## Thursday, November 21, 2013

### Geometry Problem 936: Circle, Semicircle, Diameter, Tangent, Radius, Chord, Metric Relations

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

Click on the figure below to enlarge it.

1. Enlarge circle Q with center E, such that Q→O, (Homothetic transformation)
let D→F, then EQO, COF and EDF are straight lines, with EF=7.

Since then CF is the diameter of circle O, thus ∠CED=∠CEF=90°.

Using Pythagoras theorem, we have
CE² = 4²−3² = 7
CF² = CE² + EF² = 7 + 49 = 56
So OE = OF = √14

Since ΔEQD~ΔEOF,
x/√14 = 3/7
x = (3/7)√14

2. OC^2=DC*(DC+DE)/2
QE=OC*DE/(DC+DE)=(DC*(DC+DE)/2)^(1/2)*DE/(DC+DE)
=(DC/2)^(1/2)*DE/(DC+DE)^(1/2)
=(2)^(1/2)*3/(7)^(1/2)
=(14)^(1/2)*(3/7)

3. If <BDE=a then <OED=90-a and <EOD=2a-90. By circle theorem if <EOD=2a-90 then <BCE=a-45 and <ECO=(a-45)+45=a=<EDB making CEDO cyclic. That makes <DEC=90 and CE=sqrt7 by Pythagoras. If QD cuts CE at P then Q is cicrumcenter of PED so QD=1/2*DP.

By angle bisector PE=3/sqrt7 so DP=6*sqrt(2/7) by Pythagoras. QD is half so QD=3*sqrt(2/7)