Double Simplicity

Algebra Level 5

Let $x,y$ be real numbers such that $x^2 + xy + 2y^2 = 8$ .

The greatest value that $x + y$ can attain is a real number equal to $\dfrac{a\sqrt{b}}{b}$ , where $a,b$ are positive integers such that $b$ does not divide $a$ , and $b$ is square- free.

Evaluate $a+b$ .

The answer is 15.

1 solution

Mark Hennings
Jan 30, 2017

The line $x+y=k$ meets the ellipse $x^2 + xy + 2y^2 = 8$ provided that the equation $2y^2 - ky + k^2 - 8 \; = \; (k-y)^2 + (k-y)y + 2y^8 - 8 \; = \; 0$ has solutions in $y$ . Since the discriminant of this quadratic is $64 - 7k^2$ , we see that $|k| \le \tfrac{8\sqrt{7}}{7}$ , making the answer $8 +7 = \boxed{15}$ .

a+b was asked not a+2b wasn't it?

A Former Brilliant Member - 4 years, 1 month ago

The original problem asked for $a+b$ . During a Reporting process, the author changed the question to ask for $a+2b$ . While the problem was asking for $a+2b$ I solved it and posted the solution. Calvin subsequently edited the question back (so that those who previously answered $15$ were marked as correct. I have now edited my solution!

Mark Hennings - 4 years, 1 month ago

Any other easy method ??

Satyam Tripathi - 4 years, 1 month ago

We have $(2x+y)^2 + 7y^2 = 32$ so you could write $2x+y=4\sqrt{2}\cos\theta$ and $y = \tfrac{4\sqrt{2}}{\sqrt{7}}\sin\theta$ , and hence $2(x+y) = \tfrac{4\sqrt{2}}{\sqrt{7}}\big[\sqrt{7}\cos\theta + \sin\theta\big]$ and hence $x+y = \tfrac{8}{\sqrt{7}}\cos(\theta -\alpha)$ for a suitable choice of $\alpha$ ...

Mark Hennings - 4 years, 1 month ago

Superb innovation... Geometrical approach here is awesome

space sizzlers - 4 years, 1 month ago

Double Simplicity

The answer is 15.

1 solution

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