2015 AMC 10A Problem 16

Attempt Problem 16 of the 2015 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 10A solutions, or check the answer key.

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16.

If we have that y+4=(x2)2,y+4 = (x-2)^2,x+4=(y2)2, x+4 = (y-2)^2, and xy,x \neq y, what is the value of x2+y2?x^2+y^2?

1010

1515

2020

2525

3030

Answer: B
Concepts:system of equationssymmetry (algebra)algebraic manipulation
Difficulty rating: 1420
Solution:

Adding the two equations gives us x2+y24x4y+8 x^2 + y^2 - 4x - 4y + 8 =x+y+8. = x + y + 8. We can rearrange this equation to get x2+y2=5(x+y). x^2 + y^2 = 5(x + y). We can then subtract them to get x2y24x+4y=yx. x^2 - y^2 - 4x + 4y = y - x. Once again rearranging, we can find x2y2=3(xy). x^2 - y^2 = 3(x - y). We have that xy,x \neq y, which means that we can divide both sides by xy.x - y. This gives us x+y=3 x + y = 3 x2+y2=15. x^2 + y^2 = 15.

Thus, B is the correct answer.

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