Monday, October 24, 2016

Geometry Problem 1277 Equilateral Triangle, Circumcircle, Externally Tangent Circles, Tangent Line, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1277.

Geometry Problem 1277: Equilateral Triangle, Circumcircle, Externally Tangent Circles, Tangent Line, Measurement.
at

2 comments:

  1. Peter TranOctober 24, 2016 at 7:20 PM

    https://goo.gl/photos/QdriZ4m33XjfSxQ38
    Algebra 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

    ReplyDelete
  2. Peter TranOctober 24, 2016 at 9:34 PM

    https://goo.gl/photos/8nEeJBWzkSj9nXUP7

    Apply 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

    ReplyDelete

Share your solution or comment below! Your input is valuable and may be shared with the community.