Difference between revisions of "Bretschneider's formula"

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* [[Geometry]]
 
* [[Geometry]]
  
{{stub}}
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==The Proof==
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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>
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Therefore
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:<math> 4S^2 = (pq)^2\sin^2 A + (rs)^2\sin^2 C + 2pqrs\sin A\sin C. \, </math>
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The [[cosine law]] implies that
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:<math> p^2 + q^2 -2pq\cos A = r^2 + s^2 -2rs\cos C, \, </math>
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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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:<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>
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Introducing the semiperimeter
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:<math>T = \frac{p+q+r+s}{2},</math>
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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>
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and Bretschneider's formula follows.
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NOTE TO ALL: this proof was taken from Wikipedia on December the 1st, 2006.

Revision as of 22:56, 1 December 2006

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.

See Also

The Proof

Denote the area of the quadrilateral by S. Then we have

$\begin{align} S &= \text{area of } \triangle ADB + \text{area of } \triangle BDC \
                       &= \tfrac{1}{2}pq\sin A + \tfrac{1}{2}rs\sin C 

\end{align}$ (Error compiling LaTeX. Unknown error_msg)

Therefore

$4S^2 = (pq)^2\sin^2 A + (rs)^2\sin^2 C + 2pqrs\sin A\sin C. \,$

The cosine law implies that

$p^2 + q^2 -2pq\cos A = r^2 + s^2 -2rs\cos C, \,$

because both sides equal the square of the length of the diagonal BD. This can be rewritten as

$\tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2\cos^2 A +(rs)^2\cos^2 C -2 pqrs\cos A\cos C. \,$

Substituting this in the above formula for $4S^2$ yields

$4S^2 + \tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2 + (rs)^2 - 2pqrs\cos (A+C). \,$

This can be written as

$16S^2 = (r+s+p-q)(r+s+q-p)(r+p+q-s)(s+p+q-r) - 16pqrs \cos^2 \frac{A+C}2.$

Introducing the semiperimeter

$T = \frac{p+q+r+s}{2},$

the above becomes

$16S^2 = 16(T-p)(T-q)(T-r)(T-s) - 16pqrs \cos^2 \frac{A+C}2$

and Bretschneider's formula follows. NOTE TO ALL: this proof was taken from Wikipedia on December the 1st, 2006.