Geometry Problem

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

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## Monday, November 11, 2013

### Geometry Problem 934: Square, Circle, Center, Perpendicular, Metric Relations

Labels:
circle,
metric relations,
perpendicular,
square

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Since ∡ CEA = ∡ CDA , A,B,C,D,E are concylic

ReplyDeleteLet α be ∡CDE, then ∡CAE = α ; ∡ EAD = ∡ECD = 45 - α ;

∡FCD = ∡FCE - ∡ECD = 45 - (45 - α) = α =∡CDE

So FC//DE

Further ∡ACF = 45 - ∡FCD = 45 - α

Let L be the length of the square,

By sine law on △CDE,

L/sin(135) = 1/sin(45 - α)

By sine law on △ACF

(sqrt2)L / sin(135) = x/sin(45 - α)

Hence, x = sqrt2

Let AD=a, CE=r.

ReplyDeleteSince ADEC is cyclic quadrilateral, by Ptolemy theorem,

AC×DE + AD×CE = AE×CD

√2 a + ar = a(x+r)

x = √2

a geometric solution:

ReplyDeletehttp://bleaug.free.fr/gogeometry/934.png

By construction, ABCDE are co-cyclic.

Let G be co-cyclic such that AC crosses CG in F. By construction AFG=EFC=45°

AD chord in ABCDEG circle implies ACD=AED=45°. Similarly, CD chord implies CAD=CGD=45°

Hence ED // CG and GD // AE, therefore EDGF is a parallelogram.

AF = GF.√2 = ED.√2 = √2

Very nice solution!

Delete<ACF=<DCE=45-<FCD, and <CDE=<CAF due to cyclic quadrilateral. Therefore X/CF=1/CE, X=CF/CE=sqrt2

ReplyDelete