Difference between revisions of "Ptolemy's theorem"
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<cmath>ac+bd=ef.</cmath> | <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 | + | 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> | 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> | ||
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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>. | 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 == | == 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=== | ===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>? |
Latest revision as of 20:39, 9 March 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.
Contents
Statement
Given a cyclic quadrilateral with side lengths
and diagonals
:
Proof 1
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
.
Proof 2 (inversion)
We provide a proof for the general case of Ptolemy's theorem, Ptolemy's Inequality.
Let be four points in the Euclidean plane. Taking an inversion centered at
(the point doesn't matter, it can be any of the four) with radius
, we have that
by the Triangle Inequality, with equality holding when
are collinear, i.e. when
lie on a circle containing
Additionally, by the Inversion Distance Formula, we may express the inequality as the following:
Dividing by and multiplying everything by
we get
and thus the desired.
Problems
2023 AIME I Problem 5
Square is inscribed in a circle. Point
is on this circle such that
, and
. What is the area of the square?
Solution: We may assume that is between
and
. Let
,
,
,
, and
. We have
, because
is a diagonal. Similarly,
. Therefore,
. Similarly,
.
By Ptolemy's Theorem on ,
, and therefore
. By Ptolemy's on
,
, and therefore
. By squaring both equations, we obtain
Thus, , and
. Plugging these values into
, we obtain
, and
. Now, we can solve using
and
(though using
and
yields the same solution for
).
The answer is .
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