Difference between revisions of "2014 AIME II Problems/Problem 12"
Futurewriter (talk | contribs) (→Solution 1) |
Futurewriter (talk | contribs) (→Solution 2) |
||
Line 42: | Line 42: | ||
Thus <math>\frac{3A}{2}+\frac{3B}{2}=90</math>, or <math>A+B=60</math>. | Thus <math>\frac{3A}{2}+\frac{3B}{2}=90</math>, or <math>A+B=60</math>. | ||
− | Now we know that <math>C=120</math>, so we can just use Law | + | Now we know that <math>C=120</math>, so we can just use the Law of Cosines to get <math>\boxed{399}</math> |
== See also == | == See also == |
Revision as of 22:22, 3 March 2016
Contents
Problem
Suppose that the angles of satisfy
Two sides of the triangle have lengths 10 and 13. There is a positive integer
so that the maximum possible length for the remaining side of
is
Find
Solution 1
Note that . Thus, our expression is of the form
. Let
and
.
Using the fact that , we get
, or
.
Squaring both sides, we get . Cancelling factors,
.
Expanding, .
Simplification leads to and
.
Therefore, . So
could be
or
. We eliminate
and use law of cosines to get our answer:
Solution 2
As above, we can see that
Expanding, we get
Note that , or
Thus , or
.
Now we know that , so we can just use the Law of Cosines to get
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.