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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 861.
http://img16.imageshack.us/img16/3367/problem861.pngLet r= radius of incircle and L is tangent point of GH to incircleConnect OG,OA,OE,OH,OF,OCNote that triangles AOG and HOG are right triangles and OE and OF are altitudes.Relation in right triangle AOG give OE^2=r^2= EG.AE= 7Relation in right triangle HOC give OF^2=r^2=GH.HC=2.x=7So x=7/2
Let is OE=OF=rangle DEM= angle DFM=90 degrees , why MD is diameter of circleED is perpendicular of AO and EM is perpendicular of GO. So, angle GOA=90 degrees.Similarly , angle HOC=90 degrees.From the right triangle GOA we have, OE^2=AE.EG=7 or, r^2=7 (1)From the right triangle COH we have, OF^2=CF.FH=7 o r, r^2=2x (2)From (1),(2) ,3x=7 , x=7/2http://img268.imageshack.us/img268/5052/74920469.png
Let BG=y, BH=z. Since BG/BA = BH/BC = GH/AC and BE = BFy/(y+8) = z/(z+2+x) = 3/(7+x) and y+1 = z+2xy+yz+2y = yz+8z3x+3z+6 = xz+7zxy+7y = 3y+24y+1 = z+2Solving those equations, we havex=7/2, y=16/5, z=11/5
Proiectam G si H pe AC si daca folosim Teorema lui Pitagora obtinem (x+2)(x+2)-(x-2)(x-2)=8.8-6.6,de unde x=7/2
Let BG = p so that BH = p-1 since BE = BFFrom similar trianglesp/(p+8) = (p-1)/(p+1+x) = 3/((7+x)Each ratio = 1/(7-x) where I have used the algebraic identity that if a/b = c/d then each ratio = (a-c) /(b-d)So 3(7-x) = 7+x and so x = 7/2Sumith PeirisMoratuwaSri Lanka