Difference between revisions of "2014 AMC 12B Problems/Problem 24"
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− | In <math>\triangle CFE</math> we have <math>\cos\theta = -\cos(\pi-\theta)=-7/c</math>. We use the Law of Cosines on <math>\triangle ABC</math> to get <math>60\cos\theta = 109-c^2</math>. Eliminating <math>\cos\theta</math> we get <math>c^3-109c-420=0</math> which factorizes as | + | In <math>\triangle CFE</math> we have <math>\cos\theta = -\cos(\pi-\theta)=-7/c</math>. We use the [[Law of Cosines]] on <math>\triangle ABC</math> to get <math>60\cos\theta = 109-c^2</math>. Eliminating <math>\cos\theta</math> we get <math>c^3-109c-420=0</math> which factorizes as |
− | <cmath>(c+7)(c+5)(c-12)=0.</cmath>Discarding the negative roots we have <math>c=12</math>. Thus <math>BD=AC=CE=12</math>. For <math>BE=a</math>, we use Ptolemy's theorem on cyclic quadrilateral <math>ABCE</math> to get <math>a=44/3</math>. For <math>AD=b</math>, we use Ptolemy's theorem on cyclic quadrilateral <math>ACDE</math> to get <math>b=27/2</math>. | + | <cmath>(c+7)(c+5)(c-12)=0.</cmath>Discarding the negative roots we have <math>c=12</math>. Thus <math>BD=AC=CE=12</math>. For <math>BE=a</math>, we use Ptolemy's theorem on cyclic quadrilateral <math>ABCE</math> to get <math>a=44/3</math>. For <math>AD=b</math>, we use [[Ptolemy's theorem]] on cyclic quadrilateral <math>ACDE</math> to get <math>b=27/2</math>. |
The sum of the lengths of the diagonals is <math>12+12+12+\tfrac{44}{3}+\tfrac{27}{2} = \tfrac{385}{6}</math> so the answer is <math>385 + 6 = \fbox{\textbf{(D) }391}</math> | The sum of the lengths of the diagonals is <math>12+12+12+\tfrac{44}{3}+\tfrac{27}{2} = \tfrac{385}{6}</math> so the answer is <math>385 + 6 = \fbox{\textbf{(D) }391}</math> |
Latest revision as of 08:58, 25 September 2021
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 ?
Solutions
Solution 1
Let , , and . Let be on such that . In we have . We use the Law of Cosines on to get . Eliminating we get which factorizes as Discarding the negative roots we have . Thus . For , we use Ptolemy's theorem on cyclic quadrilateral to get . For , we use Ptolemy's theorem on cyclic quadrilateral to get .
The sum of the lengths of the diagonals is so the answer is
Solution 2
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:
Or similarly into equation to check:
, being a length, must be positive, implying that . In fact, this is reasonable, since in the pentagon with apparently obtuse angles. 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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