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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1193.
Let the points of tangency on the hypotenuse be M and N. Let the parallel to BC thro O2 meet the parallel to AB thro O1 meet at LO1M = O2M = r and they are parallel to each other hence O1O2NM is a rectangle Let AM = pThen MN = 2r, CN = 5-p-2r, O1L = 3-p-r and O2L = r+p-1Tr. O1O2L is similar to Tr. ABC since corresponding sides are parallel.So 2r/5 = (3-p-r)/3 = (r+p-1)/4Solving the simultaneous equations for p and r p = 10/7 and r = 5/7Sumith PeirisMoratuwaSri Lanka
http://s22.postimg.org/3t229d6xd/pro_1193.pngLet AO1 and CO2 meet at OLet r’ and r1 are radii of incircles of triangles ABC and O1DO2Observe that O is the incenter of both triangles.r’= area of Tri. ABC/half of perimeter of ABC= 6/6=1O1DO2 and ABC are similar triangles => r1/r’= O1O2/BC= 2.r/5 => r1=2.r/5We have r’=r1+r= 7/5.r =1 => r=5/7
Name T1, T2 tg points of O1, O2 Draw from O1//to AB , from O2 // to BCThey meet at P Triangle PO1O2 similar to ABC => T2C=1,5 T1A From similarityagain => AT1=2r => r=5/7