Difference between revisions of "Ptolemy's Theorem"

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.

Definition

Given a cyclic quadrilateral $ABCD$ with side lengths ${a},{b},{c},{d}$ and diagonals ${e},{f}$ :

$ac+bd=ef$ .

Proof

Given cyclic quadrilateral $ABCD,$ extend $CD$ to $P$ such that $\angle BAD=\angle CAP.$

Since quadrilateral $ABCD$ is cyclic, $m\angle ABC+m\angle ADC=180^\circ .$ However, $\angle ADP$ is also supplementary to $\angle ADC,$ so $\angle ADP=\angle ABC$ . Hence, $\triangle ABC \sim \triangle ADP$ by AA similarity and $\frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.$

Now, note that $\angle ABD=\angle ACD$ (subtend the same arc) and $\angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAD=\angle CAP,$ so $\triangle BAD\sim \triangle CAP.$ This yields $\frac{AB}{AC}=\frac{BD}{CP}\implies CP=\frac{(AC)(BD)}{(AB)}.$

However, $CP= CD+DP.$ Substituting in our expressions for $CP$ and $DP,$ $\frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.$ Multiplying by $AB$ yields $(AC)(BD)=(AB)(CD)+(AD)(BC)$ .

Problems

Equilateral Triangle Identity

Let $\triangle ABC$ be an equilateral triangle. Let $P$ be a point on minor arc $AB$ of its circumcircle. Prove that $PC=PA+PB$ .

Solution: Draw $PA$ , $PB$ , $PC$ . By Ptolemy's Theorem applied to quadrilateral $APBC$ , we know that $PC\cdot AB=PA\cdot BC+PB\cdot AC$ . Since $AB=BC=CA=s$ , we divide both sides of the last equation by $s$ to get the result: $PC=PA+PB$ .

Regular Heptagon Identity

In a regular heptagon $ABCDEFG$ , prove that: $\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD}$ .

Solution: Let $ABCDEFG$ be the regular heptagon. Consider the quadrilateral $ABCE$ . If $a$ , $b$ , and $c$ represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of $ABCE$ are $a$ , $a$ , $b$ and $c$ ; the diagonals of $ABCE$ are $b$ and $c$ , respectively.

Now, Ptolemy's Theorem states that $ab + ac = bc$ , which is equivalent to $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$ upon multiplication by $abc$ .

1991 AIME Problems/Problem 14

A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$ , has length $31$ . Find the sum of the lengths of the three diagonals that can be drawn from $A$ .

Solution

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 $ABCD$ , $AD$ being the diameter. $AB = 2$ , $BC = 7$ , and $CD = 11$ . Construct diagonals $AC$ and $BD$ . Notice that these diagonals form right triangles. You get the following system of equations:

$(AC)(BD) = 7(AD) + 22$ (Ptolemy's Theorem)

$\n(AC)^2 = (AD)^2 - 121$ (Error compiling LaTeX. Unknown error_msg)

$(BD)^2 = (AD)^2 - 4$

Solving gives $AD = 14$

@@ Line 24: / Line 24: @@
 === Regular Heptagon Identity ===
-In a regular heptagon ''ABCDEFG'', prove that: ''1/AB = 1/AC + 1/AD''.
+In a regular heptagon <math> ABCDEFG </math>, prove that: <math> \frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD} </math>.
-Solution: Let ''ABCDEFG'' be the regular heptagon. Consider the quadrilateral ''ABCE''. If ''a'', ''b'', and ''c'' represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of ''ABCE'' are ''a'', ''a'', ''b'' and ''c''; the diagonals of ''ABCE'' are ''b'' and ''c'', 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 ''ab + ac = bc'', which is equivalent to ''1/a=1/b+1/c''.
+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 ===

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