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stoutfellowThis morning, I was - belatedly - grading the finals for my geometry class. I was rather proud of one question, as the possible responses were neatly layered. The question was worth ten points. There was a simple argument (recognizable by experimentation, if in no other way) which gave a seven-point answer. Closer attention to the wording would yield a more complete answer, worth nine points. Truly nitpicky obsessiveness would notice the small hole in the nine-point answer, yielding the correct, ten-point answer.
Sitting at the dining room table, grading the tests, I went through seven-point answer after seven-point answer. I reached the twelfth paper (out of sixteen), and hallelujah! someone found his way to the full ten-pointer.
I shouted "Yes! Bravo!" and pumped my fist in the air.
Then I went back to grading.
If you're not curious about the problem, you can stop now. The problem is below; the answer is under the cut.
First, some definitions (which had been covered in class). A triangle is nondegenerate if its vertices do not all lie on the same line. Any nondegenerate triangle has a unique circumcircle, a circle passing through all of its vertices. The center of the circumcircle is the circumcenter; it is the point of intersection of the perpendicular bisectors of the sides. Here is the problem:
Nine-point solution: add All of the points on the perpendicular bisector of AB are circumcenters.
Ten-point solution: add except for the intersection of the perpendicular bisector with the line through A perpendicular to AX, which corresponds to the degenerate case C=A.
Sitting at the dining room table, grading the tests, I went through seven-point answer after seven-point answer. I reached the twelfth paper (out of sixteen), and hallelujah! someone found his way to the full ten-pointer.
I shouted "Yes! Bravo!" and pumped my fist in the air.
Then I went back to grading.
If you're not curious about the problem, you can stop now. The problem is below; the answer is under the cut.
First, some definitions (which had been covered in class). A triangle is nondegenerate if its vertices do not all lie on the same line. Any nondegenerate triangle has a unique circumcircle, a circle passing through all of its vertices. The center of the circumcircle is the circumcenter; it is the point of intersection of the perpendicular bisectors of the sides. Here is the problem:
Let A,B,X be three noncollinear points. Let C be a variable point on the line AX. As the point C varies, how does the circumcenter of ABC vary? That is, which points are circumcenters of the triangles ABC?Seven-point solution: All of the triangles ABC share the side AB; hence their circumcenters all lie on the perpendicular bisector of AB.
Nine-point solution: add All of the points on the perpendicular bisector of AB are circumcenters.
Ten-point solution: add except for the intersection of the perpendicular bisector with the line through A perpendicular to AX, which corresponds to the degenerate case C=A.
