Could not imagine being a mathmo. I will stick to studying econ at Cambridge :)
no, entrance exam for cambridge is trash and people evaluating it are very emotional and amateur. sorry, guys
Answer to the title - Money
As someone who has actually sat STEP and been admitted to Cambridge to study maths I can tell you that is not true. Are there a lot of people from private schools? Sure. But would the department love to admit someone, regardless of background, if they are an exceptional mathematician? Absolutely.
That's just not true though. Even if you're parents are a millionaires but you're not good at maths, you're not getting in. It's as simple as that.
@@two697 but if ur parents aren't millionaires...
@@SecondQuantisation there avg fees is waaaaaay too much for most students. Even if they could perform exceptionally, it is ridiculously hard to get in if ur not European or American.
@@SecondQuantisation are u English?
1 be rich , 2 come from an upper-class family ........
A grade 8 question
One should keep in mind not to provide private information online.
Yeah.............. cambridge student wouldnt stance a chance in JEE mains
Yay, second comment!!