2016 AMC 12B Problem 10

Below is the professionally curated solution for Problem 10 of the 2016 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12B solutions, or check the answer key.

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Concepts:coordinate geometry rectangle difference of squares

Difficulty rating: 1500

10.

A quadrilateral has vertices $P(a,b),$ $Q(b,a),$ $R(-a,-b),$ and $S(-b,-a),$ where $a$ and $b$ are integers with $a\gt b\gt0.$ The area of $PQRS$ is $16.$ What is $a+b?$

$4$

$5$

$6$

$12$

$13$

Solution:

The sides $\overline{PQ}$ and $\overline{RS}$ have slope $-1,$ and $\overline{QR}$ and $\overline{PS}$ have slope $1,$ so $PQRS$ is a rectangle with sides $(a-b)\sqrt2$ and $(a+b)\sqrt2.$ Its area is $2(a-b)(a+b)=2(a^2-b^2)=16,$ so $a^2-b^2=8.$ The only perfect squares differing by $8$ are $9$ and $1,$ giving $a=3,$ $b=1,$ and $a+b=4.$

Thus, the correct answer is A.

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