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March 6, 2015

March 6, 2015

Posted by **Knights** on Saturday, February 2, 2013 at 3:44pm.

Each of the three circles in the figure below is externally tangent to the other two, and each side of the triangle is tangent to two of the circles. If each circle has radius three, then find the perimeter of the triangle.

The figure is basically an equaliteral triangle. Inside there are three circles of the same size and are congruent. The radius of each circle is 3.

PLEASE HELP!!

- PLEASE HELP TANGENTS AND CIRCLES PROBLEM GEOMETRY -
**Damon**, Saturday, February 2, 2013 at 4:04pmcircle A at bottom

circle B at upper right

circle C at upper left

side of triangle at upper right = x

so

(1/2) of the bottom is x/2, draw vertical altitude through top of triangle and center of circle A

NOW

draw line through centers of circles A and B extending beyond at both ends

It hits right side of triangle at right angle (tangent to circle perp to radius)

It hits our center altitude at 60 degree angle (30 at top of triangle so 60 at altitude intersection)

NOW

distance from intersection with altitude and right side of triangle = 2 radii, 3 + 3 + 3 = 9

SO

distance from top of triangle to that intersection = 9 sqrt 3

there is another radius length along the altitude to the center of the bottom of the triangle so the total height of the altitude = 3 + 9 sqrt 3

cos 30 = (3+9sqrt 3)/x = sqrt 3/2

x = (6+18 sqrt 3) / sqrt 3

simplify that and check my arithmetic

- PLEASE HELP TANGENTS AND CIRCLES PROBLEM GEOMETRY -
**Reiny**, Saturday, February 2, 2013 at 4:08pmJoining all the centres will produce another equilateral triangle with sides 6

So there is no problem seeing that the distance from tangent contact point to tangent contact point on the bigger triangle is 6.

let's concentrate on one end of the figure

From one of the circles draw the radii to each of the tangents.

label the centre C, the contact points A and B and the intersection of the two tangents as D. Join D and C to form the right-angled triangle ACD

angle DCA = 60°

AC = 3

so tan 60° = AD/3

AD = 3tan60=3√3

At the end of each circle we have 2 of these lengths

so the total perimeter

= 6(3√3) + 3(6)

= 18√3 + 18 units

- PLEASE HELP TANGENTS AND CIRCLES PROBLEM GEOMETRY -
**Knights**, Saturday, February 2, 2013 at 8:14pmThank you so much guys I really appreciate it!!!

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