Inverse Pythagorean theorem explained

Base Pytha- gorean triple	AC	BC	CD		AB
(3, 4, 5)	20 = 4×5	15 = 3×5	12 = 3×4		25 = 5^2
(5, 12, 13)	156 = 12×13	65 = 5×13	60 = 5×12		169 = 13^2
(8, 15, 17)	255 = 15×17	136 = 8×17	120 = 8×15		289 = 17^2
(7, 24, 25)	600 = 24×25	175 = 7×25	168 = 7×24		625 = 25^2
(20, 21, 29)	609 = 21×29	580 = 20×29	420 = 20×21		841 = 29^2
All positive integer primitive inverse-Pythagorean triples having up to three digits, with the hypotenuse for comparison<						--import fractionsdef roundm(x): return int(round(x))is_seens = setfor a in range(1, 999 + 1): for b in range(1, a + 1): r = fractions.Fraction(a, b) c2 = fractions.Fraction((a * a) * (b * b), ((a * a) + (b * b))) c = roundm(c2.numerator ** 0.5) if c2.denominator 1 and c ** 2 c2.numerator and r not in is_seens: h = roundm((a * a + b * b) ** 0.5) print('	%4s	%4s	%4s	%4s = %2s^2\n	-' % (a, b, c, h, roundm(h ** 0.5))) is_seens.add(r)-->

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

(1)/(CD ^{2}) = (1)/(AC ^{2}) + (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.

Proof

The area of triangle can be expressed in terms of either and, or and :

\begin{align} \tfrac{ 1 }{ 2 } AC ⋅ BC & = \tfrac{ 1 }{ 2 } AB ⋅ CD \\[ 4 pt] (AC ⋅ BC) ^{2} & = (AB ⋅ CD) ^{2} \\[ 4 pt] (1)/(CD ^{2}) & = (AB ^{2})/(AC ^{2} ⋅ BC ^{2}) \end{align} given, and .

Using the Pythagorean theorem,

\begin{align} (1)/(CD ^{2}) & = (BC ^{2} + AC ^{2})/(AC ^{2} ⋅ BC ^{2}) \\[ 4 pt] & = (BC ^{2})/(AC ^{2} ⋅ BC ^{2}) + (AC ^{2})/(AC ^{2} ⋅ BC ^{2}) \\[ 4 pt] \therefore (1)/(CD ^{2}) & = (1)/(AC ^{2}) + (1)/(BC ^{2}) \end{align} as above.

Note in particular:

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

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, and are each .

Substituting with and with gives

\begin{align} AC ^{2} BC ^{2} - CD ^{2} AC ^{2} - CD ^{2} BC ^{2} & = 0 \\[ 4 pt] AC ^{2} BC ^{2} & = CD ^{2} BC ^{2} + CD ^{2} AC ^{2} \\[ 4 pt] (1)/(CD ^{2}) & = (BC ^{2})/(AC ^{2} ⋅ BC ^{2}) + (AC ^{2})/(AC ^{2} ⋅ BC ^{2}) \\[ 4 pt] \therefore (1)/(CD ^{2}) & = (1)/(AC ^{2}) + (1)/(BC ^{2}) \end{align}

Inverse-Pythagorean triples can be generated using integer parameters and as follows.^[4]

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

Application

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

Notes and References

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 euclideangeometry", http://www.math.chalmers.se/~wastlund/Cosmic.pdf, pp. 4–5.
4. Web site: Diophantine equation of three variables.