Equal Parts

Construct acircle that has five regions of equal area where none of the perimeters of theregions can pass through the center of the circle.

The first thought Ihad for this problem is of a target at the fair.  The carneys would not choose a circle with five equalregions for a game, but this does make an interesting math problem.

In order to createthis target, first I considered the areas.  If I let the radius of the innermost circle be one, the areaof the circle is pi.  So the areaof the second circle is 2pi, the area of the third circle is 3 pi, the area ofthe fourth circle is four pi and the area of the fifth circle is 5 pi.

Solving backwardsusing the area of a circle we find that (dum dum dum)

To the GSP Mobile:

Createa right isosceles triangle with 2 sides of one unit.  The hypotenuse is the square root of two.  Add a leg perpendicular to thehypotenuse with a length of one. Continue this until you have drawn a spiral with the last hypotenuseequivalent to square root of 5 units.

Sides of right isosceles withunit one sides.  First draw thesegment of unit one then draw a line perpendicular to it at an endpoint.  Using the center and radius draw acircle to get the other leg to be equal. By the magic of the Pythagorean theorem, the hypotenuse must be thesquare root of two.

Draw a line perpendicular to thehypotenuse with length one.  Usingthe original side draw a circle to create a leg of one.

Connect the top point with pointY creating a hypotenuse of square root 3.

Continue this until thehypotenuse equal square root of 5

Using the original line and thehypotenuses create your circles.

It should look like thisÉ. TOthe GSP Mobile

If you move either Y or Z youcan change the circle.