Problem: Circular Window

by

Teisha, Angie, & Beth

Use compass and ruler to construct the following design of a circular window. How can it be constructed?
What fraction of the area of the circle is in the "triangular " region in the center bounded by the three arcs?



For the construction it is necessary to see the window in terms of the circles that make it up, rather than the arcs that are visible. Once we see the circles in the window the next question becomes what are their relationships with each other. Then it becomes easier to do the construction. There may be other, even easier, ways to do this construction, but following are the steps that I took.


First, construct an equilateral triangle. Look back at the picture to see if you can see the equilateral triangle in the drawing.


Secondly, I constructed the perpendicular bisectors to each line segment. At this point I really only need the midpoint of the segments, however the purpose of the lines themselves will become more apparent later.



Next, I constructed three circles with centers at A, B, and E. The radii of each these circles are equal to one-half the the length of a segment of the triangle.



Then it becomes necessary to mark the point of intersection of circle A with line AG. We do the same for each circle, marking the points of intersection of the circles with the perpendicular bisector of the segments ( I told you we would need those perpendicular bisectors). We also marked the point of intersection of the three perpendicular bisectors. From that point we will construct a larger circle with center at the intersection of the lines of bisection and tangent to each circle at the point of intersection of the circles with the perpendicular bisectors of the triangle's segments.

I removed the perpedicular bisectors to keep the picture clean and easy to work with. Can you see the window in the above figure? If not keep reading!!


The next thing I did was construct an arc on each of the smaller circles. Each arc went through the midpoints of the segments that the circle intersected and through the point of intersection between the circle and the perpendicular bisector and the circle.



Removal of the blue arcs above yields the following picture. I left the equilateral triangle in the picture for those who still did not see the triangle in the original window.



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