Inverse Pythagorean theorem

Last updated July 28, 2024

Base Pytha- gorean triple	AC	BC	CD	AB
(3, 4, 5)	20 = 4× 5	15 = 3× 5	12 = 3× 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

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]

Proof

The area of triangle $△ ABC$ can be expressed in terms of either $AC$ and $BC$ , or $AB$ and $CD$ :

{\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,

{\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.

Note in particular:

{\begin{aligned}{\tfrac {1}{2}}AC\cdot BC&={\tfrac {1}{2}}AB\cdot CD\\[4pt]CD&={\tfrac {AC\cdot BC}{AB}}\\[4pt]\end{aligned}}

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{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]

{\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 $D$ .

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References

↑ R. B. Nelsen, Proof Without Words: A Reciprocal Pythagorean Theorem, Mathematics Magazine, 82, December 2009, p. 370
↑ The upside-down Pythagorean theorem, Jennifer Richinick, The Mathematical Gazette, Vol. 92, No. 524 (July 2008), pp. 313-316
↑ Johan Wästlund, "Summing inverse squares by euclidean geometry", http://www.math.chalmers.se/~wastlund/Cosmic.pdf, pp. 4–5.
↑ "Diophantine equation of three variables".

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[1] R. B. Nelsen, Proof Without Words: A Reciprocal Pythagorean Theorem, Mathematics Magazine, 82, December 2009, p. 370

[2] The upside-down Pythagorean theorem, Jennifer Richinick, The Mathematical Gazette, Vol. 92, No. 524 (July 2008), pp. 313-316

[3] Johan Wästlund, "Summing inverse squares by euclidean geometry", http://www.math.chalmers.se/~wastlund/Cosmic.pdf, pp. 4–5.

[4] "Diophantine equation of three variables".

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