Inverse Pythagorean theorem



In geometry, the inverse Pythagorean theorem (also known as the reciprocal Pythagorean theorem or the upside down Pythagorean theorem ) is as follows:


 * Let $A$, $B$ be the endpoints of the hypotenuse of a right triangle $△ABC$. Let $D$ be the foot of a perpendicular dropped from $C$, the vertex of the right angle, to the hypotenuse. Then
 * $$ \frac 1 {CD^2} = \frac 1 {AC^2} + \frac 1 {BC^2}.$$

This theorem should not be confused with proposition 48 in book 1 of Euclid's Elements, the converse of the Pythagorean theorem, which states that if the square on one side of a triangle is equal to the sum of the squares on the other two sides then the other two sides contain a right angle.

Proof
The area of triangle $△ABC$ can be expressed in terms of either $AC$ and $BC$, or $AB$ and $CD$:
 * $$\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 $CD &gt; 0$, $AC &gt; 0$ and $BC &gt; 0$.

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, $a$ and $b$ are each $CD$.

Substituting $x$ with $AC$ and $y$ with $BC$ 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 $t$ and $u$ 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 $A$ and $B$, the theorem and the inverse-square law imply that the light intensity at $C$ is the same as when a single lamp is placed at $D$.