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Inverse Pythagorean theorem

## Summary

In geometry, the inverse Pythagorean theorem is as follows:[1]

Comparison of the inverse Pythagorean theorem with the Pythagorean theorem using the smallest positive integer inverse-Pythagorean triple in the table below
Base
Pytha-
gorean
triple
AC BC CD AB
(3, 4, 5) 20 = 5 15 = 5 12 = 4 25 = 52
(5, 12, 13) 156 = 12×13 65 = 5×13 60 = 5×12 169 = 132
(8, 15, 17) 255 = 15×17 136 = 8×17 120 = 8×15 289 = 172
(7, 24, 25) 600 = 24×25 175 = 7×25 168 = 7×24 625 = 252
(20, 21, 29) 609 = 21×29 580 = 20×29 420 = 20×21 841 = 292
All positive integer primitive inverse-Pythagorean triples having up to
three digits, with the hypotenuse for comparison
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
${\displaystyle {\frac {1}{CD^{2}}}={\frac {1}{AC^{2}}}+{\frac {1}{BC^{2}}}.}$

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:

{\displaystyle {\begin{aligned}{\tfrac {1}{2}}AC\cdot BC&={\tfrac {1}{2}}AB\cdot CD\\(AC\cdot BC)^{2}&=(AB\cdot CD)^{2}\\{\frac {1}{CD^{2}}}&={\frac {AB^{2}}{AC^{2}\cdot BC^{2}}}\end{aligned}}}

given CD > 0, AC > 0 and BC > 0.

Using the Pythagorean theorem,

{\displaystyle {\begin{aligned}{\frac {1}{CD^{2}}}&={\frac {BC^{2}+AC^{2}}{AC^{2}\cdot BC^{2}}}\\&={\frac {BC^{2}}{AC^{2}\cdot BC^{2}}}+{\frac {AC^{2}}{AC^{2}\cdot BC^{2}}}\\\quad \therefore \;\;{\frac {1}{CD^{2}}}&={\frac {1}{AC^{2}}}+{\frac {1}{BC^{2}}}\end{aligned}}}

as above.

## Special case of the cruciform curve

The cruciform curve or cross curve is a quartic plane curve given by the equation

${\displaystyle x^{2}y^{2}-b^{2}x^{2}-a^{2}y^{2}=0}$

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

{\displaystyle {\begin{aligned}AC^{2}BC^{2}-CD^{2}AC^{2}-CD^{2}BC^{2}&=0\\AC^{2}BC^{2}&=CD^{2}BC^{2}+CD^{2}AC^{2}\\{\frac {1}{CD^{2}}}&={\frac {BC^{2}}{AC^{2}\cdot BC^{2}}}+{\frac {AC^{2}}{AC^{2}\cdot BC^{2}}}\\\therefore \;\;{\frac {1}{CD^{2}}}&={\frac {1}{AC^{2}}}+{\frac {1}{BC^{2}}}\end{aligned}}}

Inverse-Pythagorean triples can be generated using integer parameters t and u as follows.[2]

{\displaystyle {\begin{aligned}AC&=(t^{2}+u^{2})(t^{2}-u^{2})\\BC&=2tu(t^{2}+u^{2})\\CD&=2tu(t^{2}-u^{2})\end{aligned}}}

## Application

If two identical lamps are placed at A and B, the theorem and the inverse-square law imply that the amount of light received at C is the same as when a single lamp is placed at D.