inverse Pythagorean theorem
{{Short description|Relation between the side lengths and altitude of a right triangle}}
File:inverse_pythagorean_theorem.svg
| class="wikitable" style="text-align:right; float:right; clear:right; margin-left:1ex;"
! Base | ||||
| (3, {{fsp}}4, {{fsp}}5) | 20 = {{fsp}}4×{{fsp}}5 | 15 = {{fsp}}3×{{fsp}}5 | 12 = {{fsp}}3×{{fsp}}4 | 25 = {{fsp}}52 |
| (5, 12, 13) | 156 = 12×13 | 65 = {{fsp}}5×13 | 60 = {{fsp}}5×12 | 169 = 132 |
| (8, 15, 17) | 255 = 15×17 | 136 = {{fsp}}8×17 | 120 = {{fsp}}8×15 | 289 = 172 |
| (7, 24, 25) | 600 = 24×25 | 175 = {{fsp}}7×25 | 168 = {{fsp}}7×24 | 625 = 252 |
| (20, 21, 29) | 609 = 21×29 | 580 = 20×29 | 420 = 20×21 | 841 = 292 |
| colspan="6" style="text-align:left;"|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 theoremR. B. Nelsen, Proof Without Words: A Reciprocal Pythagorean Theorem, Mathematics Magazine, 82, December 2009, p. 370 or the upside down Pythagorean theoremThe upside-down Pythagorean theorem, Jennifer Richinick, The Mathematical Gazette, Vol. 92, No. 524 (July 2008), pp. 313-316) is as follows:Johan Wästlund, "Summing inverse squares by euclidean
geometry", http://www.math.chalmers.se/~wastlund/Cosmic.pdf, pp. 4–5.
:Let {{mvar|A}}, {{mvar|B}} be the endpoints of the hypotenuse of a right triangle {{math|△ABC}}. Let {{mvar|D}} be the foot of a perpendicular dropped from {{mvar|C}}, the vertex of the right angle, to the hypotenuse. Then
::
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 {{math|△ABC}} can be expressed in terms of either {{mvar|AC}} and {{mvar|BC}}, or {{mvar|AB}} and {{mvar|CD}}:
:
\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{align}
given {{math|CD > 0}}, {{math|AC > 0}} and {{math|BC > 0}}.
Using the Pythagorean theorem,
:
\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{align}
as above.
Note in particular:
:
\tfrac{1}{2} AC \cdot BC &= \tfrac{1}{2} AB \cdot CD \\[4pt]
CD &= \tfrac{AC \cdot BC}{AB} \\[4pt]
\end{align}
Special case of the cruciform curve
The cruciform curve or cross curve is a quartic plane curve given by the equation
:
where the two parameters determining the shape of the curve, {{mvar|a}} and {{mvar|b}} are each {{mvar|CD}}.
Substituting {{mvar|x}} with {{mvar|AC}} and {{mvar|y}} with {{mvar|BC}} gives
:
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{align}
Inverse-Pythagorean triples can be generated using integer parameters {{mvar|t}} and {{mvar|u}} as follows.{{Cite web|url=http://math.stackexchange.com/a/2688836|title=Diophantine equation of three variables}}
:
AC &= (t^2 + u^2)(t^2 - u^2) \\
BC &= 2tu(t^2 + u^2) \\
CD &= 2tu(t^2 - u^2)
\end{align}
Application
If two identical lamps are placed at {{mvar|A}} and {{mvar|B}}, the theorem and the inverse-square law imply that the light intensity at {{mvar|C}} is the same as when a single lamp is placed at {{mvar|D}}.
See also
- {{Annotated link|Geometric mean theorem}}
- {{Annotated link|Pythagorean theorem}}