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Difference between revisions of "Bretschneider's formula"
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Suppose we have a [[quadrilateral]] with [[edge]]s of length <math>a,b,c,d</math> (in that order) and [[diagonal]]s of length <math>p, q</math>. '''Bretschneider's formula''' states that the [[area]]
 
Suppose we have a [[quadrilateral]] with [[edge]]s of length <math>a,b,c,d</math> (in that order) and [[diagonal]]s of length <math>p, q</math>. '''Bretschneider's formula''' states that the [[area]]
−
<math>[ABCD]=\frac{1}{4}*\sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}</math>.
+
<math>[ABCD]=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}</math>.
  
 
It can be derived with [[vector]] [[geometry]].
 
It can be derived with [[vector]] [[geometry]].
  
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==See Also==
+
==Proof==
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* [[Brahmagupta's formula]]
+
 
−
* [[Geometry]]
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Suppose a quadrilateral has sides <math>\vec{a}, \vec{b}, \vec{c}, \vec{d}</math> such that <math>\vec{a} + \vec{b} + \vec{c} + \vec{d} = \vec{0}</math> and that the diagonals of the quadrilateral are <math>\vec{p} = \vec{b} + \vec{c} = -\vec{a} - \vec{d}</math> and <math>\vec{q} = \vec{a} + \vec{b} = -\vec{c} - \vec{d}</math>. The area of any such quadrilateral is <math>\frac{1}{2} |\vec{p} \times \vec{q}|</math>.
 +
 
 +
 
 +
<math>K = \frac{1}{2} |\vec{p} \times \vec{q}| </math>
 +
 
 +
[[Lagrange's Identity]] states that <math>|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2=|\vec{a}\times\vec{b}|^2 \implies \sqrt{|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2}=|\vec{a}\times\vec{b}|</math>. Therefore:
 +
 
 +
<math>K = \frac{1}{2} \sqrt{|\vec{p}|^2|\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2} </math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - (2 \vec{p} \cdot \vec{q})^2} </math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 (\vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b})]^2} </math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{b} \cdot (\vec{a} + \vec{b}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2} </math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot (\vec{c} + \vec{d}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2} </math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d} + 2 \vec{a} \cdot \vec{c} + 2 \vec{b} \cdot \vec{c}]^2}</math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{a} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d}]^2} </math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - ([(\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - \vec{a}\cdot\vec{a} - \vec{c}\cdot\vec{c}] - [(\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d}) - \vec{b}\cdot\vec{b} - \vec{d}\cdot\vec{d}])^2} </math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + (\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - (\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d})]^2} </math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |\vec{b} + \vec{d}|^2]^2} </math>
 +
 
 +
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |-(\vec{a} + \vec{c})|^2]^2} </math>
  
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==The Proof==
+
<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2]^2} </math>
−
Denote the area of the quadrilateral by ''S''. Then we have
 
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:<math> \begin{align} S &= \text{area of } \triangle ADB + \text{area of } \triangle BDC \\
 
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                        &= \tfrac{1}{2}pq\sin A + \tfrac{1}{2}rs\sin C
 
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\end{align} </math>
 
  
−
Therefore
+
Then if <math>a, b, c, d</math> represent <math>|\vec{a}|, |\vec{b}|, |\vec{c}|, |\vec{d}|</math> (and are thus the side lengths) while <math>p, q</math> represent <math>|\vec{p}|, |\vec{q}|</math> (and are thus the diagonal lengths), the area of a quadrilateral is:
−
:<math> 4S^2 = (pq)^2\sin^2 A + (rs)^2\sin^2 C + 2pqrs\sin A\sin C. \, </math>
 
  
−
The [[cosine law]] implies that
+
<cmath> K = \frac{1}{4} \sqrt{4p^2q^2 - (b^2 + d^2 - a^2 - c^2)^2} </cmath>
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:<math> p^2 + q^2 -2pq\cos A = r^2 + s^2 -2rs\cos C, \, </math>
 
−
because both sides equal the square of the length of the diagonal ''BD''. This can be rewritten as
 
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:<math>\tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2\cos^2 A +(rs)^2\cos^2 C -2 pqrs\cos A\cos C. \,</math>
 
  
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Substituting this in the above formula for <math>4S^2</math> yields
 
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:<math>4S^2 + \tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2 + (rs)^2 - 2pqrs\cos (A+C). \, </math>
 
  
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This can be written as
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==See Also==
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:<math>16S^2 = (r+s+p-q)(r+s+q-p)(r+p+q-s)(s+p+q-r) - 16pqrs \cos^2 \frac{A+C}2. </math>
+
* [[Brahmagupta's formula]]
 +
* [[Geometry]]
  
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Introducing the semiperimeter
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[[Category:Geometry]]
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:<math>T = \frac{p+q+r+s}{2},</math>
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[[Category:Theorems]]
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the above becomes
 
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:<math>16S^2 = 16(T-p)(T-q)(T-r)(T-s) - 16pqrs \cos^2 \frac{A+C}2</math>
 
−
and Bretschneider's formula follows.
 
−
NOTE TO ALL: this proof was taken from Wikipedia on December the 1st, 2006.
 

Latest revision as of 03:51, 12 February 2021

Suppose we have a quadrilateral with edges of length $a,b,c,d$ (in that order) and diagonals of length $p, q$. Bretschneider's formula states that the area $[ABCD]=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}$.

It can be derived with vector geometry.

Proof

Suppose a quadrilateral has sides $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ such that $\vec{a} + \vec{b} + \vec{c} + \vec{d} = \vec{0}$ and that the diagonals of the quadrilateral are $\vec{p} = \vec{b} + \vec{c} = -\vec{a} - \vec{d}$ and $\vec{q} = \vec{a} + \vec{b} = -\vec{c} - \vec{d}$. The area of any such quadrilateral is $\frac{1}{2} |\vec{p} \times \vec{q}|$.


$K = \frac{1}{2} |\vec{p} \times \vec{q}|$

Lagrange's Identity states that $|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2=|\vec{a}\times\vec{b}|^2 \implies \sqrt{|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2}=|\vec{a}\times\vec{b}|$. Therefore:

$K = \frac{1}{2} \sqrt{|\vec{p}|^2|\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - (2 \vec{p} \cdot \vec{q})^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 (\vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b})]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{b} \cdot (\vec{a} + \vec{b}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot (\vec{c} + \vec{d}) + 2 \vec{c} \cdot (\vec{a} + \vec{b})]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [-2 \vec{b} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d} + 2 \vec{a} \cdot \vec{c} + 2 \vec{b} \cdot \vec{c}]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 \vec{a} \cdot \vec{c} - 2 \vec{b} \cdot \vec{d}]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - ([(\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - \vec{a}\cdot\vec{a} - \vec{c}\cdot\vec{c}] - [(\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d}) - \vec{b}\cdot\vec{b} - \vec{d}\cdot\vec{d}])^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + (\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - (\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d})]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |\vec{b} + \vec{d}|^2]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2 + |\vec{a} + \vec{c}|^2 - |-(\vec{a} + \vec{c})|^2]^2}$

$= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{c}|^2]^2}$

Then if $a, b, c, d$ represent $|\vec{a}|, |\vec{b}|, |\vec{c}|, |\vec{d}|$ (and are thus the side lengths) while $p, q$ represent $|\vec{p}|, |\vec{q}|$ (and are thus the diagonal lengths), the area of a quadrilateral is:

\[K = \frac{1}{4} \sqrt{4p^2q^2 - (b^2 + d^2 - a^2 - c^2)^2}\]


See Also

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