Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Monday, October 24, 2016
Geometry Problem 1277 Equilateral Triangle, Circumcircle, Externally Tangent Circles, Tangent Line, Measurement
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https://goo.gl/photos/QdriZ4m33XjfSxQ38
ReplyDeleteAlgebra solution:
Let R and r are radius of circles O and O1
Let θ= ∠ (TOB) and ∠ (TOA)= θ+120 and ∠ (TOC)= θ-120 ( see sketch)
In triangle BOO1 we have O1B^2=OO1^2+OB^2-2.OO1.OB.cos(θ)
BB1^2=b^2=O1B^2-r^2=(R+r)^2+R^2-r^2-2R(R+r).cos(θ) …. (1)
Replace cos(θ)= 2. cos(θ/2)^2-1 in (1) and simplify it we get
BB1= b= sqrt(4R(R+r)).sin(θ/2)
Similarly AA1=a= sqrt(4R(R+r)).sin(θ/2+60)
And CC1=c= sqrt(4R(R+r)).sin(θ/2-60)
Verify that sin(θ/2+60)= sin(θ/2-60)+ sin(θ/2)
So a=b+c
https://goo.gl/photos/8nEeJBWzkSj9nXUP7
ReplyDeleteApply Casey’s theorem, generalized Ptolemy’s theorem for the case 4 circles tangent externally to a central circle. ( see sketch)
In our case
Central circle: circle O, radius R
Surrounding circles: circle A, tangent to circle O , radius=0; circle B, radius= 0; circle C radius=0 and circle O1, radius r.
Per sketch as shown
x = BC=m
y=AA1
d=CC1
c=BB1
b=AB=m
a=AC=m
per Casey theorem we have x.y= a.c+ b.d or m.AA1=m.BB1+m.CC1
So AA1=BB1+CC1