Triangle Arcs
TriangleArcs
In the above figure, let DeltaABC be a right triangle, arcs AP and AQ be segments of circles centered at C and B respectively, and define
a = BC
(1)
b = CA=CP
(2)
c = BA=BQ.
(3)
Then
| PQ^2=2BP·QC. |
(4)
|
This can be seen by letting x=BP, y=PQ, and z=QC and then solving the equations
x+y = c
(5)
y+z = b
(6)
x+y+z = a
(7)
to obtain
x = BP=a-b
(8)
y = PQ=-a+b+c
(9)
z = QC=a-c.
(10)
Plugging in the above gives
| y^2-2xz=b^2+c^2-a^2=0 |
(11)
|
by the Pythagorean theorem, so plugging in a=sqrt(b^2+c^2), the figure yields the algebraic identity
| (b+c-sqrt(b^2+c^2))^2=2(sqrt(b^2+c^2)-b)(sqrt(b^2+c^2)-c). |
(12)
|
The area of intersection formed (inside the triangle) by the circular sectors determined by arcs is given by
![画像: A=1/2[b^2cos^(-1)(b/a)+c^2cos^(-1)(c/a)-bc].](/index.cgi/contrast/https://mathworld.wolfram.com/images/equations/TriangleArcs/NumberedEquation4.svg){kind=link}
See also
Arc, Circle-Circle Intersection, Lens, TriangleExplore with Wolfram|Alpha
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References
Berndt, B. C. Ramanujan's Notebooks, Part IV. New York: Springer-Verlag, pp. 8-9, 1994.Dharmarajan, T. and Srinivasan, P. K. An Introduction to Creativity of Ramanujan, Part III. Madras, India: Assoc. Math. Teachers, pp. 11-13, 1987.Referenced on Wolfram|Alpha
Triangle ArcsCite this as:
Weisstein, Eric W. "Triangle Arcs." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/TriangleArcs.html