Putnam Exam | 2006: B1
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- čas přidán 8. 09. 2024
- We look at a solution to question B1 from the 2006 William Lowell Putnam Mathematics Competition. This problem has a nice hint of algebraic geometry.
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The factoriztion is also the special case of the formula a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) for c=-1.
wait why didnt i think of this earlier now i feel so dumb
Did the same thing.
I did not know this formula! So, question:
a^3 + b^3 = (a + b) (a^2 + b^2 - ab)
a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc -ca)
Are these two special cases of an even more general formula?
@@35571113 I think last one is wrong
@@user-xu1oo8zw1v Oh... You are right, I meant to copy the formula from the original post, and missed a term. Now the question is not as interesting.
It is less 'magic' if you first notice by inspection that -1,-1 is an obvious solution.
Then, since the geometry is unchanged by translation, set x->x-1, y->y-1
That makes the equation x3+y3-3x2+3xy-3y2 = 0.
But we can factor
x3+y3 = (x+y)(x2-xy+y2), so it is easy to factor the lhs:
(x+y-3)(x2-xy+y2) = 0.
Since the discriminant of the quadratic is < 0, the only real solution is 0,0 for the quadratic
Sure, that is easy to see for this polynomial, but what are you going to do if the solution isn't that easy to spot?
just a little fact:
x^2-xy+y^2+x+y+1
=(x^2+2x+1)+(y^2+2y+1)-(xy+x+y+1)
=(x+1)^2+(y+1)^2-(x+1)(y+1)
=a^2-ab+b^2
>=0
with equality happening iff a=0,b=0
which is equivalent to x=-1,y=-1.
That was the first thing that came to my mind as well!
Just took my first “Putnam” (unofficial, as it was online) as a freshman in college, and it was a really fun experience. Your videos and solutions are great and I’ll probably be watching many more in the future, so thank you
wow gamer doin maffs, really amazing
Yeah. I give myself ‘an official’ unofficial putnam every year. At home. I hadn’t really known about Putnam in college so I went theough my eligibility years without using any tokens to do the exam.
Let's gooooo omicrongaminggggg
Thank you for making and sharing this interesting video. The presented equation is also the equation of the Folium of Descartes for a = -1. The general equation is x^3 + y^3 -3axy = 0. The Folium’s asymptote is x + y + a = 0. The top of the loop coordinates is (3a/2 , 3a/2).
there is also a constant term in the question! I checked the graph and for any value there is an asymptote .. didn't knew it has a name :D Thanks
The graph of x^3 + 3xy + y^3 = 1 degenerating to a point and a line is beautiful and fantastic!
Michael Penn, I'd like to thank you for helping me with analysis. You either know exactly how my brain works or you are an extremely talented professor.
As a retired university physics lecturer I can assure you it's the latter.
Hey, I'm a mathematical layman, but find maths relaxing. Cool that you climb and backflip & stuff.. I'm learning Parkour. Subscribed.
wow your videos are so good the like to dislike ratio is undefined.
Sadly a 9y/o disliked it😭
Jinx!
Very very very clever ratio!!!
Now I'll never know the like/dislike ratio, thanks Susan 😔
I remember taking that exam. I walked out angry at myself for not knowing how to do it. After all these years, you cleared it up for me.
Impressively lucid description of a general method- congratulations.
One slightly different approach is to exploit the fact that the graph is clearly symmetric around y=x and rotate the "image" by 45 degrees. This is accomplished by a substitution x = u-v, y=u+v.
Then the problem becomes (u-v)^3 + (u+v)^3 - 3(u-v)(u+v) = 1, which can be trivially manipulated into
3v^2(2u-1) + 2u^3 + 3u^2 -1 = 0. But 2u^3 + 3u^2 - 1 factors into (2u-1)(u+1)^2. So we have
(2u-1)(3v^2 + (u+1)^2) = 0. Thus either u=1/2 and v is free, or v=0 and u=-1. The area of the triangle in the u-v coordinates is 3*sqrt(3)/4, which needs to be multiplied by 2 for the final answer (because 2 is the determinant of the [[1, -1], [1,1]] matrix that we use to transform u-v coordinates back to x-y)
That's an awesome method. Reading the comments I saw how people came up with so many solutions but I find yours to be the most beautiful.
When I solved this question long ago, I used some known formulas from conic sections.
You solved it just from basic maths and symmetry. That is awesome 🤩🤩🤩
Another way to see that x^2 - xy + y^2 + x + y + 1 = 0 only has the solution x=y=-1:
x^2 - xy + y^2 + x + y + 1 = 1/2 * ((x-y)^2 + (x+1)^2 + (y+1)^2)
Which can of course only be zero if all the squared quantities are zero.
How'd you get the idea to factorize in such a fashion?
3
16. Find all solutions of x^(n+1) − (x +1)ⁿ = 2001 in positive integers x and n. (A-5, Putnam 2001)
This is a very interesting questions
A5? It's certainly gonna be fun then!
(I'm assuming that the +1 is in the exponent of 'x' as well)
@@manswind3417 Yes 😀...thanks for pointing it out
It was a copy paste problem I didn't notice before
This problem has some hidden beauty in it: Rewrite x^3+3xy+y^3=1 by moving 3xy to the right side, and adding missing "cubic" terms to both sides to get x^3+3(x^2)y+3x(y^2)+y^3=1-3xy+3(x^2)y+3x(y^2). This simplifies to (x+y)^3=1=3xy((x+y)-1). Substituting u=x+y and v=x-y, noting xy=(u^2 - v^2)/4, gives u^3=1+(3/4)(u^ 2- v^2)(u-1). Now if u=1, the right side simplifies to give u^3=1 and we get one part of the solution: x+y=1. If u is not =1, we can rewrite the expression as (u^3 - 1)/(u-1) = (3/4)(u^2 - v^2). Using the fact (a^3 - 1)/(a - 1) = a^2 +a+1, and multiplying both sides by 4, we have 4(u^2) +4u+4 = 3(u^2) - 3(v^2). Collecting u terms on the left gives u^2 +4u+4 = -3(v^2). But this is the same as (u+2)^2 = -3(v^2). For real values, this is possible only if u+2 = 0 and v=0, which means x+y = -2 and x-y = 0, giving the isolated point (-1,-1) as a solution.
This was absolutely amazing. Thank you, from Italy.
I solved the problem in the last summer and it took some time. Eventually I realized, because of (x, y) - (y, x) symmetry of solutions that the solutions must be symmetric around line y = x. However when I tried to find the solutions along the lines which are perpendicular to y = x, there weren't many. Which was against my intuition since for each x or y there are possibly 3 solutions. Then the factorization was easy to find.
Great video. I have a nice problem from Polish Math Olympiad from 2009/2010:
Let a, b are rational such that:
a^3 + 4a^2*b = 4*a^2 + b^4.
Show that sqrt(a) - 1 is the square of rational number.
Greetings!
A very cool problem because many different tools are used and the solution is clean.
These are unbelievable videos Michael. Production quality is great (the color tone too)
18:08
hello again
@@i.am.jihoonk How are you today?
pretty noice
He's back! :O
heh. I punched the equation into desmos. You can see the little ellipse forming around -1, -1 if you rock the -1 value back and forth around -1. With it equal to -1 desmos doesn't show anything at -1,-1 at all. Interesting curve.
Great factoring technique! I love your videos! 🤩🤩🤩
Without directly Factoring: By symmetry exchange with x and y, we know that a line of symmetry of the equilateral triangle falls along y=x. Lines of symmetry of an equilateral triangle must contain a vertex and the equation only allows (-1,-1) and (1/2,1/2) as candidates. A base must be perpendicular to the line of symmetry, so x+y=c for some real c for the base. Plugging in y=c-x to the cubic yields a quadratic. All x values are allowable if c=1. If c1, discriminant requires c=-2 but only (-1,-1) falls on the base after solving the resulting quadratic. A base must have at least two points so c=-2 is not allowed. So, we know the base must be x+y=1, a vertex must be (-1,-1) since (1/2,1/2) would be collinear with the base. The point (1/2,1/2) is where the altitude from that vertex intersects the base. The length of the altitude between vertex and midpoint is 3/sqrt(2) by the Pythagorean theorem. 3/sqrt(2) =r sqrt(3)/2 by trigonometry which implies r=sqrt(6), r side length of the equilateral triangle. So the other vertices are solutions to (x+1)^2+(y+1)^2=6 and x+y=1. Substitute then solve with the quadratic formula.
The linear factor can be got by noting that the cubic polynomial can be written as (x+y)^3 + 3xy - 3 x y^2 - 3 x^2 y -1 = (x+y)^3 + 3 xy (1 - (x + y)) -1 which is clearly zero when x + y = 1. That makes the step of finding the coefficients of the quadratic a bit easier.
Wow, I loved this one! Solved it straight away (before watching the video ofc). Wonder if anyone went the same way?
[ S P O I L E R S ]
let G be the graph of the thing, then (x, y) ∈ G (y, x) ∈ G. Also there are two unique points (x, x) ∈ G.
Wait, maybe some whole line of type y = k - x (perpendicular to y=x) belongs to G? It does! Then the polynomial x³+3xy+y³-1 is divisible by x+y-1
For the remainder, it looks neat but not factorizable. It should be equal to 0, so what about its mins and maxes? Oh, it actually has a strict minimum of 0 in (-1, -1), that's nice. Leaves us with a single point.
Thus, G = { line } U { point }, the rest is arithmetics.
When there's symmetry between x and y, I like to change basis to highlight that symmetry. If we denote u=x+y, v=x-y, then:
x=(u+v)/2 and y=(u-v)/2
The problem x^3 + y^3 +3xy =1 then becomes:
1/4*u^3 + 3/4*uv^2 +3/4*(u^2-v^2) - 1 = 0;
*4 and collecting v terms together gives:
u^3 +3u^2 -4 + 3uv^2-3v^2 = 0;
(u-1)(u^2+4u+4 + 3v^2) = 0
(u-1)((u+2)^2 + 3v^2)=0;
This has solutions at u-1=0, or x+y=1, and at u=-2 and v=0. v=0 means x=y, and u=-2 means x+y=2x=-2, so x=y=-1
This was a really nice one! Super cool to see you do it using some intuition rather than having to rely on knowing the standard form of a conic section.
I first spotted (1,0) and (0,1) as solutions, and noticed (1/2, 1/2) as well. This led me to check that (x,1-x) is on the curve, meaning that there could only be one point off the curve for the statement to be true. (-1,-1) worked, giving the triangle's height as h=sqrt(2)(3/2) and A=h*h/sqrt(3).
I got as far as noticing (1,0), (0,1) and (1/2,1/2) are solutions. These are collinear which is kind of a pointer. In general cases a line cuts a cubic curve in three points. In this case the middle intersection would give an asymmetry. So (I didn't quite get) the line x+y-1=0 must fall entirely on the curve, providing a linear factor. I tried the factorisation anyway, by computer, result! Nice problem, great brain exercise EDIT: Using the symmetry substitute x=u+v, y=u-v. The curve is (2u-1)(u^2+2u+1+3v^2). The quadratic term is (u+1)^2+3v^2, giving a point ellipse
I got so close. On my own I got the line x+y=1 and the curve x^2 - xy + y^2 + x + y + 1 = 0 but couldn't find any solutions. In previous problems I had remembered I could make a quadratic in y, but it didn't occur to me here. Oh well.
You really got the competition meta heuristics going.
Once again, fair and square. So nice. Big like! 👍👍
Oooooh, the factorization, of course!
To find the nature of that quadratic, I eliminated the x and y terms by shifting the axis. I assumed we'd shift the axis to (h,k). Then I equated coefficients of x and y to 0 (in the shifted equation) and got h=k=-1. Then the equation became x²+y²-xy=0. Using the AM-GM inequality, we can see that this is impossible (except for the shifted origin). So the only possible point which the quadratic satisfies is (-1,-1). Solving after this point was easy.
Absolutely LOVED IT!!!!
ingenious solution
We can also find length of altitude directly by formula
Wow this is amazing.
A big thanks to you
What a nice problem and solution!
x^3 + 3xy + y^3 = 1 -> (x+y)^3 - 1 = 3x^2y + 3y^2x - 3xy -> (x+y-1)((x+y)^2 + x + y + 1) = 3xy(x+y-1) than "cancel" x+y-1 and proceed as in the video
I don't how my mind told me that I should try substituting x = X-1 and y = Y-1 into the equation after a few minutes of thinking. That worked out very well and I was able to factorise the remaining expression using basic algebra I knew already. I don't know I would have done this otherwise. Maybe by substituting values and seeing that a line is formed?
Hi,
For fun:
1 "so on and so forth",
1 "and so on and so forth",
1 "ok, great",
1 "let's may be go head and".
You are really clever
This was so satisfying
We can use a well-known formula
x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)=(1/2)(x+y+z)[(x-y)²+(y-z)²+(z-x)²]
to obtain x³+y³＋(-1)³-3xy(-1)=(x+y-1)(x²+y²+1-xy+y+x)=(1/2)(x+y-1)[(x-y)²+(y+1)²+(-1-x)²].
i like this color correction
This one was also presented by ProfOmar youtube channel.
What if used (x-x0)^2+(y-y0)^2=0 to specify the single point? That looks nice also, but don't know what it is...
Isn't equilateral triangle always unique, when there is a line and a point?
I checked on "Desmos" the graph of x^3 + 3xy + y^3 = 1 and it showed only the straight line but not the point (-1,-1). Why?
I would like to know an answer to a question: how to conclude from implicit derivation that y'(x) is essentially a constant. There is so much symmetry going on that it seems plausible to achieve. I wasn't able to do that when I tried.
By implicit differentiation with respect to x you get 3x^2+(3xy'+3y)+3y^2y'=0 (middle terms grouped to illustrate product rule).
Factor into (x^2+y)+y'(x+y^2)=0
Apply symmetry argument. Since f(x,y)=f(y,x), x^2+y=x+y^2. Call this quantity z.
Rewrite as z+y'z=0
Factor to get z(1+y')=0.
z is variable (it must take on any value generated by x+y^2), so it cannot be the constant 0.
So 1+y'=0 or y'=-1.
@@chunkypunkyboy Very neatly done, thanks.
@@chunkypunkyboy and of course a symmetric solution to x² + y = 0 is (-1, -1), which is the spare point and leaves the derivative indeterminate. hm, wonder why the origin doesn't work
Yes, but if x^3 + x^2*y+ x*y^2 + y^3 = 1 and y = f(x), what is the integral from 0 to 1 of f(x)?
nice problem
I am glad it was one of the easiest questions 😅
Usefull Indentity
x^3+y^3+z^3 - 3xyz=((x+y+z)/2)*((x-y)^2+(y-z)^2+(z-x)^2)
Professor,
Can you make a lecture series on Sets and Logic?
There is a proof writing playlist including and logic operators
@@goodplacetostop2973 Can you share the link?
@@saniap234 I’ve tried 5 times earlier but the comment was removed by anti-spam filters 😤 I’ll try again
@@saniap234 Here : czcams.com/play/PL22w63XsKjqykuLOimt7N59e6wn_aE0wd.html
@@goodplacetostop2973 Thanks a lot for sharing. I saw it
meu simulado ta focado sim, olha a porra dessa questão!
ne mano KKKKKKKKK
os caras nao tem dó
Out of curiosity... I graphed this curve on Geogebra. Can someone help me understand why it displays the same as x+y-1=0 (just one of the factors)?
After watching a little further, I see that the other factor only has 1 point in the Real plane... so I guess the point (being infinitely small) is not represented in the graphing calculator.
quite amazing honestly.
Genial, 🐾🎸
How is this possible? When i type the equation in desmos its a straight line with the equation y=-x+1
I guess Desmos doesn't only show curves and not unique points (unless you tell it explicitely), so you don't get to see the point (-1,-1) solution of x2 - xy +y2 + x + y + 1 =0
I've seen this problem copied in Ukraian university olympiad in 2018(
???
“Okay cool....” 😂
Damn, that was pretty
Syber math
I feel like a detective after watching this one 😄
ez
This is "easy"? Ok how were the other problems 🥺
Well I could directly think as x+y=1 because cubing it we would directly get the curves equation 😁😁😁😁😁
it doesnt
@@farmer4121 u can cube and see , I am just talking of the curves eqn not soln
@@vibhanarayan9668 You really can't. Cubing x+y=1 in no way gets you the equation of the original curve.
@@TJStellmach (x+y)^3= x^3+3xy(x+y)+y^3=x^3+3xy+y^3=1, so in the particular case that x+y=1 cubing it does actually give you the original equation of the curve, although in general cubing x+y doesn't give you the equation because x+y is not always 1
I am glad to get in touch with you. I am a person who creates and posts math related videos.
I make videos not only for ordinary mathematical problems but also for mathematical puzzle problems. In addition, I am doing to cite the math problems of overseas videos and create and introduce similar themes in my own way.
< However, it's only been about a month since I started posting videos. >
This time, as a special project of the video channel I am creating, I am planning the following.
"I am planning to introducing the math channel of overseas videos, to explain the problems with using your method and/or my own method to solve, and creating and introducing similar and revised problems. Also, for answers to my own problems, I will explain the videos at a later date. (I plan to upload.) "
Therefore, first of all, I would like to inform you that I will use the problems used in your video and quoting a part of the explanation. Instead, I'll show you your video channel in my video.
I would like to use this problem: x^3+3xy+y^3=1
and another problem: ∑(k= from 1 to m) (k!) = n^2
If you do not admit that your videos will be quoted due to the copyright, please inform me so.
(If quoting is prohibited, please contact me as soon as possible. I will cut that part when creating the video ...)
Also, regarding my video channel, if you can promote it in your video, feel free to quote it.
Then, please understand this matter.
18:08
THE TIME WIZARD
@@goodplacetostop2973 HOW
WTF
@@chongweifongchongweifong7180 He went in the playlists and he found unlisted videos before they are "officially" published
@@goodplacetostop2973 forsenPls ARGH forsenPls UGGHH gachiHYPER THE TIME WIZARD gachiHYPER