Really Irrational Equations

Algebra Level 5

There exist a real number p p

such that the following system of equations

1 1 + 2 x 2 + 1 1 + 2 y 2 = 2 1 + 2 x y \dfrac { 1 }{ \sqrt { 1+2{ x }^{ 2 } } } +\dfrac { 1 }{ \sqrt { 1+2{ y }^{ 2 } } } =\dfrac { 2 }{ \sqrt { 1+2xy } }

x x ( 1 2 x ) + y ( 1 2 y ) = p x\sqrt { x(1-2x) } +\sqrt { y(1-2y) } =p

has only one real value x x as a root, which can be expressed as

x = a + b c x=\dfrac { a+\sqrt { b } }{ c }

where a , b , c a, b, c are all non-zero integers and b b is square-free. Find a + b + c a+b+c

Details and Assumptions :

  • There's no typo in this question.

See Victor Loh's Irrational Equations


The answer is 48.

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1 solution

Michael Mendrin
Aug 17, 2014

For
p = 1 64 117 + 165 33 = 0.450383... p=\dfrac { 1 }{ 64 } \sqrt { -117+165\sqrt { 33 } } =0.450383...

the only real root is

x = 1 + 33 16 x=\dfrac { -1+\sqrt { 33 } }{ 16 } .

Thus, the answer is 48 48 .

Hello,

Sonnhard.

Fatrick Chao - 6 years, 9 months ago

If we look from the first equation we can say that x=y is a solution, which make the second equation after we change y by x : -2x^4-3x^3+x-p^2=0 the problem here is what is the value of p can make the equation have 1 solution so we need simply to find the max of this equation -2x^4-3x^3+x by using the derivative of that fonction so we'll have -8x^3 -9x^2+1=0 that's how you can find p

Mehdi Elkouhlani - 6 years, 3 months ago

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Good insight.

Calvin Lin Staff - 6 years, 3 months ago

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