2012 AMC 12B Problem 20

Attempt Problem 20 of the 2012 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 2012 AMC 12B solutions, or check the answer key.

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20.

A trapezoid has side lengths 3,3, 5,5, 7,7, and 11.11. The sum of all the possible areas of the trapezoid can be written in the form of r1n1+r2n2+r3,r_1\sqrt{n_1} + r_2\sqrt{n_2} + r_3, where r1,r_1, r2,r_2, and r3r_3 are rational numbers and n1n_1 and n2n_2 are positive integers not divisible by the square of a prime. What is the greatest integer less than or equal to r1+r2+r3+n1+n2?r_1 + r_2 + r_3 + n_1 + n_2?

5757

5959

6161

6363

6565

Answer: D
Concepts:trapezoidHeron’s Formulacasework
Difficulty rating: 2150
Small Hint:

Slide one parallel side over: a trapezoid with parallel sides a<ca\lt c and legs b,db,d gives a triangle with sides b,d,cab,d,c-a

Big Hint:

The triangle inequality forces the longer parallel side to be 11;11; apply Heron’s formula in the cases a=3,5,7a=3,5,7

Solution:

For a trapezoid with parallel sides a<ca\lt c and legs b,d,b,d, translating a leg forms a triangle with sides b,b, d,d, and ca.c-a. The triangle inequality forces the longer parallel side to be c=11.c=11.

If a=3,a=3, the triangle has sides 5,7,85,7,8 with area 103,10\sqrt3, and the trapezoid has area 3523.\tfrac{35}{2}\sqrt3. If a=5,a=5, the triangle has sides 3,6,73,6,7 with area 45,4\sqrt5, giving trapezoid area 3235.\tfrac{32}{3}\sqrt5. If a=7,a=7, the triangle has sides 3,4,5,3,4,5, a right triangle, giving trapezoid area 27.27.

The total is 3523+3235+27,\tfrac{35}{2}\sqrt3+\tfrac{32}{3}\sqrt5+27, so r1+r2+r3+n1+n2=352+323+27+3+5=63+16. \begin{gathered} r_1+r_2+r_3+n_1+n_2 \\ = \tfrac{35}{2}+\tfrac{32}{3}+27+3+5 \\ = 63+\tfrac16. \end{gathered}

The greatest integer at most this value is 63.63.

Thus, the correct answer is D.

Problem 19#19
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