In geometry, the inverse Pythagorean theorem states that for a right triangle ABC with right angle at C and D as the foot of the perpendicular from C to the hypotenuse AB, the relationship 1/CD² = 1/AC² + 1/BC² holds. This differs from the well-known Pythagorean theorem, proven in Euclid's Elements. The table shows primitive inverse-Pythagorean triples like (3, 4, 5), where products such as AC = 4×5 = 20 and BC = 3×5 = 15 satisfy the theorem, providing integer examples with hypotenuse values for comparison. These triples illustrate the harmonic relation between the segments formed by the altitude on the hypotenuse in right triangles.
Proof
The area of triangle △ABC can be expressed in terms of either AC and BC, or AB and CD:
1 2 A C ⋅ B C = 1 2 A B ⋅ C D ( A C ⋅ B C ) 2 = ( A B ⋅ C D ) 2 1 C D 2 = A B 2 A C 2 ⋅ B C 2 {\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,
1 C D 2 = B C 2 + A C 2 A C 2 ⋅ B C 2 = B C 2 A C 2 ⋅ B C 2 + A C 2 A C 2 ⋅ B C 2 ∴ 1 C D 2 = 1 A C 2 + 1 B C 2 {\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.
Note in particular:
1 2 A C ⋅ B C = 1 2 A B ⋅ C D C D = A C ⋅ B C A B {\displaystyle {\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 {\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
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 1 C D 2 = B C 2 A C 2 ⋅ B C 2 + A C 2 A C 2 ⋅ B C 2 ∴ 1 C D 2 = 1 A C 2 + 1 B C 2 {\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
A C = ( t 2 + u 2 ) ( t 2 − u 2 ) B C = 2 t u ( t 2 + u 2 ) C D = 2 t u ( t 2 − u 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 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
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. http://www.math.chalmers.se/~wastlund/Cosmic.pdf ↩
"Diophantine equation of three variables". http://math.stackexchange.com/a/2688836 ↩