Fit that Curve, Part 2

Geometry Level 5

Find the largest positive integer $n$ such that we can fit a conic through any $n$ points in $\mathbb{R}^2$ .

Clarification : A conic is a curve in $\mathbb{R}^2$ given by an equation of the form $c_1+c_2x+c_3y+c_4x^2+c_5xy+c_6y^2$ $=0$ , where at least one of the coefficients $c_k$ is nonzero.

The answer is 5.

1 solution

Otto Bretscher
Mar 30, 2016

If we plug $n$ points into the given formula of a conic, we end up with $n$ homogeneous linear equations in 6 variables, $c_1,...,c_{6}$ . If $n<6$ , we have more variables than equations, and there will be non-trivial solutions (with at least one of the $c_k$ being nonzero), thus giving us a conic through 5 or fewer points.

However, a little computation shows that it is impossible to fit a conic through the 6 points $(0,0), (1,0), (2,0), (0,1), (0,2)$ and $(1,1)$ .

Thus the answer is $\boxed{5}$ .

What's that ccomputation?

Saurabh Chaturvedi - 5 years, 2 months ago

The coefficients of 1, $x$ , and $x^2$ must be zero since the curve runs through $(0,0),(1,0), (2,0)$ . Likewise the coefficients of $y$ and $y^2$ must be zero since the curve runs through $(0,0),(0,1), (0,2)$ . Finally the coefficient of $xy$ must be zero since the curve runs through $(1,1)$ . Thus there is no such conic.

Otto Bretscher - 5 years, 2 months ago

@Otto Bretscher Sir, could you please help me out and think about this........if we have 5 arbitrary non collinear points A,B,C,D and E......and now, the lines joining A-B , B-C , C-D , D-A are named L1 = 0 , L2 = 0 , L3 = 0 , and L4 = 0 respectively, we find that the general equation of the conic is (L1) (L3) + a (L2)*(L4) = 0............where a is a constant determined by the fifth point given to us..................Does this approach work??? Or did I just get lucky???

Aaghaz Mahajan - 2 years, 7 months ago

Yes, this approach works for conics. Can you generalise the method to cubics and curves of even higher order?

Otto Bretscher - 2 years, 7 months ago

Well Sir, I tried and found out that for a cubic and a quartic, 9 and 14 points are needed.....(quartic took a bit of time!!)......so, if the general pattern holds, I think that for a degree n curve, a minimum of n*(n+3)/2 points are needed........I still need to prove whether the pattern holds tho.......!!!

Aaghaz Mahajan - 2 years, 7 months ago

@Aaghaz Mahajan – The customary way to do this, at least since the time of Euler, is by solving linear equations (that's the approach I took in my solution). Since the equation of a curve of degree $n$ contains $\frac{(n+1)(n+2)}{2}$ parameters, you can always fit an $n$ th degree curve through one fewer points since you have fewer homogeneous linear equations than unknowns.

To make the solution complete, you also have to provide an example of $\frac{(n+1)(n+2)}{2}$ points in the plane through which you cannot but a curve (six points in the case of a conic).

Otto Bretscher - 2 years, 7 months ago

@Otto Bretscher – @Otto Bretscher Yes Sir!!!!! I took the same approach which you have stated here.......!!! I was just confused whether this is a good method or not......because, even I found 9 points for a cubic by making linear equations......!! But I am so happy right now!!! Thanks a lot sir!!! :)

Also, the example for 6 points in a plane can be found very easily, so all in all the solution is completed!!!

Aaghaz Mahajan - 2 years, 7 months ago

@Aaghaz Mahajan – Well, show us how you find the six points, please, just to make the solution complete.

Otto Bretscher - 2 years, 7 months ago

@Otto Bretscher – Well sir, this was my approach..............We see that a conic needs 5 things to be determined exactly, viz. Eccentricity, Scaling factor, Rotation, absicca and the ordinate of the center..........So, 5 points will suffice to find a conic.......Now, if we take any 5 points lying on a conic, we can select the sixth point not lying on it, which won't give an equation for the curve........

Aaghaz Mahajan - 2 years, 7 months ago

@Aaghaz Mahajan – This works only if we know that there is only one conic through the five points.

I agree that in the simple case of a conic this is all pretty obvious, but it gets more interesting in the case of a cubic. How many cubics run through nine given points? My country man Euler and Cramer had a spirited discussion about that issue ("Cramer's Paradox"), which contributed a lot to the development of linear algebra. See here

Otto Bretscher - 2 years, 7 months ago

@Otto Bretscher – @Otto Bretscher Yes Sir, I agree!!!! And I read about Cramer's theorem and Cramer's paradoz after your reply!!

Sir, on the other hand, can I ask you for a small favor??? Could you share your email??? As I have some doubts regarding mathematical papers.........You see, I saw that you are a professor and also, you have published books, so I thought I could take your help.........Please tell me whether I should communicate further on email or this website only........Thanks in anticipation!!

Aaghaz Mahajan - 2 years, 7 months ago

@Aaghaz Mahajan – Yes, of course, feel free to write at otto@post.harvard.edu . If we are addressing specific questions related to Brilliant, it may be better to write here, so that others can participate.

Otto Bretscher - 2 years, 7 months ago

@Otto Bretscher – Thank you Sir........I have sent you an email...........!!

Aaghaz Mahajan - 2 years, 7 months ago

@Otto Bretscher – Umm.... Sir???

Aaghaz Mahajan - 2 years, 7 months ago

Is the formula correct tho??? I mean, is this problem solved by someone already??

Aaghaz Mahajan - 2 years, 7 months ago

Fit that Curve, Part 2

The answer is 5.

1 solution

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