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Onerous And Complicated Take a look at Query – Thoughts Your Selections


Because of Adarsh for the suggestion! I’m instructed this was on India’s JEE Superior examination, and I’ve additionally seen it within the Harvard-MIT math competitors.

Let x, y be actual numbers satisfying:

56x + 33y = –y/(x2 + y2)
33x – 56y = x/(x2 + y2)

Discover the worth of

|x| + |y|

As traditional, watch the video for an answer.

Onerous And Complicated Take a look at Query

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“All will likely be effectively if you happen to use your thoughts on your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of recreation concept and arithmetic. MindYourDecisions now has over 1,000 free articles with no adverts because of neighborhood assist! Assist out and get early entry to posts with a pledge on Patreon.

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Reply To Onerous And Complicated Take a look at Query

(Just about all posts are transcribed rapidly after I make the movies for them–please let me know if there are any typos/errors and I’ll right them, thanks).

For a posh downside like this, the trick is to make use of complicated numbers! Let z = x + iy. Then z* = xiy in order that zz* = x2 + y2. We now have the system of equations:

56x + 33y = –y/(x2 + y2)
33x – 56y = x/(x2 + y2)

Multiply the primary by i and add to the second to get:

33x – 56y + 56ix + 33iy = (xy)/(x2 + y2)
x(33 + 56i) + y(33i – 56) = z*/(zz*)
x(33 + 56i) + iy(33 + 56i) = 1/z
(33 + 56i)(x + iy) = 1/z
(33 + 56i)z = 1/z
z2 = 1/(33 + 56i)

Now we need to discover a quantity a + bi whose sq. is 33 + 56i.

33 + 56i
= (a + bi)2
= a2b2 + 2abi

Equating actual and imaginary coefficients provides:

33 = a2b2
56 = 2ab
28 = ab

We are able to check out components of 28, and we are going to discover a appropriate pair a = 7 and b = 4:

72 – 42 = 33
7×4 = 28

Thus 33 + 56i = (7 + 4i)2, and now we have:

z2 = 1/(33 + 56i)
z = ± 1/(7 + 4i)

Multiplying the numerator and denominator by the conjugate 7 – 4i provides:

z
= ± (7 – 4i)/((7 + 4i)(7 – 4i))
= ± (7 – 4i)/65

Since z = x + iy, now we have two pairs of actual options:

(x, y) = (7/65, -4/65)
(x, y) = (-7/65, 4/65)

In both case now we have:

|x| + |y|
= 7/65 + 4/65
= 11/65

And that’s the reply!

References

Math StackExchange
https://math.stackexchange.com/questions/3684126/pair-of-real-number-satisfying-56x33y-fracyx2y2-and-33x56y-fracx

Harvard MIT competitors archive 2008 options (downside 28)
https://hmmt-archive.s3.amazonaws.com/tournaments/2008/feb/guts/options.pdf

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