Geometry Problem

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

Click the figure below to see the complete problem 700.

## Monday, December 5, 2011

### Problem 700: Equilateral Triangle, Circle, Circular Segment, Midpoint, Metric Relations

Labels:
circle,
circular segment,
equilateral,
metric relations,
midpoint,
triangle

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http://img37.imageshack.us/img37/9728/problem700.png

ReplyDeleteFD cut AB and AO at G and H ( see picture)

Note that AD=AG=DG=6

OA= AG*2/√3 =12/√3

In triangle OHF , applying Pythagoras Theorem we get HF=3. √5

So x=HF-HD=3(√5-1)

Peter Tran

Extend ED to cut AC at X and circle at Y

ReplyDeleteEDXY ∥ AB

So DE = 6 ; and XY = x (by symmetry)

ED.DY = AD.DC

x(6 + x) = 36

x²+ 6x = 36

(x + 3)² = 45

x = 3(√5 - 1)