Difference between revisions of "Ptolemy's theorem"

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'''Ptolemy's theorem''' gives a relationship between the side lengths and the diagonals of a [[cyclic quadrilateral]]; it is the [[equality condition | equality case]] of [[Ptolemy's Inequality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures.
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#REDIRECT [[Ptolemy's Theorem]]
 
 
== Statement ==
 
 
 
Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>{a},{b},{c},{d}</math> and [[diagonal]]s <math>{e},{f}</math>:
 
 
 
<cmath>ac+bd=ef.</cmath>
 
 
 
== Proof 1 ==
 
 
 
Given cyclic quadrilateral <math>ABCD,</math> extend <math>CD</math> to <math>P</math> such that <math>\angle BAC=\angle DAP.</math>
 
 
 
Since quadrilateral <math>ABCD</math> is cyclic, <math>m\angle ABC+m\angle ADC=180^\circ .</math> However, <math>\angle ADP</math> is also supplementary to <math>\angle ADC,</math> so <math>\angle ADP=\angle ABC</math>. Hence, <math>\triangle ABC \sim \triangle ADP</math> by AA similarity and <math>\frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.</math>
 
 
 
Now, note that <math>\angle ABD=\angle ACD </math> (subtend the same arc) and <math>\angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAD=\angle CAP,</math> so <math>\triangle BAD\sim \triangle CAP.</math> This yields <math>\frac{AB}{AC}=\frac{BD}{CP}\implies CP=\frac{(AC)(BD)}{(AB)}.</math>
 
 
 
However, <math>CP= CD+DP.</math> Substituting in our expressions for  <math>CP</math> and  <math>DP,</math>  <math> \frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.</math> Multiplying by <math>AB</math> yields  <math>(AC)(BD)=(AB)(CD)+(AD)(BC)</math>.
 
 
 
 
 
== Proof 2 (inversion) ==
 
 
 
We provide a proof for the general case of Ptolemy's theorem, Ptolemy's Inequality.
 
 
 
Let <math>A,B,C,D</math> be four points in the Euclidean plane. Taking an inversion centered at <math>D</math> (the point doesn't matter, it can be any of the four) with radius <math>r</math>, we have that <math>A^*B^*+B^*C^*\geq A^*C^*</math> by the Triangle Inequality, with equality holding when <math>A^*, B^*, C^*</math> are collinear, i.e. when <math>A,B,C</math> lie on a circle containing <math>D.</math> Additionally, by the Inversion Distance Formula, we may express the inequality as the following:
 
 
 
<cmath>\frac{r^2}{AD\cdot BD}\cdot AB + \frac{r^2}{BD\cdot CD}\cdot BC \geq \frac{r^2}{AD\cdot CD}\cdot AC.</cmath>
 
 
 
Dividing by <math>r^2</math> and multiplying everything by <math>AD\cdot BD \cdot CD,</math> we get <math>AB\cdot CD + BC\cdot AD \geq AC\cdot BD,</math> and thus the desired. <math>_\blacksquare</math>
 
 
 
== Problems ==
 
===2023 AIME I Problem 5===
 
Square <math>ABCD</math> is inscribed in a circle. Point <math>P</math> is on this circle such that <math>AP \cdot CP = 56</math>, and <math>BP \cdot DP = 90</math>. What is the area of the square?
 
 
 
Solution: We may assume that <math>P</math> is between <math>B</math> and <math>C</math>. Let <math>PA = a</math>, <math>PB = b</math>, <math>PC = C</math>, <math>PD = d</math>, and <math>AB = s</math>. We have <math>a^2 + c^2 = AC^2 = 2s^2</math>, because <math>AC</math> is a diagonal. Similarly, <math>b^2 + d^2 = 2s^2</math>. Therefore, <math>(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112</math>. Similarly, <math>(b+d)^2 = 2s^2 + 180</math>.
 
 
 
By Ptolemy's Theorem on <math>PCDA</math>, <math>as + cs = ds\sqrt{2}</math>, and therefore <math>a + c = d\sqrt{2}</math>. By Ptolemy's on <math>PBAD</math>, <math>bs + ds = as\sqrt{2}</math>, and therefore <math>b + d = a\sqrt{2}</math>. By squaring both equations, we obtain
 
 
 
<cmath>2d^2 = (a+c)^2 = 2s^2 + 112</cmath>
 
<cmath>2a^2 = (b+d)^2 = 2s^2 + 180.</cmath>
 
 
 
Thus, <math>a^2 = s^2 + 90</math>, and <math>d^2 = s^2 + 56</math>. Plugging these values into <math>a^2 + c^2 = b^2 + d^2 = 2s^2</math>, we obtain <math>c^2 = s^2 - 90</math>, and <math>b^2 = s^2 - 56</math>. Now, we can solve using <math>a</math> and <math>c</math> (though using <math>b</math> and <math>d</math> yields the same solution for <math>s</math>).
 
 
 
<cmath>(\sqrt{s^2 + 90})(\sqrt{s^2 - 90}) = ac = 56</cmath>
 
<cmath>(s^2 + 90)(s^2 - 90) = 56^2</cmath>
 
<cmath>s^4 = 90^2 + 56^2 = 106^2</cmath>
 
<cmath>s^2 = 106.</cmath>
 
 
 
The answer is <math>\boxed{106}</math>.
 
 
 
===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>?
 
 
 
<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>
 
 
 
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 ===
 
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>.
 
 
 
Solution: Draw <math>PA</math>, <math>PB</math>, <math>PC</math>. By Ptolemy's theorem applied to quadrilateral <math>APBC</math>, we know that <math>PC\cdot AB=PA\cdot BC+PB\cdot AC</math>. Since <math>AB=BC=CA=s</math>, we divide both sides of the last equation by <math>s</math> to get the result: <math>PC=PA+PB</math>.
 
 
 
=== Regular Heptagon Identity ===
 
In a regular heptagon <math> ABCDEFG </math>, prove that: <math> \frac{1}{AB}=\frac{1}{AC}+\frac{1}{AE} </math>.
 
 
 
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 division by <math> abc </math>.
 
 
 
=== 1991 AIME Problems/Problem 14 ===
 
A hexagon is inscribed in a circle. Five of the sides have length <math>81</math> and the sixth, denoted by <math>\overline{AB}</math>, has length <math>31</math>. Find the sum of the lengths of the three diagonals that can be drawn from <math>A</math>.
 
 
 
[[1991_AIME_Problems/Problem_14#Solution|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 <math>ABCD</math>, <math>AD</math> being the diameter. <math>AB = 2</math>, <math>BC = 7</math>, and <math>CD = 11</math>. Construct diagonals <math>AC</math> and <math>BD</math>. Notice that these diagonals form right triangles. You get the following system of equations:
 
 
 
<math>(AC)(BD) = 7(AD) + 22</math> (Ptolemy's theorem)
 
 
 
<math>\text(AC)^2 = (AD)^2 - 121</math>
 
 
 
<math>(BD)^2 = (AD)^2 - 4</math>
 
 
 
Solving gives <math>AD = 14</math>
 
 
 
== See also ==
 
* [[Geometry]]
 
* [[Cyclic quadrilateral]]
 
 
 
[[Category:Geometry]]
 
[[Category:Theorems]]
 

Revision as of 08:35, 8 December 2024

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