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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 1222.
Drop a perpendicular BH to AC. With the usual notation BC = a, and AC = b = DB. CH = a/2 and BH = sqrt3 a/2Let AD = cFrom similar triangles c/(c+b) = b/(b+a/2) which yields ac = 2b^2.....(1)Applying Pythagoras, (b+c)^2 -(b+a/2)^2 = 3a^2/4Substituting for a from (1) 3b^4/c^2 = (b+c)^2 - (b+b^2/c)^2Simplifying the algebra and factorizing(c^3-2b^3)(c+2b) = 0c+2b = 0 is not a permissible solutionHence c^3 = 2b^3Sumith PeirisMoratuwaSri Lanka
let AC=BD= x , AD=y and DC=hApplying sine rule to triangle BDC=> sin(30)/x = sin(B)/h=> sin(B) = h/2x -------------- (1)Applying sine rule to triangle ABC => sin(B)/x = sin(120)/(x+y)=> sin(B) = sqrt(3).(x)/2.(x+y) ----------------- (2)(1) = (2)=> h/2x = sqrt(3).(x)/2.(x+y)=> sqrt(y2-x2)/2.x = sqrt(3).(x)/2.(x+y) (Applying pythogrous to tri ADC => h2 = y2-x2)=> sqrt(3).x2 = (sqrt(y2-x2)).(x+y)Squaring on both sides=> 3.x4 = (y2-x2).(x2+y2+2xy)=> 3.x4 = y4-x4+2xy3-2x3y=> 3 = y4/x4 - 1 + 2y3/x3 - 2y/xlet p = y/x=> p4-2p+2p3-4 = 0=> p(p3-2)+2(p3-2)=0=> (p+2)(p3-2) = 0Since p!= -2 => p3 = 2 => y3/x3 = 2