Difference between revisions of "Ptolemy's Theorem"
Ishankhare (talk | contribs) (→Problems) |
Ishankhare (talk | contribs) (→2004 AMC 10B Problems/Problem 24) |
||
Line 18: | Line 18: | ||
== Problems == | == Problems == | ||
− | ===2004 AMC 10B | + | ===2004 AMC 10B Problem 24=== |
In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC=9</math>. Point <math>D</math> is on the circumscribed circle of the triangle so that <math>AD</math> bisects angle <math>BAC</math>. What is the value of <math>AD/CD</math>? | In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC=9</math>. Point <math>D</math> is on the circumscribed circle of the triangle so that <math>AD</math> bisects angle <math>BAC</math>. What is the value of <math>AD/CD</math>? | ||
Line 24: | Line 24: | ||
Solution: Set <math>\overline{BD}</math>'s length as <math>x</math>. <math>CD</math>'s length must also be <math>x</math> since <math>\angle BAD</math> and <math>\angle DAC</math> intercept arcs of equal length(because <math>\angle BAD =\angle DAC</math>). Using Ptolemy's Theorem, <math>7x+8x=9(AD)</math>. The ratio is <math>\boxed{\frac{5}{3}}\implies(B)</math> | Solution: Set <math>\overline{BD}</math>'s length as <math>x</math>. <math>CD</math>'s length must also be <math>x</math> since <math>\angle BAD</math> and <math>\angle DAC</math> intercept arcs of equal length(because <math>\angle BAD =\angle DAC</math>). Using Ptolemy's Theorem, <math>7x+8x=9(AD)</math>. The ratio is <math>\boxed{\frac{5}{3}}\implies(B)</math> | ||
+ | |||
=== Equilateral Triangle Identity === | === Equilateral Triangle Identity === | ||
Let <math>\triangle ABC</math> be an equilateral triangle. Let <math>P</math> be a point on minor arc <math>AB</math> of its circumcircle. Prove that <math>PC=PA+PB</math>. | Let <math>\triangle ABC</math> be an equilateral triangle. Let <math>P</math> be a point on minor arc <math>AB</math> of its circumcircle. Prove that <math>PC=PA+PB</math>. |
Revision as of 15:29, 28 April 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
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
2004 AMC 10B Problem 24
In triangle we have , , . Point is on the circumscribed circle of the triangle so that bisects angle . What is the value of ?
Solution: Set 's length as . 's length must also be since and intercept arcs of equal length(because ). Using Ptolemy's Theorem, . The ratio is
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