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Math Olympiad | A Nice Algebra Problem | How to solve for 'X' and 'Y' in this problem?
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- Опубликовано: 23 дек 2023
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Good solid algebra. Yes a bit slow, but for someone like myself who has not done much serious math since student days (half a century ago now), better to include all the steps rather than skip some. And yes, the obvious real solutions from simple inspection are x=2, y=1 and x=1, y=2, however that's not the point of the exercise. Whether it's a good Math Olympiad problem or not, I'm afraid I can't judge.
thanks
You're welcome!
x=2.y=1
x=1,y=2
You can use the identity
_uₘ₊ₙ = uₘuₙ - uₘ₋ₙ(xy)ⁿ_
where _uₙ = xⁿ + yⁿ_
Thus: _u₀ = 2, u₁ = 3, u₅ = 33_
_u₂ = u₁² - u₀(xy)¹ = 9 - 2xy_
_u₃ = u₂u₁ - u₁(xy)¹ = 3(9 - 2xy) - 3xy = 27 - 9xy_
_u₅ = 33 = u₃u₂ - u₁(xy)² = (9 - 2xy)(27 - 9xy) - 3(xy)²_
∴ _243 - 135xy + 15(xy)² = 33_
⇒ _(xy)² - 9xy + 14 = 0_
⇒ _(xy - 2)(xy - 7) = 0_
∴ _xy = 2_ or _7_
Note that _x_ and _y_ are the roots of
_t² - (x + y)t + xy = 0_
Case (i): _xy = 2_
_t² - 3t + 2 = 0_
⇒ _(t - 1)(t - 2) = 0_
∴ _x, y = 1, 2_ in either order.
Case (ii): _xy = 7_
_t² - 3t + 7 = 0_
⇒ _t = ½[3 ± √(9 - 28)]_
∴ _x, y = ½(3 + √19), ½(3 - √19)_ in either order.
I graph it in demos , only 2 real sol
Edit : Sadly at the last part you messed up 9-28 is -19 not 19 so 2 complex roots
By "looking at it and thinking for a second," x=2, y=1 and x=1, y=2 are solutions.
(1;2) (2;1)
x = 2, y = 1 or vice cersa
you can see it must be 2 and 1 immediately.
Комплексные корни не интересны.
@@user-lf1bm8fm2g has to be integers, both greater than 1. simple.
@@neilmccafferty5886Doesn’t need to be integer
this has sipleset way since the numbers of X AND Ymyst be integer X+Y=3 RESULT X=1 and Y=2 X5+y5=1+2^5=1+32=33
(1, 2), (2, 1). They €R².
X= 2/1, Y=2/1.
Очень простое уровнение, почему так мучается?
Надо немножко подумать,потом мучатся,точно !!!
33+3+33+3=72
66+6=72
(66/72) + (6/72)=1=100%
Это же устно решается, сразу в глаза бросается, что х=1, у=2 или х=2, у=1.
Hо система может иметь др. решения.
We have an equation of grade 5. But you have just four roots or solutions por x and y
While it might be nice to see how to work out another solution, it's possible any additional solution would be extraneous, or a duplicate of an existing solution.
But most importantly, any valid solution for the problem as a whole, has to work in both equations, and another solution to the "x⁵ + y⁵=33" equation, if it exists, might not work in the "x + y = 3" equation.
Such system has 5 solutions or less.
X:2 y:1 or x:1 y:2
X = 1 , Y = 2
or
X = 2 , Y = 1
x=1, y=2 or x=2, y=1
Who is your audience? Your audience is people who can appreciate and learn what you are showing them. Such people must have mastered high school algebra, and have presumably mastered grade school arithmetic, preparing them to learn the more advanced techniques and principles you are showing them. They--we--don't need to see basic numerical division belabored. They--we--don't need the detailed steps of factoring a quadratic. With the factors given, we need about three seconds to verify them, then we can move on.
The value in these solutions is the overall scheme, recognizing the larger patterns that we can use to pry the problem apart. But you bury those steps in the minutia that we have already mastered, making it hard to find the precious grains of gold mixed with the dross of first year algebra and fourth grade arithmetic.
These are brilliant solutions. The presentations would be so much better if you demonstrated the brilliance and moved routinely through the routine mechanics.
X=2 et Y=1
Can anyone explain why there are only 4 roots, not 5?
An degree n equation doesn’t need to have n distinct root , some can be the same so the amount of root is not more than n
@@Verdiw I know that so which one is the double root?🌲
Porque si sustituyes y=3-x
Te va a quedar x^5+(3-x)^5-33=0 y el término quíntico se irá, quedando un polinomio de grado 4
Supposed to be 5 roots
If you set x = 3-y using equation 2 then the highest term of x^5 is - y^5, which cancels out with y^5 in equation 1. This reduces the problem to solving a quartic and results in four roots.
5 or less.
X^5 + Y^5 # (X + Y)^5
Х=1, У=2 ; У=2, Х=1.🙉🙉🙉🙉🙉
・・これを日本の大学入試は
5分以内で全てやらにゃあかん。
このスピードでは既に不合格。
Sorry, but I find your video disappointing for 3 reasons:
1. You did not specify in which domain you are solving, integers? real numbers? complex numbers?. You solved for complex, but you did not specify in advance. In math alympiads, problems are always specified more precisely. You omitted important information. This is an error often commited by a number of people doing this kind of videos, and I find it hugly annoying.
2. Your solution is correct but the problem can be solved more easily. In this kind of problems, there are often more efficient approaches than the brute force method. The colculations get easier when you realise that that x5+x5=(x+y)(x4-x3y+x2y2-xy3+y4) and continue from that. You then only need 4th and second powers of (x+y) which makes the fomulas a little bit simpler and thus less error prone.
3. This comment is mino: state your sources. Like xxx math olympiad problem from yyyy where xxx is the orangisation/country or whatever and dddd is the year.
For poitn 3, I have the impression that many youtubers just copy the problems from other youtube videos as while there are so many interessing math olympiad problems, you oftend find the same problems over and over again.
Evaluate before doing all of your shit. If x+y = integers. and x⁵+y⁵= integer then x and y is integer then evaluate b²-4ac, if it's negative, stop. No point of continuing, i^odd integers. Will always has I as result.
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