Minimum Angle ProofDate: 07/05/2001 at 21:25:38 From: Pieta Subject: Geometry involving incentres of triangles Label the point of intersection of the angle bisectors of triangle ABC as Q. Let M be the midpoint of side BC. It is given that MQ = QA. Find the minimum value of angle MQA. I have tried to let the two lengths MQ and QA be the radii of a circle to see if angle MQA is subtended by any of the sides of the circle. I have also tried to use similar triangles and triangle congruences, but I do not seem to be making any progress. I have also tried to use the rule a/sinA = b/sinB, and I have tried to make use of the fact that the radius is absinC/( a+ b+ c). Please help! :) Pieta Date: 07/08/2001 at 08:44:19 From: Doctor Jaffee Subject: Re: Geometry involving incentres of triangles Hi Pieta, This problem was challenging, and it took me a while to solve it, but I came up with a solution that works. Here is how I did it. Draw triangle ABC and assume that side AB is shorter than side AC. If the opposite were true the proof would be similar. Notice that triangle ABQ is congruent to triangle MBQ. I'll leave that up to you to prove. The result of this congruency is that we now know that AB = BM, and since M is the midpoint of BC, we can conclude that AB = (1/2)AC. sin C sin A According to the Law of Sines, ----- = ----- AB 2AB which means sin C = (1/2)sin A Now, if you let the measure of angle A = 90 degrees, you can calculate the exact measure of angle AQM. If you let the measure of angle A be more or less than 90 degrees (call the measure 90 + x or 90 - x), then you can prove that the measure of angle AQM must be more than its measure when angle A is a right angle. I have provided you with a bare outline of the proof. Try to fill in the details and if you want to check your solution with me, write back. If you are having difficulties, let me know and show me what you have done so far, and I'll try to help you some more. Good luck. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/ |
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