Difference between revisions of "1977 AHSME Problems/Problem 15"
(Created page with "== Problem 15 == <asy> size(120); real t = 2/sqrt(3); real x = 1 + sqrt(3); pair A = t*dir(90), D = x*A; pair B = t*dir(210), E = x*B; pair C = t*dir(330), F = x*C; draw(D--E...") |
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| − | Draw perpendicular lines from the radii to the circles to the sides of the triangle, and lines from the radii of the circles to the vertices of the triangle. | + | Draw perpendicular lines from the radii to the circles to the sides of the triangle, and lines from the radii of the circles to the vertices of the triangle. Then 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 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 13:40, 21 November 2016
Problem 15
![[asy] size(120); real t = 2/sqrt(3); real x = 1 + sqrt(3); pair A = t*dir(90), D = x*A; pair B = t*dir(210), E = x*B; pair C = t*dir(330), F = x*C; draw(D--E--F--cycle); draw(Circle(A, 1)); draw(Circle(B, 1)); draw(Circle(C, 1)); //Credit to MSTang for the diagram [/asy]](http://latex.artofproblemsolving.com/0/9/e/09eb9cd5e511702149aa9236e78030f2f0300893.png)
Each of the three circles in the adjoining figure 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 the perimeter of the triangle is
Solution
Solution by e_power_pi_times_i
Draw perpendicular lines from the radii to the circles to the sides of the triangle, and lines from the radii of the circles to the vertices of the triangle. Then because the triangle is equilateral, the lines divide the big triangle into a small triangle, three rectangles, and six small 
triangles. The length of one side is 
. The perimeter is 
.
