Difference between revisions of "1977 AHSME Problems/Problem 15"

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draw((1 + sqrt(3))*dir(90)--(1 + sqrt(3))*dir(90 + 120)--(1 + sqrt(3))*dir(90 + 240)--cycle);
 
draw((1 + sqrt(3))*dir(90)--(1 + sqrt(3))*dir(90 + 120)--(1 + sqrt(3))*dir(90 + 240)--cycle);
 
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[/asy]
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[img]https://latex.artofproblemsolving.com/5/6/c/56c2e4a15c4c4654523f9b7ac15535647d8bce47.png[/img]
  
 
==Solution==
 
==Solution==

Revision as of 15:39, 19 September 2020

Three circles are drawn, so that each circle is externally tangent to the other two circles. Each circle has a radius of $3.$ A triangle is then constructed, such that each side of the triangle is tangent to two circles, as shown below. Find the perimeter of the triangle.

Diagram

Diagram by Aryamans

[asy] unitsize(1 cm);

draw(Circle(dir(90),sqrt(3)/2)); draw(Circle(dir(90 + 120),sqrt(3)/2)); draw(Circle(dir(90 + 240),sqrt(3)/2)); draw((1 + sqrt(3))*dir(90)--(1 + sqrt(3))*dir(90 + 120)--(1 + sqrt(3))*dir(90 + 240)--cycle); [/asy]

[img]https://latex.artofproblemsolving.com/5/6/c/56c2e4a15c4c4654523f9b7ac15535647d8bce47.png[/img]

Solution

Solution by e_power_pi_times_i

Draw perpendicular lines from the radii of the circles to the sides of the triangle, and lines from the radii of the circles to the vertices of the triangle. Because the triangle is equilateral, the lines divide the big triangle into a small triangle, three rectangles, and six small $30-60-90$ triangles. The length of one side is $2(3\sqrt{3})+6 = 6\sqrt{3}+6$. The perimeter is $\boxed{\textbf{(D) }18\sqrt{3}+18}$.

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