Difference between revisions of "Ptolemy's theorem"

Latest revision as of 23:28, 19 September 2024

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.

Statement

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

$ac+bd=ef.$

Proof 1

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

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 $(AC)(BD)=(AB)(CD)+(AD)(BC)$ .

Proof 2 (inversion)

We provide a proof for the general case of Ptolemy's theorem, Ptolemy's Inequality.

Let $A,B,C,D$ be four points in the Euclidean plane. Taking an inversion centered at $D$ (the point doesn't matter, it can be any of the four) with radius $r$ , we have that $A^*B^*+B^*C^*\geq A^*C^*$ by the Triangle Inequality, with equality holding when $A^*, B^*, C^*$ are collinear, i.e. when $A,B,C$ lie on a circle containing $D.$ Additionally, by the Inversion Distance Formula, we may express the inequality as the following:

$\frac{r^2}{AD\cdot BD}\cdot AB + \frac{r^2}{BD\cdot CD}\cdot BC \geq \frac{r^2}{AD\cdot CD}\cdot AC.$

Dividing by $r^2$ and multiplying everything by $AD\cdot BD \cdot CD,$ we get $AB\cdot CD + BC\cdot AD \geq AC\cdot BD,$ and thus the desired. $_\blacksquare$

Problems

2023 AIME I Problem 5

Square $ABCD$ is inscribed in a circle. Point $P$ is on this circle such that $AP \cdot CP = 56$ , and $BP \cdot DP = 90$ . What is the area of the square?

Solution: We may assume that $P$ is between $B$ and $C$ . Let $PA = a$ , $PB = b$ , $PC = C$ , $PD = d$ , and $AB = s$ . We have $a^2 + c^2 = AC^2 = 2s^2$ , because $AC$ is a diagonal. Similarly, $b^2 + d^2 = 2s^2$ . Therefore, $(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112$ . Similarly, $(b+d)^2 = 2s^2 + 180$ .

By Ptolemy's Theorem on $PCDA$ , $as + cs = ds\sqrt{2}$ , and therefore $a + c = d\sqrt{2}$ . By Ptolemy's on $PBAD$ , $bs + ds = as\sqrt{2}$ , and therefore $b + d = a\sqrt{2}$ . By squaring both equations, we obtain

$2d^2 = (a+c)^2 = 2s^2 + 112$ $2a^2 = (b+d)^2 = 2s^2 + 180.$

Thus, $a^2 = s^2 + 90$ , and $d^2 = s^2 + 56$ . Plugging these values into $a^2 + c^2 = b^2 + d^2 = 2s^2$ , we obtain $c^2 = s^2 - 90$ , and $b^2 = s^2 - 56$ . Now, we can solve using $a$ and $c$ (though using $b$ and $d$ yields the same solution for $s$ ).

$(\sqrt{s^2 + 90})(\sqrt{s^2 - 90}) = ac = 56$ $(s^2 + 90)(s^2 - 90) = 56^2$ $s^4 = 90^2 + 56^2 = 106^2$ $s^2 = 106.$

The answer is $\boxed{106}$ .

2004 AMC 10B Problem 24

In triangle $ABC$ we have $AB=7$ , $AC=8$ , $BC=9$ . Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects angle $BAC$ . What is the value of $AD/CD$ ?

$\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}$

Solution: Set $\overline{BD}$ 's length as $x$ . $CD$ 's length must also be $x$ since $\angle BAD$ and $\angle DAC$ intercept arcs of equal length(because $\angle BAD =\angle DAC$ ). Using Ptolemy's theorem, $7x+8x=9(AD)$ . The ratio is $\boxed{\frac{5}{3}}\implies(B)$

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: $\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AE}$ .

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 division 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)

$\text(AC)^2 = (AD)^2 - 121$

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

Solving gives $AD = 14$

@@ Line 7: / Line 7: @@
 <cmath>ac+bd=ef.</cmath>
-== Proof ==
+== Proof 1 ==
-Given cyclic quadrilateral <math>ABCD,</math> extend <math>CD</math> to <math>P</math> such that <math>\angle BAD=\angle CAP.</math>
+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 APD=\angle ABC</math>. Hence, <math>\triangle ABC \sim \triangle APD</math> by AA similarity and <math>\frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.</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>?

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