All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
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.

No Solution Yet Submitted by Gamer    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution can this be simplified? -- spoiler | Comment 1 of 3

Call the corner on the interior of the circle C. Call the intersections of the sides of the square with the circle A and B.  Triangle ABC is a right triangle with AB as the hypotenuse.  AB is also a chord of the circle, and the segment of the circle that it defines, along with the right triangle ABC, form the desired area.

The right triangle has area a*b/2.

The hypotenuse has length sqrt(a^2+b^2), and that is also the chord of the circle segment.

Half the arc of the segment is given by arcsin(sqrt(a^2+b^2)/(2*r)), so the whole arc is 2*arcsin(sqrt(a^2+b^2)/(2*r)).  This makes the sector of the circle defined by that arc have area pi*r^2 * 2*arcsin(sqrt(a^2+b^2)/(2*r)) / (2*pi) = r^2 * arcsin(sqrt(a^2+b^2)/(2*r)).  To get the area of the segment we need to subtract the area of the central triangle from that of the sector.  This triangle has sides r, r and sqrt(a^2+b^2), so Heron's formula can be used:

Let s = (2*r+sqrt(a^2+b^2))/2 = r + sqrt(a^2+b^2)/2

This triangle's area is sqrt(s*(s-r)^2*(s-sqrt(a^2+b^2))).

So the area of the segment is r^2 * arcsin(sqrt(a^2+b^2)/(2*r)) - sqrt(s*(s-r)^2*(s-sqrt(a^2+b^2))).

That makes the area of the whole overlap:

r^2 * arcsin(sqrt(a^2+b^2)/(2*r)) - sqrt(s*(s-r)^2*(s-sqrt(a^2+b^2))) + a*b/2.

This has been checked for reasonableness by having a program perform these calculations when (1) a=r and b=r, where the result should be one quarter the area of the full circle, and (2) a = 2*r and b=0, where the result should be half the area of the full circle:

r        a        b             area
5        5        5             19.63495408493621
5        10       0             39.26990816987242

pi = ATN(1) * 4
r = 5
a = 5: b = 5
GOSUB calc
a = 10: b = 0
GOSUB calc

s = r + SQR(a ^ 2 + b ^ 2) / 2
area = r ^ 2 * arcsin(SQR(a ^ 2 + b ^ 2) / (2 * r)) - SQR(s * (s - r) ^ 2 * (s - SQR(a ^ 2 + b ^ 2))) + a * b / 2
PRINT r, a, b, area

FUNCTION arcsin (x)
 IF ABS(x) = 1 THEN
   arcsin = pi / 2 * SGN(x)
   arcsin = ATN(x / SQR(1 - x * x))


  Posted by Charlie on 2006-08-09 10:47:16
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information