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
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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