2016 AMC 12B Problem 10

Attempt Problem 10 of the 2016 AMC 12B 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 2016 AMC 12B solutions, or check the answer key.

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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

16.

What is

a+b?

4

5

6

12

13

Check the slopes:

\overline{PQ}

and

\overline{RS}

have slope

-1,

while

\overline{QR}

and

\overline{PS}

have slope

1,

PQRS

is a rectangle

Its sides are

(a-b)\sqrt2

and

(a+b)\sqrt2,

so the area is

2(a^2-b^2)=16

The sides

\overline{PQ}

and

\overline{RS}

have slope

-1,

and

\overline{QR}

and

\overline{PS}

have slope

1,

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,

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.

Problem 10 in other years