A difficult geometry problem for smart mathematicians!? - XP Math - Forums

 XP Math - Forums A difficult geometry problem for smart mathematicians!?

 08-02-2007 #1 Amir Guest   Posts: n/a A difficult geometry problem for smart mathematicians!? Suppose M is the center of a circle and A and B two points on the circle circumference. (AB is not a diameter)The tangent lines on the circle from A and B intersect in C.CM intersects with the circle in D.The tangent line of circle from D, intersects with AC and BC in E and F respectively. ***************************Please show that the area of ADBM is the geometric mean of the areas of ABM and ACBM.Please see this image:http://m1.freeshare.us/156fs850081.jpg
 08-02-2007 #2 Bumpy Shafer Guest   Posts: n/a I assume that the geometric mean of the two areas is like the mean of a set of numbers.In this case maybe ADBM = ABM + ACBM / 2Ok, Im no mathematician, but I did a similar problem to this a while back. You need to get the formulas Area of a sector: 0.5 r^2*X (where X is the angle)Area of a triangle: 0.5 (base)(height) and do subsitution. I'll have a look at it now and see how I get on. Ouch tough one.ABM = pi*r^2 - 0.5 r^2*X
 08-02-2007 #3 fernando_007 Guest   Posts: n/a Mark intersection of AB and MC with G. Then from the similarity of MAC and MAG follows:MA/MC = MG/MA, and since MD = MA follows:MD/MC = MG/MD. If you multiply all by AG you getMD*AG)/(MC*AG) = (MG*AG)/(MD*AG).MD*AG = 2*AMD = ADBMMC*AG = 2*AMC = ACBMMG*AG = 2*AMG = ABMTherefore: ABM * ACBM = ADBM^2.

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