Difference between revisions of "2014 AMC 12B Problems/Problem 24"
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<math>c</math>, being a length, must be positive, implying that <math>c=12</math>. Plugging this back into equations <math>(1)</math> and <math>(2)</math> we find that <math>a = \frac{44}{3}</math> and <math>b= \frac{135}{10}=\frac{27}{2}</math>. | <math>c</math>, being a length, must be positive, implying that <math>c=12</math>. Plugging this back into equations <math>(1)</math> and <math>(2)</math> we find that <math>a = \frac{44}{3}</math> and <math>b= \frac{135}{10}=\frac{27}{2}</math>. | ||
− | We desire <math>3c+a+b = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = \frac{385}{6}</math>, so it follows that the answer is <math>385 + 6 = | + | We desire <math>3c+a+b = 3\cdot 12 + \frac{44}{3} + \frac{27}{2} = \frac{385}{6}</math>, so it follows that the answer is <math>385 + 6 = \fbox{\textbf{(E) }421\qquad}</math>. |
== See also == | == See also == | ||
{{AMC12 box|year=2014|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2014|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 07:50, 31 January 2015
Problem
Let be a pentagon inscribed in a circle such that
,
, and
. The sum of the lengths of all diagonals of
is equal to
, where
and
are relatively prime positive integers. What is
?
Solution
Let denote the length of a diagonal opposite adjacent sides of length
and
,
for sides
and
, and
for sides
and
. Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain:
Using equations and
, we obtain:
and
Plugging into equation , we find that:
, being a length, must be positive, implying that
. Plugging this back into equations
and
we find that
and
.
We desire , so it follows that the answer is
.
See also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |
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