Inverse Pythagorean theorem (original) (raw)

From Wikipedia, the free encyclopedia

Relation between the side lengths and altitude of a right triangle

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

1 C D 2 = 1 A C 2 + 1 B C 2 . {\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.

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

[edit]

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}}}

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.

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

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".