what i did is, make formula 1 + formula 2. AND make formula 1 - formula 2. AND as you did at the beginning, made formula 1 mutiple by formula 2. so in the end you have x^4-6x^2y^2+y^4=0 and x^4-14x^2y^2+y^4=-1 (if x+y and x-y both are not 0). then you have x^2y^2=1/8 and x^2+y^2=1. can get same result without using imaginary. anyway you solution is awesome!
Great solution! I think it would also be helpful to provide some visual intuition about the relationship between x+iy and y+ix, and how that relates to complex conjugates.
After getting x/y, one can just put it into equation x=3x^2y-y^3 and find y^2. There will be answer in radical form, like (sqrt(2)-1)/(sqrt(4-2sqrt(2)) (one of possible solution) which is the same as cos(3pi/8).
I've recently seen the following problem (German national Maths Olympiad 2020), which seems really easy, though I couldn't figure out a way to prove the left inequality: Show that the equation x(x+1)(x+2)...(x+2020)-1=0 has exactly one positive real solution a. Furthermore prove the following inequality: 1/(2020!+2020)
The key to both is the intermediate value theorem (IVT). en.wikipedia.org/wiki/Intermediate_value_theorem (It is a calculus theorem but the intuition is there: that continuous functions don't "skip" over any values as you traverse it over a continuous interval). Let's start with the first part. Let f(x) = x(x+1)(x+2)...(x+2020) - 1. Then we wish to show that a positive real a for which f(a) = 0 exists, and furthermore that this positive real solution is in fact unique. Notice that f(0) is negative, and f(1) is positive. Also f is continuous. Thus, by the IVT there must be an a between 0 and 1 (0 < a < 1) for which f(a) = 0. To show its uniqueness, notice that for values of x > 0, f(x) is clearly a strictly increasing function. What this means in the context of our problem is that positive reals less than a are LESS THAN f(a) = 0, and positive reals greater than a are GREATER THAN f(a) = 0. Thus, no other positive real but a can be a zero of f(x). As for the inequality, note that in the first part of the question we showed that 0 < a < 1, using the intermediate value theorem, since f(0) was negative and f(1) was positive. Other than being nice numbers to work with the choice of 0 and 1 as inputs into f is completely arbitrary. Rather, in general if we could show that f(n) < 0 (negative) and f(m) > 0 (positive), then we have n < a < m. Thus, if we could show that f(left_num_in_the_ineq) is negative, while the f(right_num_in_the_ineq) is positive, then that would be sufficient to show that left_num < a < right_num. Try it out!
@@abububen6148 thank you for the answer! I actually did exactly that, using IVT, however it's not that trivial to show that f(left habd side of inequality) is less than 0, is it? I ended up with huge products, not looking too promising...
There are actually 9 solutions, not 5. You discarded 4 purely imaginary solutions which are obtained by multiplying x by i and y by −i in each of the four nonzero real solutions pairs.
x+y=(x-y)(x^2+4xy+y^2) x-y=-(x+y)(x^2-4xy+y^2) (x+y)(x-y)[(x^2+y^2)^2-(4xy)^2+1)=0 x=y => x=+-1/kök2 olur ki denklemlerden birini sağlamaz. x=-y =>2y^2=-1 olur ki buradan y reel sayısı bulunamaz. O halde (x^2+y^2)^2-(4xy)^2+1=0 olacaktır. Bu eşitlikte x=sinm, y=sinm dönüşümü yapılırsa (sin2m)^2=1/2 çıkar ki buradan da |sin2m|=1/kök2 => m=(pi/8)+(pi*n/4) (n bir tam sayıdır.) O halde x=sin(pi/8)+pi*n/4) ve y=cos(pi/8)+pi*n/4) çıkar.
Something similar without using complex numbers would be using the compound angle formula for tan(3t) where tan(t)=y/x
Thanks for these elegant questions with their solutions it helped me a lot during my Olympiad preparation
what i did is, make formula 1 + formula 2. AND make formula 1 - formula 2. AND as you did at the beginning, made formula 1 mutiple by formula 2. so in the end you have x^4-6x^2y^2+y^4=0 and x^4-14x^2y^2+y^4=-1 (if x+y and x-y both are not 0). then you have x^2y^2=1/8 and x^2+y^2=1. can get same result without using imaginary. anyway you solution is awesome!
Great solution! I think it would also be helpful to provide some visual intuition about the relationship between x+iy and y+ix, and how that relates to complex conjugates.
After getting x/y, one can just put it into equation x=3x^2y-y^3 and find y^2. There will be answer in radical form, like (sqrt(2)-1)/(sqrt(4-2sqrt(2)) (one of possible solution) which is the same as cos(3pi/8).
Thank you, I love this problem.
Undoubtedly ,very good solution . Keep it up!!!
I've recently seen the following problem (German national Maths Olympiad 2020), which seems really easy, though I couldn't figure out a way to prove the left inequality:
Show that the equation x(x+1)(x+2)...(x+2020)-1=0 has exactly one positive real solution a. Furthermore prove the following inequality: 1/(2020!+2020)
damn i also struggle with left hand side inequality, i hope to see solution of this aswell now
The key to both is the intermediate value theorem (IVT).
en.wikipedia.org/wiki/Intermediate_value_theorem
(It is a calculus theorem but the intuition is there: that continuous functions don't "skip" over any values as you traverse it over a continuous interval).
Let's start with the first part. Let f(x) = x(x+1)(x+2)...(x+2020) - 1.
Then we wish to show that a positive real a for which f(a) = 0 exists, and furthermore that this positive real solution is in fact unique.
Notice that f(0) is negative, and f(1) is positive. Also f is continuous.
Thus, by the IVT there must be an a between 0 and 1 (0 < a < 1) for which f(a) = 0.
To show its uniqueness, notice that for values of x > 0, f(x) is clearly a strictly increasing function. What this means in the context of our problem is that positive reals less than a are LESS THAN f(a) = 0, and positive reals greater than a are GREATER THAN f(a) = 0. Thus, no other positive real but a can be a zero of f(x).
As for the inequality, note that in the first part of the question we showed that 0 < a < 1, using the intermediate value theorem, since f(0) was negative and f(1) was positive. Other than being nice numbers to work with the choice of 0 and 1 as inputs into f is completely arbitrary. Rather, in general if we could show that f(n) < 0 (negative) and f(m) > 0 (positive), then we have n < a < m.
Thus, if we could show that f(left_num_in_the_ineq) is negative, while the f(right_num_in_the_ineq) is positive, then that would be sufficient to show that
left_num < a < right_num. Try it out!
@@abububen6148 thank you for the answer! I actually did exactly that, using IVT, however it's not that trivial to show that f(left habd side of inequality) is less than 0, is it? I ended up with huge products, not looking too promising...
@@fix5072 exactly same problem, i tried to do it in many ways- solution must be either super smart and elegant or super obscure.
Something seems a thing with x and y on left, I can feel it 🤣
⭐️
Noice!! There is also the sub y=kx that works
When you see this you know its something complex.
you have forgotren (x,y)=(0,0)
I solved it using y=mx
Yes,this is the general way to solve that system of equations.
There are actually 9 solutions, not 5. You discarded 4 purely imaginary solutions which are obtained by multiplying x by i and y by −i in each of the four nonzero real solutions pairs.
x+y=(x-y)(x^2+4xy+y^2)
x-y=-(x+y)(x^2-4xy+y^2)
(x+y)(x-y)[(x^2+y^2)^2-(4xy)^2+1)=0
x=y => x=+-1/kök2 olur ki denklemlerden birini sağlamaz.
x=-y =>2y^2=-1 olur ki buradan y reel sayısı bulunamaz.
O halde (x^2+y^2)^2-(4xy)^2+1=0 olacaktır.
Bu eşitlikte x=sinm, y=sinm dönüşümü yapılırsa (sin2m)^2=1/2 çıkar ki buradan da |sin2m|=1/kök2 => m=(pi/8)+(pi*n/4) (n bir tam sayıdır.)
O halde x=sin(pi/8)+pi*n/4) ve y=cos(pi/8)+pi*n/4) çıkar.
x + y = (x-y)^3