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Geometry ProblemPost your solution in the comment box belowLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 924.
AB and CD intersect at M, MFOG the quadrilateral is inscribed in a circle of radius r, EO crosses FG in N, EN = 1/2R, EO = EF = EG = EM = r, NG = r.sq3 / 2, FG = r SQ3, HJ = 2r, r = 12/sq3, HJ = 3 24/sq.Reply
Problem 924. To Anonymous: What is the reason for your statements EN = 1/2R,EO = EF = EG = EM = r, NG = r.sq3 / 2,Thanks
Anonymous. I'm sorry I was wrong, I took E to the center point of the arc FEG but it is not. We say that P is the midpoint of the arc FG then PG = PF = PO = r and necessarily PM=r. N represents the medpoint and meeting point between FG and PO, then NG = r.sq3 / 2, FG = r.sq3, HJ = 2r, r = 12/sq3, HJ = 24/sq3 Reply
Problem 924. To Anonymous: What is the reason for your statement PG = PF = PO = r Thanks
http://img585.imageshack.us/img585/2808/4xgl.pngDefine angles α and β as shown on the sketchSince G is the midpoint of CD => GC=GD=GE∠ (OEC)= ∠ (OEJ)+ ∠ (CEJ) = ½(180- α)+90= 180-1/2. α∠ (CEG)= ∠ (OEC)- ∠ (OEG)=180-1/2. α -1/2(180- β)=90-1/2. α +1/2. β∠ (OGC)= ∠ (OGE)+ ∠ (EGC)= 90½(180- β)+180-2(90-1/2. α +1/2. β)=90β =2/3. α => ∠ (EOG)=2/3∠ (EOJ)Similarly we also have ∠ (EOF)=2/3. ∠ (EOH)Add above 2 expression we have ∠ (FOG)=2/3.(HOJ)= 120 degreesSo HJ= 2.OG= 8.SQRT(3)
<FEA=<FAE=a, <BFE=<FHE=2a, and similarly <GDE=<GED=b, <EGC=<EJG=2b. By cyclic quadrilateral theorem, <FEG+<FHE+<GJE=(a+b+90)+(2a)+(2b)=180 so 2a+2b=<FHG=60. By law of sines,sin(60)*HJ=12, so HJ=8*sqrt3