Inverse Pythagorean theorem

Comparison of the inverse Pythagorean theorem with the Pythagorean theorem Template:Nowrap positive integer inverse-Pythagorean triple in the table below.

Base Pytha- gorean triple	AC	BC	CD		AB
(3, Template:Fsp4, Template:Fsp5)	20 = Template:Fsp4×Template:Fsp5	15 = Template:Fsp3×Template:Fsp5	12 = Template:Fsp3×Template:Fsp4		25 = Template:Fsp52
(5, 12, 13)	156 = 12×13	65 = Template:Fsp5×13	60 = Template:Fsp5×12		169 = 132
(8, 15, 17)	255 = 15×17	136 = Template:Fsp8×17	120 = Template:Fsp8×15		289 = 172
(7, 24, 25)	600 = 24×25	175 = Template:Fsp7×25	168 = Template:Fsp7×24		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 or the upside down Pythagorean theorem) is as follows:

Let Template:Mvar, Template:Mvar be the endpoints of the hypotenuse of a right triangle Template:Math. Let Template:Mvar be the foot of a perpendicular dropped from Template:Mvar, the vertex of the right angle, to the hypotenuse. Then

$\frac{1}{C{D}^2}=\frac{1}{A{C}^2}+\frac{1}{B{C}^2}.$

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 Template:Math can be expressed in terms of either Template:Mvar and Template:Mvar, or Template:Mvar and Template:Mvar:

$\begin{array}{cc}\frac{1}{2}AC\cdot BC& =\frac{1}{2}AB\cdot CD\\ (AC\cdot BC{)}^2& =(AB\cdot CD{)}^2\\ \frac{1}{C{D}^2}& =\frac{A{B}^2}{A{C}^2\cdot B{C}^2}\end{array}$

given Template:Math, Template:Math and Template:Math.

Using the Pythagorean theorem,

$\begin{array}{cc}\frac{1}{C{D}^2}& =\frac{B{C}^2+A{C}^2}{A{C}^2\cdot B{C}^2}\\ & =\frac{B{C}^2}{A{C}^2\cdot B{C}^2}+\frac{A{C}^2}{A{C}^2\cdot B{C}^2}\\ \therefore \frac{1}{C{D}^2}& =\frac{1}{A{C}^2}+\frac{1}{B{C}^2}\end{array}$

as above.

Note in particular:

$\begin{array}{cc}\frac{1}{2}AC\cdot BC& =\frac{1}{2}AB\cdot CD\\ CD& =\frac{AC\cdot BC}{AB}\\ \end{array}$

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, Template:Mvar and Template:Mvar are each Template:Mvar.

Substituting Template:Mvar with Template:Mvar and Template:Mvar with Template:Mvar gives

$\begin{array}{cc}A{C}^{2}B{C}^2-C{D}^{2}A{C}^2-C{D}^{2}B{C}^2& =0\\ A{C}^{2}B{C}^2& =C{D}^{2}B{C}^2+C{D}^{2}A{C}^2\\ \frac{1}{C{D}^2}& =\frac{B{C}^2}{A{C}^2\cdot B{C}^2}+\frac{A{C}^2}{A{C}^2\cdot B{C}^2}\\ \therefore \frac{1}{C{D}^2}& =\frac{1}{A{C}^2}+\frac{1}{B{C}^2}\end{array}$

Inverse-Pythagorean triples can be generated using integer parameters Template:Mvar and Template:Mvar as follows.

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

Application

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

See also

• Template:Annotated link
• Template:Annotated link

References

Template:Geometry-stub