Geometry Problem

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Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 924.

## Thursday, September 19, 2013

### Geometry Problem 924: Quadrilateral, Diagonal, Perpendicular, Circle, Tangent, Midpoint, Metric Relations

Labels:
circle,
diagonal,
metric relations,
midpoint,
perpendicular,
quadrilateral,
tangent

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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.

ReplyDeleteReply

Problem 924. To Anonymous: What is the reason for your statements

DeleteEN = 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

ReplyDeleteProblem 924. To Anonymous: What is the reason for your statement

DeletePG = PF = PO = r

Thanks

http://img585.imageshack.us/img585/2808/4xgl.png

ReplyDeleteDefine angles α and β as shown on the sketch

Since 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)= 90

½(180- β)+180-2(90-1/2. α +1/2. β)=90

β =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 degrees

So 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,

ReplyDeletesin(60)*HJ=12, so HJ=8*sqrt3