Difference between revisions of "2014 AMC 12B Problems/Problem 24"
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MA("\pi-\theta",C,E0,A,0.1); | MA("\pi-\theta",C,E0,A,0.1); | ||
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− | In <math>\triangle CFE</math> we have <math>\cos\theta = -\cos(\pi-\theta)=-7/c</math>. We use | + | 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>. | ||
Revision as of 08:50, 25 September 2021
Contents
[hide]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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.