2009 AMC 12A Problem 10 — Solution | LIVE by Po-Shen Loh

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2009 AMC 12A Problem 10

Attempt Problem 10 of the 2009 AMC 12A 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 2009 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

In quadrilateral $ABCD,$ $AB = 5,$ $BC = 17,$ $CD = 5,$ $DA = 9,$ and $BD$ is an integer. What is $BD?$

$11$

$12$

$13$

$14$

$15$

Answer: C

Concepts:triangle inequality bounding to limit cases

Difficulty rating: 1500

Solution:

In $\triangle BCD,$ the triangle inequality gives $BD + CD \gt BC,$ so $BD + 5 \gt 17$ and $BD \gt 12.$

In $\triangle ABD,$ $AB + DA \gt BD,$ so $BD \lt 5 + 9 = 14.$

The only integer with $12 \lt BD \lt 14$ is $13.$

Thus, the correct answer is C.

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Problem 10 in Other Years

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