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Difference between revisions of "Bretschneider's formula"

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==Proof==
 
==Proof==
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''This article needs a proof. Please help us out by [http://www.artofproblemsolving.com/Wiki/index.php?title=Bretschneider%27s_formula&action=edit adding one]''.
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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>.
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<math>K = \frac{1}{2} |\vec{p} \times \vec{q}| </math>
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<math> K^2 = \frac{1}{4} |\vec{p} \times \vec{q}|^2 </math>
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<math>= \frac{1}{4} (\vec{p} \times \vec{q}) \cdot (\vec{p} \times \vec{q}) </math>
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[[Lagrange's Identity]] states that <math>(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})</math>. Therefore:
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<math>K^2 = \frac{1}{4} [ (\vec{p} \cdot \vec{p})(\vec{q} \cdot \vec{q}) - (\vec{p} \cdot \vec{q})(\vec{p} \cdot \vec{q}) ] </math>
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<math>= \frac{1}{4} [|\vec{p}|^2|\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2] </math>
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<math>K = \frac{1}{2} \sqrt{|\vec{p}|^2|\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2} </math>
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<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - (2 \vec{p} \cdot \vec{q})^2} </math>
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<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [2 (\vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b})]^2} </math>
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<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>
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<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>
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<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>
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<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>
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<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>
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<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{d}|^2 + (\vec{a} + \vec{c})\cdot(\vec{a} + \vec{c}) - (\vec{b} + \vec{d})\cdot(\vec{b} + \vec{d})]^2} </math>
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<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{d}|^2 + |\vec{a} + \vec{c}|^2 - |\vec{b} + \vec{d}|^2]^2} </math>
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<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{d}|^2 + |\vec{a} + \vec{c}|^2 - |-(\vec{a} + \vec{c})|^2]^2} </math>
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<math>= \frac{1}{4} \sqrt{4 |\vec{p}|^2|\vec{q}|^2 - [|\vec{b}|^2 + |\vec{d}|^2 - |\vec{a}|^2 - |\vec{d}|^2]^2} </math>
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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:
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<cmath> K = \frac{1}{4} \sqrt{4p^2q^2 - (b^2 + d^2 - a^2 - c^2)^2} </cmath>
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==See Also==
 
==See Also==
 
* [[Brahmagupta's formula]]
 
* [[Brahmagupta's formula]]

Revision as of 15:26, 3 January 2012

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

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


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

$K^2 = \frac{1}{4} |\vec{p} \times \vec{q}|^2$

$= \frac{1}{4} (\vec{p} \times \vec{q}) \cdot (\vec{p} \times \vec{q})$


Lagrange's Identity states that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$. Therefore:


$K^2 = \frac{1}{4} [ (\vec{p} \cdot \vec{p})(\vec{q} \cdot \vec{q}) - (\vec{p} \cdot \vec{q})(\vec{p} \cdot \vec{q}) ]$

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

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

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

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

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

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

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

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

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

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

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

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

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

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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