## Monday, November 11, 2013

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

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

Click on the figure below to enlarge it.

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

Let 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

√2 a + ar = a(x+r)
x = √2

3. a geometric solution:
http://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

1. Very nice solution!

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