Difference between revisions of "Talk:2001 AIME I Problems/Problem 7"
Newton1988 (talk | contribs) (Alternative solution) |
Newton1988 (talk | contribs) |
||
| Line 3: | Line 3: | ||
Since the point is the center (call it P) of the inscribed circle, it must be the intersection of all three angle bisectors. Drawing the bisector AP, to where it intersects BC, we shall call this intersection F. | Since the point is the center (call it P) of the inscribed circle, it must be the intersection of all three angle bisectors. Drawing the bisector AP, to where it intersects BC, we shall call this intersection F. | ||
| − | Using the angle bisector theorem, we know the | + | Using the angle bisector theorem, we know the ratio BF:CF is 21:22, thus we shall assign a weight of 22 to point B and a weight of 21 to point C, giving F a weight of 43. In the same manner, using another bisector, we find that A has a weight of 20. So, now we know P has a weight of 63, and the ratio of FP:PA is 20:43. Therefore, the smaller similar triangle ADE is 43/63 the height of the original triangle ABC. So, DE is 43/63 the size of BC. Multiplying this ratio by the length of BC, we find DE is 860/63 = m/n. Therefore, m+n=923. |
Latest revision as of 20:58, 24 July 2008
A more elegant solution exists using the fact that an angle bisector divides the opposite leg in direct proportion to the ratio of the adjacent legs, and mass point geometry. Essentially, it goes something like:
Since the point is the center (call it P) of the inscribed circle, it must be the intersection of all three angle bisectors. Drawing the bisector AP, to where it intersects BC, we shall call this intersection F. Using the angle bisector theorem, we know the ratio BF:CF is 21:22, thus we shall assign a weight of 22 to point B and a weight of 21 to point C, giving F a weight of 43. In the same manner, using another bisector, we find that A has a weight of 20. So, now we know P has a weight of 63, and the ratio of FP:PA is 20:43. Therefore, the smaller similar triangle ADE is 43/63 the height of the original triangle ABC. So, DE is 43/63 the size of BC. Multiplying this ratio by the length of BC, we find DE is 860/63 = m/n. Therefore, m+n=923.
