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Circular Logic III (Posted on 2006-08-09) Difficulty: 4 of 5
Suppose you had a square intersecting a circle like in Circular Logic, where one vertex is inside the circle and three are outside.

If the vertex inside the circle was not located at the center, but anywhere inside the circle, such that segments ending at that vertex were of length a and b and the circle had radius r, what would be the area of their intersection be?

Note: A geometric solution is expected, without using calculus.

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Puzzle Thoughts Comment 4 of 4 |
The required  area of their intersection
= r^2 * arcsin(v(a^2+b^2)/(2*r)) - sqrt(s*(s-r)^2*(s-v(a^2+b^2))) + a*b/2.

  Posted by K Sengupta on 2024-05-13 11:47:27
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