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

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2009 AMC 10A Problem 12

Below is the professionally curated solution for Problem 12 of the 2009 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:triangle inequality bounding to limit cases

Difficulty rating: 1220

12.

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

$11$

$12$

$13$

$14$

$15$

Solution:

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

In $\triangle ABD,$ it gives $5 + 9 \gt BD,$ so $BD \lt 14.$

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

Thus, the correct answer is C.

← Problem 11 Full Exam Problem 13 →

Problem 12 in Other Years

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