2009 AMC 10A Problem 12

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

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

12.

In quadrilateral ABCD,ABCD, AB=5,AB = 5, BC=17,BC = 17, CD=5,CD = 5, DA=9,DA = 9, and BDBD is an integer. What is BD?BD?
1111
1212
1313
1414
1515
Answer: C
Concepts:triangle inequalitybounding to limit cases
Small hint:
Apply the triangle inequality to △BCD\triangle BCD and △ABD\triangle ABD separately
Big hint:
The diagonal BDBD must satisfy 12<BD<1412 \lt BD \lt 14
Solution:
In △BCD,\triangle BCD, the triangle inequality gives 5+BD>17,5 + BD \gt 17, so BD>12.BD \gt 12.
In △ABD,\triangle ABD, it gives 5+9>BD,5 + 9 \gt BD, so BD<14.BD \lt 14.
The only integer with 12<BD<1412 \lt BD \lt 14 is BD=13.BD = 13.
Thus, the correct answer is C.
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