Geometry Problem

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

Click the figure below to see the complete problem 970.

## Tuesday, January 28, 2014

### Geometry Problem 970: Equilateral Triangle, Circumcircle, Secant, Metric Relations

Labels:
circumcircle,
equilateral,
metric relations,
secant,
triangle

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Let BD=y.

ReplyDeleteDE×DA=DB×DC

y(x+y)=12

In ΔABD, using cosine law,

36=x²+y²+xy=x²+y(x+y)=x²+12

x²=24

x=2√6

Let BD = p

ReplyDeleteFrom similar triangles p(p+x) = 2(2+4) = 12...(1)

From Pythagoras AD^ 2 = 3x^2/4 + (p+x/2)^2 ....(2)

(2) - (1) ; x^2 = AD^2 - 12 = 24

so x = sqrt 24 = 2sqrt6

AEBC is a cyclic quad from that 3 similar triangles fall out: ABE, CDE and ABD. Note angles AEC = CEB = DEB = 60 degrees.

ReplyDeleteConsider just ABE and ABD since they're similar from those we get 4/x = x/6

and x^2 = 24 => x = 2 sqrt(6)

Let BD=y

ReplyDeleteAs DE.DA=DB.DC

=> 12=y.(x+y)=> y^2+xy=12

Let F be the foot of perpendicular from A to BC

AF^2+DF^2=DA^2

=>3/4x^2+x^2/4+y^2+xy=36

=>x^2=24