Geometry Problem

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

Click the figure below to see the complete problem 699.

## Monday, December 5, 2011

### Problem 699: Equilateral Triangle, Circle, Circular Segment, Midpoint, Metric Relations

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

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http://img821.imageshack.us/img821/6871/problem699.png

ReplyDeleteBD cut EF at G ( see picture)

Note that DG=OG (angle ABC=60)

Triangle ABG similar to tri. EDG with ratio =BG/DG=3

So x= AB/3= 4

Peter Tran

DAC = 30, AED = 120 imply ADE = 30

ReplyDeleteSo AE = ED = x

FC = DF = x (by symmetry)

So x = AE = EF = FC = AC/3 = 4

Let radius of the circle be r,

ReplyDeleter^2+r^2-2r^2cos120=144

r=4sqrt3

BD perpendicular to AC and intersects at H

BH=sqrt(12^2-6^2)=6sqrt3

DH=DB-BH-2r-6sqrt3=2sqrt3

x^2=EH^2+DH^2

x^2=4+12

x=4

< ADC = 120

ReplyDeleteSince D is the mid point of arc AC, < CAD = < ACD = 30

Since < DEF = 60, AE = DE = x

Similarly CF = x

So 3x = 12 and x = 4

Sumith Peiris

Moratuwa

Sri Lanka