A Math Forum Project

Geometry Forum - Problem of the Week Solutions - Circles, Chords, and Polygons, March 13-17, 1995 Annie says: Only two correct solutions this week, but they're very good. Sean especially explains some things really well, and Samantha's answer is concise and to the point. These ideas are a good way to find the center of a circle - not a simple task without knowing this part. You can come close by guessing, or playing around with a compass, but this'll do it for you every time (barring that human error that Sean mentions). Sean Nichols Hello, This is Sean Nichols at Burnaby South Secondary School in Burnaby, BC, Canada. My solution to the Problem of the Week for March 13 - 17 is as follows. The first part of the solution is a almost given in the answer; given a circle and a chord, by definition, both endpoints of the chord lie on the circle itself. Therefore, again by definition, the center of the circle is equidistant from these two endpoints. All points equidistant from the endpoints of a segment lie on the perpendicular bisector of the segment. So, the center of the circle _must_ lie on the perpendicular bisector, and the perpendicular bisector goes through the center. The solution to the first part leads us to the second. In order for a circle to be circumscribed around a polygon, all vertices of the polygon must lie on the circle. (i.e: all sides of the polygon are chords of the circle.) So, it would follow that the center of the circle must lie on the perpendicular bisector of each and every side. For this to happen, all perpendicular bisectors must intersect at a single point (this point is the only location through which they all pass). In fact, that is how the circumcenter of a polygon is normally found, by finding the intersection of the perpendicular bisectors. Normally, finding the point of intersection of only two will suffice. Then, draw a circle center this point which passes through one of the vertices, and see if it passes through all vertices. If so, then a circumcircle can be drawn about the polygon. If not, then no such circle can be drawn, unless the calculations are off somewhere. Conversely, if you have to find this without drawing the circle itself, you can just draw the bisectors which look least likely, visually, to pass through the point of intersection, and draw them to find out. It only takes one which doesn't to disqualify the whole polygon from having a circumcircle. Of course, using this method, if one bisector passes _close_ to the point of intersection, but does not cross it exactly, it could be either that in fact the polygon has no circumcircle, or that there was a slight human error in drawing the bisector (such things are inevitable). If the latter case is true, the only way to find out would be to actually draw the circle. Thanks, - Sean Nichols Samantha Brenner Samantha Brenner, grade 9, Fairfield HS Draw a radius from each end of the chord, AB, to the center, O. Draw a segment, OC, perpendicular to AB. Triangle AOB is isosceles, therefore, the altitude from O to AB is also a median. This makes OC a perpendicular bisector of AB. Since a segment has exactly one perpendicular bisector, the perpendicular bisector of the chord AB goes through the center. Given any polygon, you may not be able to circumscribe a circle around it. In some polygons it would be impossible for all the vertices to touch the circle. You could figure out if a polygon could be circumscribed if you drew all perpendicular bisectors of each side and they intersected at one point. That one point would be the center of the circle. That would mean the sides of the polygon would be chords of the circle. Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page