Difference between revisions of "Bretschneider's formula"

Revision as of 09:59, 21 January 2020

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

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

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:

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

@@ Line 11: / Line 11: @@
 <math>K = \frac{1}{2} |\vec{p} \times \vec{q}| </math>
-<math> K^2 = \frac{1}{4} |\vec{p} \times \vec{q}|^2 </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>= \frac{1}{4} (\vec{p} \times \vec{q}) \cdot (\vec{p} \times \vec{q}) </math>
-[[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:
-<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>
-<math>= \frac{1}{4} [|\vec{p}|^2|\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2] </math>
 <math>K = \frac{1}{2} \sqrt{|\vec{p}|^2|\vec{q}|^2 - (\vec{p} \cdot \vec{q})^2} </math>

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Difference between revisions of "Bretschneider's formula"

Revision as of 09:59, 21 January 2020

Proof

See Also