Inverse Pythagorean theorem

Comparison of the inverse Pythagorean theorem with the Pythagorean theoremusing the smallestpositive integer inverse-Pythagorean triple in the table below.

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

Ingeometry,theinverse Pythagorean theorem(also known as thereciprocal Pythagorean theoremor theupside down Pythagorean theorem) is as follows:

LetA,Bbe the endpoints of thehypotenuseof aright triangle△ABC.LetDbe the foot ofa perpendiculardropped fromC,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 ofEuclid'sElements,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△ABCcan be expressed in terms of eitherACandBC,orABandCD:

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

givenCD> 0,AC> 0andBC> 0.

Using thePythagorean 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

Thecruciform curveor cross curve is aquartic plane curvegiven by the equation

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

where the twoparametersdetermining the shape of the curve,aandbare eachCD.

SubstitutingxwithACandywithBCgives

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

Inverse-Pythagorean triples can be generated using integer parameterstanduas follows.

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

Application

If two identical lamps are placed atAandB,the theorem and theinverse-square lawimply that the light intensity atCis the same as when a single lamp is placed atD.

See also

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