Inverse Pythagorean theorem

The area of triangle can be expressed in terms of either and , or and : :\begin{align} \tfrac{1}{2} AC \cdot BC &= \tfrac{1}{2} AB \cdot CD \\[4pt] (AC \cdot BC)^2 &= (AB \cdot CD)^2 \\[4pt] \frac{1}{CD^2} &= \frac{AB^2}{AC^2 \cdot BC^2} \end{align} given , and . Using the Pythagorean theorem, :\begin{align} \frac{1}{CD^2} &= \frac{BC^2 + AC^2}{AC^2 \cdot BC^2} \\[4pt] &= \frac{BC^2}{AC^2 \cdot BC^2} + \frac{AC^2}{AC^2 \cdot BC^2} \\[4pt] \quad \therefore \;\; \frac{1}{CD^2} &= \frac{ 1 }{AC^2} + \frac{1}{BC^2} \end{align} as above. Note in particular: :\begin{align} \tfrac{1}{2} AC \cdot BC &= \tfrac{1}{2} AB \cdot CD \\[4pt] CD &= \tfrac{AC \cdot BC}{AB} \\[4pt] \end{align} ==Special case of the cruciform curve==

The cruciform curve or cross curve is a quartic plane curve given by the equation :x^2 y^2 - b^2 x^2 - a^2 y^2 = 0 where the two parameters determining the shape of the curve, and are each . Substituting with and with gives :\begin{align} AC^2 BC^2 - CD^2 AC^2 - CD^2 BC^2 &= 0 \\[4pt] AC^2 BC^2 &= CD^2 BC^2 + CD^2 AC^2 \\[4pt] \frac{1}{CD^2} &= \frac{BC^2}{AC^2 \cdot BC^2} + \frac{AC^2}{AC^2 \cdot BC^2} \\[4pt] \therefore \;\; \frac{1}{CD^2} &= \frac{1}{AC^2} + \frac{1}{BC^2} \end{align} Inverse-Pythagorean triples can be generated using integer parameters and as follows. :\begin{align} AC &= (t^2 + u^2)(t^2 - u^2) \\ BC &= 2tu(t^2 + u^2) \\ CD &= 2tu(t^2 - u^2) \end{align} ==Application==

If two identical lamps are placed at and , the theorem and the inverse-square law imply that the light intensity at is the same as when a single lamp is placed at . ==See also==