Difference between revisions of "Ptolemy's Theorem"
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Solution: Let <math> ABCDEFG </math> be the regular heptagon. Consider the quadrilateral <math> ABCE </math>. If <math> a </math>, <math> b </math>, and <math> c </math> represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of <math> ABCE </math> are <math> a </math>, <math> a </math>, <math> b </math> and <math> c </math>; the diagonals of <math> ABCE </math> are <math> b </math> and <math> c </math>, respectively. | Solution: Let <math> ABCDEFG </math> be the regular heptagon. Consider the quadrilateral <math> ABCE </math>. If <math> a </math>, <math> b </math>, and <math> c </math> represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of <math> ABCE </math> are <math> a </math>, <math> a </math>, <math> b </math> and <math> c </math>; the diagonals of <math> ABCE </math> are <math> b </math> and <math> c </math>, respectively. | ||
− | Now, Ptolemy's Theorem states that <math> ab + ac = bc </math>, which is equivalent to <math> \frac{1}{a}=\frac{1}{b}+\frac{1}{c} </math> upon | + | Now, Ptolemy's Theorem states that <math> ab + ac = bc </math>, which is equivalent to <math> \frac{1}{a}=\frac{1}{b}+\frac{1}{c} </math> upon division by <math> abc </math>. |
=== 1991 AIME Problems/Problem 14 === | === 1991 AIME Problems/Problem 14 === |
Revision as of 08:07, 11 February 2016
Ptolemy's Theorem gives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of Ptolemy's Inequality. Ptolemy's Theorem frequently shows up as an intermediate step in problems involving inscribed figures.
Contents
[hide]Statement
Given a cyclic quadrilateral with side lengths
and diagonals
:
Proof
Given cyclic quadrilateral extend
to
such that
Since quadrilateral is cyclic,
However,
is also supplementary to
so
. Hence,
by AA similarity and
Now, note that (subtend the same arc) and
so
This yields
However, Substituting in our expressions for
and
Multiplying by
yields
.
Problems
Equilateral Triangle Identity
Let be an equilateral triangle. Let
be a point on minor arc
of its circumcircle. Prove that
.
Solution: Draw ,
,
. By Ptolemy's Theorem applied to quadrilateral
, we know that
. Since
, we divide both sides of the last equation by
to get the result:
.
Regular Heptagon Identity
In a regular heptagon , prove that:
.
Solution: Let be the regular heptagon. Consider the quadrilateral
. If
,
, and
represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of
are
,
,
and
; the diagonals of
are
and
, respectively.
Now, Ptolemy's Theorem states that , which is equivalent to
upon division by
.
1991 AIME Problems/Problem 14
A hexagon is inscribed in a circle. Five of the sides have length and the sixth, denoted by
, has length
. Find the sum of the lengths of the three diagonals that can be drawn from
.
Cyclic Hexagon
A hexagon with sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle. Find the diameter of the circle.
Solution: Consider half of the circle, with the quadrilateral ,
being the diameter.
,
, and
. Construct diagonals
and
. Notice that these diagonals form right triangles. You get the following system of equations:
(Ptolemy's Theorem)
Solving gives