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

Labels:
circle,
circumcircle,
equilateral,
measurement,
tangent,
triangle

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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