Difference between revisions of "Ptolemy's Theorem"
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=== Regular Heptagon Identity === | === Regular Heptagon Identity === | ||
− | In a regular heptagon | + | In a regular heptagon <math> ABCDEFG </math>, prove that: <math> \frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD} </math>. |
− | Solution: Let | + | 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 | + | 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 multiplication by <math> abc </math>. |
=== 1991 AIME Problems/Problem 14 === | === 1991 AIME Problems/Problem 14 === |
Revision as of 11:42, 22 January 2014
Ptolemy's Theorem gives a relationship between the side lengths and the diagonals of a cyclic quadrilateral; it is the equality case of the Ptolemy Inequality. Ptolemy's Theorem frequently shows up as an intermediate step in problems involving inscribed figures.
Contents
[hide]Definition
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 multiplication 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)
$\n(AC)^2 = (AD)^2 - 121$ (Error compiling LaTeX. Unknown error_msg)
Solving gives