Difference between revisions of "Ptolemy's Theorem"
m (→2004 AMC 10B Problem 24) |
Wesleythang (talk | contribs) m (→2004 AMC 10B Problem 24) |
||
Line 19: | Line 19: | ||
== Problems == | == Problems == | ||
===2004 AMC 10B Problem 24=== | ===2004 AMC 10B Problem 24=== | ||
− | In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC= | + | 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>? |
<math>\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}</math> | <math>\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}</math> |
Revision as of 00:42, 16 August 2017
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