Geometry Problem

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

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## Thursday, November 21, 2013

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

Labels:
chord,
circle,
diameter,
metric relations,
radius,
semicircle,
tangent

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Enlarge circle Q with center E, such that Q→O, (Homothetic transformation)

ReplyDeletelet 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

OC^2=DC*(DC+DE)/2

ReplyDeleteQE=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)

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.

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

Super!

ReplyDeleteTrigonometry Solution

ReplyDeleteDraw the Common Tangent at E, XEY

If < XEC = p, then < COE = 2p = < OQD, so < QED = p = < XEC whence < CED = 90

So Cos(2p) = 3/4. From Tr. QED, cos (p) = 3/2x

Now cos (2p) = 2cos^2 (p) - 1

Hence 9/(2x^2) - 1 = 3/4

Solving, x = 3.sqrt(2/7)

Sumith Peiris

Moratuwa

Sri Lanka