Comparison of the inverse Pythagorean theorem with the Pythagorean theorem using the smallest positive integer inverse-Pythagorean triple in the table below.

In geometry, the inverse Pythagorean theorem (also known as the reciprocal Pythagorean theorem^[1] or the upside down Pythagorean theorem^[2]) is as follows:^[3]

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:

${\displaystyle {\begin{aligned}{\tfrac {1}{2}}AC\cdot BC&={\tfrac {1}{2}}AB\cdot CD\\[4pt](AC\cdot BC)^{2}&=(AB\cdot CD)^{2}\\[4pt]{\frac {1}{CD^{2}}}&={\frac {AB^{2}}{AC^{2}\cdot BC^{2}}}\end{aligned}}}$

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

Using the Pythagorean theorem,

${\displaystyle {\begin{aligned}{\frac {1}{CD^{2}}}&={\frac {BC^{2}+AC^{2}}{AC^{2}\cdot BC^{2}}}\\[4pt]&={\frac {BC^{2}}{AC^{2}\cdot BC^{2}}}+{\frac {AC^{2}}{AC^{2}\cdot BC^{2}}}\\[4pt]\quad \therefore \;\;{\frac {1}{CD^{2}}}&={\frac {1}{AC^{2}}}+{\frac {1}{BC^{2}}}\end{aligned}}}$

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

${\displaystyle {\begin{aligned}AC^{2}BC^{2}-CD^{2}AC^{2}-CD^{2}BC^{2}&=0\\[4pt]AC^{2}BC^{2}&=CD^{2}BC^{2}+CD^{2}AC^{2}\\[4pt]{\frac {1}{CD^{2}}}&={\frac {BC^{2}}{AC^{2}\cdot BC^{2}}}+{\frac {AC^{2}}{AC^{2}\cdot BC^{2}}}\\[4pt]\therefore \;\;{\frac {1}{CD^{2}}}&={\frac {1}{AC^{2}}}+{\frac {1}{BC^{2}}}\end{aligned}}}$

Inverse-Pythagorean triples can be generated using integer parameters t and u as follows.^[4]

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

Application

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

See also

• Geometric mean theorem – Theorem about right triangles
• Pythagorean theorem – Relation between sides of a right triangle

References