Solving This System of Equations in 2 Ways | Harvard-MIT Mathematics Tournament 2000

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@DaveyJonesLocka 2 жыл бұрын

I was totally expecting the first technique. By contrast, that second technique blew my mind. It is so simple, yet so elegant. I can’t believe it wasn’t part of my equation solving arsenal. Well, it is now!

@user-mc4vd1cx9b 2 жыл бұрын

Check IMO2019 shortlist's A3

@mcwulf25 2 жыл бұрын

I did it another way. Factorised the second eqn (x+y)(x^4-..........+y ^4) = 82 First brackets is 2. So x^4 - x^3.y + x^2 y^2 - x.y^3 + y^4 = 41 Subtract from this (x+y)^4, expanded, which is just 16, and we get after factorising out -5xy -5xy(x^2 + xy + y^2) = 25 The brackets is just 4-xy after substituting in the square of eqn 1. The rest is like solution 1 where we solve for xy and substitute into eqn 1.

@wise_math 2 жыл бұрын

Hello there mcwulf, consider look to my channel too for similar math olympiad problems. Thanks and regards.

@Skyler827 2 жыл бұрын

You introduced a new variable in the second method, but it seems like it would be easier to just define y= 2-x and put that in the second equation and just solve for x.

@bowlteajuicesandlemon Жыл бұрын

It's doesn't seem that much easier to me. You have to factor a cubic polynomial, you can take the factor (x-y) out to make it quartic, while this solution is much quicker.

@reeb3687 9 ай бұрын

it just looks so nice when the a^5 cancels

@themathsgeek8528 Жыл бұрын

The second method is something I use often so it was nice to see it here!

@keinKlarname 2 жыл бұрын

I really like the 2nd approach.

@wise_math 2 жыл бұрын

Hello there! consider look to my channel too for similar math olympiad problems. Thanks and regards.

@rlouisw 2 жыл бұрын

I did it the first way. The second method is very clever, and it's why I watch videos like this. My brain doesn't often accept the method that makes the problem harder, but in some cases that's the easiest thing to do.

@fedorlozben6344 2 жыл бұрын

The second one was as so unusual! Very interesting substitution

@star_ms 2 жыл бұрын

This is going in my list of tools. Thanks

@wes9627 8 ай бұрын

I only know one way to solve this problem. Substitute x=1+z and y=1-z into the given equation and rearrange to (z+1)^5-(z-1)^5-82=0. Noting that odd powers of z cancel out and using Pascal's Triangle: 1 5 10 10 5 1, we get 2[5z^4+10z^2+1]-82=0 or z^4+2x^2-8=0, which has roots z^2=(-2±6)/2=2 or -4 or z=±√2 or ±2i. It follows that x=1+z=1±√2 or 1±2i and y=1-z=1∓√2 or 1∓2i.

@ilayday1 Ай бұрын

Why do u substitute it like that?

@Amoeby 2 жыл бұрын

Why did you reject complex solutions? They are quite fitting the system. (1+2i)^5 = 41+38i and (1-2i)^5 = 41-38i. So their sum is equal to 82 and sum of the 1+2i and 1-2i is equal to 2.

@ropenutter6321 2 жыл бұрын

It's because the question asked for real number solutions. He says it in the first 5 seconds of the video.

@sdspivey 2 жыл бұрын

@@ropenutter6321 Complex numbers ARE real. Just as real as negative numbers.

@ropenutter6321 2 жыл бұрын

@@sdspivey They certainly do exist BUT they are not real numbers as in they aren't in the set of real numbers, just as 1/2 is not an integer complex numbers do not belong to R but they do belong to C.

@Amoeby 2 жыл бұрын

@@ropenutter6321 oh, yeah, I missed that. That explains everything.

@charliebrett7510 2 жыл бұрын

4:00 where do you get z^2 - 2z + 5 from?

@badribishaldas9627 2 жыл бұрын

Wonderful approach

@wise_math 2 жыл бұрын

Hello there Badri! consider look to my channel too for similar math olympiad problems. Thanks and regards.

@willbishop1355 2 жыл бұрын

Nice problem. I did it the first way, solving for xy. I also noticed that xy has to be negative, because if x and y are two positive numbers that add up to 2, then x^5 + y^5 cannot be larger than 32. So we can reject the xy = 5 solution right away and assume xy = -1.

@LouisLeCrack 2 жыл бұрын

Why can't x^5+y^5 be larger than 32?

@cristianionita8359 2 жыл бұрын

i wrote y=2-x, obtained a quartic and applied newton raphson for it. starting from 0, after 3 or 4 iterations i got something like -0.41421 which looked oddly similar to sqrt(2) after the decimal point.

@ranshen1486 2 жыл бұрын

The quartic can be factored into the form (y^2+ay+b)*(y^2+cy+d).

@cristianionita8359 2 жыл бұрын

@@ranshen1486 nice, good to know

@GillAgainsIsland12 2 жыл бұрын

That second method was very clever. Conjugates. Of course.

@genosingh 2 жыл бұрын

I decided to represent x^5+y^5 as product of x+y and x^4+y^4 and subtracting products appropriately and this allows us to substitute xy=z and then solve the quadratic, and plug into the first eqn to solve for x and y.

@robyzr7421 2 жыл бұрын

Ok until 1.38 time. Then putting (x^2 +y^2) = (x +y) ^2 - 2xy and simplifying I find this : (x+y) ^5=(x^5+y^5) - 5x^2y^2(x+y) +5xy(x+y)^3 and so xy =-1 v xy = 5. Then find x and y by a simple system : x^2 + y^2 = 4 - 2xy.. =6 and x + y = 2.... x =1+ 2^0,5 V y = 1- 2^0,5