Circles, chords, perpendicular bisectors - March 18-22, 1996
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Last weekend I introduced some teachers to The Geometer's Sketchpad. We were actually doing an Internet workshop, but we like to throw a little geometry in to keep them on their toes. I had them do a construction that made me think of this problem.First, take a circle with a chord drawn - any chord. Prove that the perpendicular bisector of the chord goes through the center of the circle.
Now, using your newfound knowledge, explain how to decide, given any polygon, whether you can circumscribe a circle around it.
Annie says: Solutions
The hard part of this problem was actually proving the first part of the question. A number of students stated theorems that showed that the perpendicular bisector of a chord goes through the center of the circle, but I had asked for a proof. The second part showed why you need to draw a lot of pictures - a number of people said that you can circumscribe a circle around any polygon, which certainly isn't the case! You need to draw things besides squares and triangles to really test a question like this.
I had some conversations with people about what constitutes a proof. A couple of folks said that any line through the center of a circle bisects a chord, but it was a little tougher to show it the other way around. While some things are certainly true both ways, not everything is, so it's important to make sure you are going about things in the right direction (for example, all dobermans are dogs, but not all dogs are dobermans).
Following are highlights. The names of all the people who submitted correct solutions and most of the solutions are also available.
Christine Francescani Grade: 10 School: Martin County High SchoolTo begin, I drew chord AC and perpendicularly bisected it with segment FB. Segments AD and DC are congruent. Segment BD is congruent to itself, and therefore triangles ABD and CBD are congruent. Because segments AB and CB are congruent, their chords are congruent. The arcs made by angles ABD and CBD (AF and CF) are congruent because their angles are congruent. Arc BA plus AF equals arc BC plus CF. The arcs BAF and BCF are congruent. Because they cover the entire circumference of the circle, they form two semicircles. Therefore, segment FB, the perpendicular bisector of chord AC, goes through the center of the circle. Since the perpendicular bisector of any chord goes through the center of the circle, a polygon can be circumscribed by a circle only if the perpendicular bisectors of each side of the polygon converge at a single point, or the midpoint of the circle (see figure X). If the bisectors do not converge, the polygon cannot be circumscribed in the circle.
Christen Bogenrief, 9 Martin County High School Mr. Limber Stuart, Florida To solve the first part of the problem I made a proof. First I drew a circle and labeled it "Q" and drew a chord called PB. Here is my proof: 1) PB is a chord of circle Q A) given 2) PQ and QB are congruent A) they are radii of the same circle so they have to be congruent 3) QE is equal to itself A) reflexive 4) triangle PQB is an isosceles triangle A) ITT 5) The angle bisector of PQB is also the perpendicular bisector of triangle PQB A) the bisector of the vertex angle of an isosceles triangle is perpendicular the base at its midpoint (corollary of ITT) 6) The perpendicular bisector goes through the center of the circle A) vertex is center of the circle The second question was harder but I think I got the answer: no, you can't circumscribe a circle around any polygon. The way you find out if you can is to draw the polygon. You then draw all the perpendicular bisectors of the polygon and if they meet in one place then that place is the center of the circle and you can circumscribe it.
Amy Forster Age 11, Grade 7 Wilkins/Forster family Crooked Tree Point Cygnet, Tasmania, Australia Proof that the perpendicular bisector of a chord goes through the centre of the circle. Let AB be the chord Let X be the middle point of the chord Let XY be the perpendicular bisector of AB Let Y be a point on the same side of the chord as the middle of the circle. If I join Y to A and to B there are now two triangles, triangle YXA and triangle YXB. The line AX = the length of line BX because the chord is bisected by point X. XY is the same length in each triangle because the line is shared by both triangles. Angle AXY = angle BXY = 90 degrees (because XY is a perpendicular bisector of AB) If two sides of triangle YXA are the same as two sides of triangle YXB and the angle between the 2 sides is the same, then the third sides must be the same as well. Since AY must equal BY, then point Y must always be an equal distance from A and B. When joining the points where Y can be so that it is an equal distance from A and B, the centre of the circle must be included since it is one of those points (making AY and BY radii of the circle). Therefore a line which is a perpendicular bisector of a chord will pass through the centre of the circle. . . . . . . . . . . . . . . . . . . . . . . How to decide whether you can circumscribe a circle around any given polygon: I found from the dictionary that for a circle to circumscribe a polygon, all the vertices of the polygon must lie on the circle. Using the ideas from the first proof (above), I realised that for a given polygon, a circle could be drawn so it passes through two adjacent vertices and the side of the polygon which connects those two vertices creates a chord to that circle. The perpendicular bisector of that chord must then pass through the centre of the circle drawn. I tried drawing perpendicular bisectors to each side of a polygon and found that if they all meet at one point, then a single circle can be drawn with its centre at that point, which will circumscribe the whole polygon, i.e. pass through each vertex of the polygon. So to test whether I can circumscribe a circle around any given polygon, I would construct perpendicular bisectors to each side of the polygon to see if all those lines meet at one point, in which case the polygon can be circumscribed.
Joshua Zucker Math Tutor, Education Program for Gifted Youth, Stanford Univ. Given a circle with chord AB. A point is on the perpendicular bisector of AB if and only if it is equidistant from the endpoints. Since the endpoints of the chord are on the circle, they are both a distance r from the center of the circle, and hence the center of the circle is on the perpendicular bisector. To prove that fact about perpendicular bisectors, let M be the midpoint of AB. If X is on the perpendicular bisector of AB, then triangles AMX and BMX are congruent by SAS because AM = BM, MX = MX, and angle AMX and angle BMX are equal (both 90 degrees). Hence AX = BX because corresponding parts of congruent triangles are equal. Conversely, if X is equidistant from the endpoints, then AX = BX, still MX = MX, and the angles are still 90 degrees. Since the hypotenuse and one leg are equal, the triangles are congruent (by HL, or by the Pythagorean theorem). (Explain how to decide whether you can circumscribe a circle around a particular polygon.) If a circle can be circumscribed about the polygon, then its center O would be an equal distance from all the vertices, and the sides of the polygon would all be chords of the circle. Hence, using the previous part of the problem, the perpendicular bisectors of the sides of the polygon would all pass through the point O. So, one way to test would be to construct all the perpendicular bisectors of the sides and see if they all meet at one point. Actually, we only need to do n-1 of the sides, since that suffices to prove that all the vertices are equidistant from the center. If the vertices of the polygon are P_1, P_2, ..., P_n, then if the perpendicular bisectors of sides P_1P_2 and P_2P_3 meet in a point X, then X is equidistant from P_1, P_2, and P_3. Continuing, we see that n-1 perpendicular bisectors intersecting at one point X are enough to prove that n points are all equidistant from X.
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