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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Details: Click on the figure below.
Let AO = a and OF = b so that FC = a-b and AC = sqrt2.aab = 30......(1)From areas of similar triangles(a^2+b^2)/(a-b)^2 = 15/7 which simplifies to a^2+b^2 = 225/2 ...,(2) using (1)Now use Ptolemy in cyclic quadrilateral AOBC, noting that BC = (a-b)a/sqrt(a^2+b^2) using Pythagoras and we get after much simplification x = 2ab/sqrt2(a^2+b^2) = 60/sqrt225 (from (2)) = 4Sumith PeirisMoratuwaSri Lanka
To SumithOn what relation is included x
Note that AF = 15/sqrt2Using Ptolemy in AOBCsqrt2.a.x + a.BC = AB.aNow note that BC = sqrt2. a(a-b)/15and AB = 15/sqrt2 + b.(a-b)/sqrt2/15Now substitute these in the Ptolemy equation
Full Ptolemy isn't really necc. once you derive AF = 15/sqrt(2). Then since its cyclic OBF is similar to ACF and x / AC = OF / AF or x = AC * OF / AF. AC = sqrt(2) * a since its a 45-45-90 and OF = 30 / a using the AFO's area.So AC * OF = 30 * sqrt(2) and 30 * sqrt(2) / (15 / sqrt(2)) = 4.
AC = a \/2 (a sqrt2)