Difference between revisions of "2014 AMC 12B Problems/Problem 24"
Isabelchen (talk | contribs) (→Solution 4) |
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Let <math>BE = a</math>, <math>AC = CE = BD = b</math> | Let <math>BE = a</math>, <math>AC = CE = BD = b</math> | ||
− | By [Ptolemy's theorem] for quadrilateral <math>ABCE</math>, <math>3c + 140 = ab</math>, <math>a = 3 + \frac{140}{b}</math> | + | By [[Ptolemy's theorem]] for quadrilateral <math>ABCE</math>, <math>3c + 140 = ab</math>, <math>a = 3 + \frac{140}{b}</math> |
− | By [Ptolemy's theorem] for quadrilateral <math>ACDE</math>, <math>3a + 100 = b^2</math> | + | By [[Ptolemy's theorem]] for quadrilateral <math>ACDE</math>, <math>3a + 100 = b^2</math> |
<math>3(3 + \frac{140}{b}) + 100 = b^2</math>, <math>b^3 - 109 b -420 = 0</math>, <math>(b-12)(b+7)(b+5) = 0</math>, <math>b = 12</math> | <math>3(3 + \frac{140}{b}) + 100 = b^2</math>, <math>b^3 - 109 b -420 = 0</math>, <math>(b-12)(b+7)(b+5) = 0</math>, <math>b = 12</math> | ||
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<math>a = 3 + \frac{140}{12} = \frac{44}{3}</math> | <math>a = 3 + \frac{140}{12} = \frac{44}{3}</math> | ||
− | By [Ptolemy's theorem] for quadrilateral <math>ABDE</math>, <math>AD \cdot a = 14b + 30</math>, <math>AD = \frac{27}{2}</math> | + | By [[Ptolemy's theorem]] for quadrilateral <math>ABDE</math>, <math>AD \cdot a = 14b + 30</math>, <math>AD = \frac{27}{2}</math> |
<math>12 + 12 + 12 + \frac{44}{3} + \frac{27}{2} = \boxed{\textbf{(D) }391}</math> | <math>12 + 12 + 12 + \frac{44}{3} + \frac{27}{2} = \boxed{\textbf{(D) }391}</math> |
Revision as of 06:55, 1 January 2023
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
?
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=1-2vT_GIceA
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
Solution 3 (Ptolemy's but Quicker)
Let us set to be
and
to be
and
to be
. It follow from applying Ptolemy's Theorem on
to get
. Applying Ptolemy's on
gives
; and applying Ptolemy's on
gives
. So, we have the have the following system of equations:
From , we have
. Isolating the x gives
. By setting
and
equal, we have
. Manipulating it gives
. Finally, plugging back into
gives
. Plugging in the
as well gives
It is impossible for for
; that means
. That means
and
.
Thus, the sum of all diagonals is , which implies our answer is
.
~sml1809
Solution 4
Let ,
By Ptolemy's theorem for quadrilateral ,
,
By Ptolemy's theorem for quadrilateral ,
,
,
,
By Ptolemy's theorem for quadrilateral ,
,
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.