Difference between revisions of "2014 AMC 12B Problems/Problem 24"
(→Solution) |
(→Solution) |
||
Line 10: | Line 10: | ||
==Solution== | ==Solution== | ||
− | + | Let <math>a</math> denote the length of a diagonal opposite adjacent sides of length <math>14</math> and <math>3</math>, <math>b</math> for sides <math>14</math> and <math>10</math>, and <math>c</math> for sides <math>3</math> and <math>10</math>. Using Ptolemy's Theorem on the five possible quadrilaterals in the configuration, we obtain: | |
+ | |||
+ | <cmath> | ||
+ | \begin{align} | ||
+ | c^2 &= 3a+100 \\ | ||
+ | c^2 &= 10b+9 \\ | ||
+ | 14c &= 30+ab \\ | ||
+ | ac &= 3c+140\\ | ||
+ | bc &= 10c+42 | ||
+ | \end{align} | ||
+ | </cmath> | ||
+ | |||
+ | Using equations <math>(1)</math> and <math>(2)</math>, we obtain: | ||
+ | |||
+ | <cmath> | ||
+ | a = \frac{c^2-100}{3} | ||
+ | </cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath> | ||
+ | b = \frac{c^2-9}{10} | ||
+ | </cmath> | ||
+ | |||
+ | Plugging into equation <math>(4)</math>, we find that: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{c^2-100}{3}c &= 3c + 140\\ | ||
+ | \frac{c^3-100c}{3} &= 3c + 140\\ | ||
+ | c^3-100c &= 9c + 420\\ | ||
+ | c^3-109c-420 &=0\\ | ||
+ | (c-12)(c+7)(c+5)&=0 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | <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\dot 12 + \frac{44}{3} + \frac{27}{2} = \frac{385}{6}</math>, so it follows that the answer is <math>385 + 6 = 391</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 07:46, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.