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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

 A=1/2[b^2cos^(-1)(b/a)+c^2cos^(-1)(c/a)-bc].
(13)

See also

Arc, Circle-Circle Intersection, Lens, Triangle

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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 Arcs

Cite this as:

Weisstein, Eric W. "Triangle Arcs." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/TriangleArcs.html

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