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

m ([asy] and [/asy] changed to <asy> and </asy>.)
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Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
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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 <math>30-60-90</math> triangles. The Longer length of the rectangles is 6 because it is the radius of one circle + the radius of another (<math>3+3</math>). The length of one side is <math>2(3\sqrt{3})+6 = 6\sqrt{3}+6</math>. The perimeter is <math>\boxed{\textbf{(D) }18\sqrt{3}+18}</math>.
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Draw perpendicular lines from their if goe[ 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 <math>30-60-90</math> triangles. The Longer length of the rectangles is 6 because it is the radius of one circle + the radius of another (<math>3+3</math>). The length of one side is <math>2(3\sqrt{3})+6 = 6\sqrt{3}+6</math>. The perimeter is <math>\boxed{\textbf{(D) }18\sqrt{3}+18}</math>.

Revision as of 20:30, 1 May 2024

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

[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]

Or visit here: https://latex.artofproblemsolving.com/5/6/c/56c2e4a15c4c4654523f9b7ac15535647d8bce47.png

Solution

Solution by e_power_pi_times_i

Draw perpendicular lines from their if goe[ 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 Longer length of the rectangles is 6 because it is the radius of one circle + the radius of another ($3+3$). 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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