Difference between revisions of "2016 AMC 12B Problems/Problem 10"
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Solution by I_Dont_Do_Math | Solution by I_Dont_Do_Math | ||
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| + | ==Solution 2== | ||
| + | Solution by e_power_pi_times_i | ||
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| + | By the Shoelace Theorem, the area of the quadrilateral is <math>2a^2 - 2b^2</math>, so <math>a^2 - b^2 = 8</math>. Since <math>a</math> and <math>b</math> are integers, <math>a = 3</math> and <math>b = 1</math>, so <math>a + b = \boxed{\textbf{(A)}\ 4}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2016|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:51, 22 February 2016
Contents
Problem
A quadrilateral has vertices 
, 
, 
, and 
, where 
and 
are integers with 
. The area of 
is 
. What is 
?
Solution
By distance formula we have 
. SImplifying we get 
. Thus and 
have to be a factor of 8. The only way for them to be factors of 
and remain integers is if 
and 
. So the answer is 
Solution by I_Dont_Do_Math
Solution 2
Solution by e_power_pi_times_i
By the Shoelace Theorem, the area of the quadrilateral is 
, so 
. Since and are integers, 
and 
, so 
.
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
