Luxembourg - Math Olympiad Question You should know this trick
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- čas přidán 27. 08. 2024
- Math Olympiad Question
Maths Olympiads are held all around the world to recognise students who excel in maths.
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Luxembourg - Math Olympiad Question You should know this trick
Excellent
sqrt9 = 3, sqrt8 = 2sqrt2
3 - 2sqrt(2). want to make it some squared quantity to get rid of the secondary sqrt.
lets say x = sqrt2, x^2 = 2. we create both of these terms, and are left with + 1 for the constant.
x^2 - 2x + 1, which is a perfect square:
= (x-1)^2
--> sqrt(2) - 1
Let's try to expand this further.
Want (A + B*sqrt(K))^2 = M + Usqrt(K); Where M, U, and K are given.
= (A^2 + KB^2) + 2ABsqrt(K)
Match coefficients:
A^2 + KB^2 = M
2AB = U
--> AB = U/2
--> B = [U/2]/A = U/2A (Isolate for B)
Now plug B back into eq 1:
A^2 + K* (U/2A)^2 - M = 0
A^4 - MA^2 + KU^2 / 4 = 0 [Multiply out by A^2]
A^2 = [M +/- sqrt(M^2 - KU^2)]/2 [So if M^2 - KU^2 < 0, we only have imaginary answers, which doesn't seem to work here]
Our best bet is taking the '+' version given that it is a squared value, and must be positive:
A^2 = [M + sqrt(M^2 - KU^2)]/2
A = sqrt[(M + sqrt(M^2 - KU^2))/2] (Let's just take the positive version here for ease of use also)
B = U/2A
----------------------------------------------------------------------------
We test this out with our starting problem:
-- A = sqrt[(M + sqrt(M^2 - KU^2))/2]
-- B = U/2A
-- A + B*sqrt(K) = sqrt(M + Usqrt(K)),
Ie, our result of A + Bsqrt(K) should be equal to our initial result, with
sqrt [3 - 2sqrt(2)] = sqrt(2) - 1
M = 3, U = -2, K = 2
A = sqrt[(3 + sqrt(9 - 2*4))/2]
= sqrt[(3 + 1)/2]
= sqrt(4/2)
= sqrt(2)
B = -2 / 2sqrt(2)
= -1/sqrt2
A + B*sqrt(K)
= sqrt2 - 1
Great channel
All right