Equal Parts

By Melissa Madsen

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

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.