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

Revision as of 00:20, 18 May 2020

Solution

Solution by e_power_pi_times_i (Picture would be helpful.)

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 $30-60-90$ 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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-== 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>
-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
-<math>\textbf{(A) }36+9\sqrt{2}\qquad
-\textbf{(B) }36+6\sqrt{3}\qquad
-\textbf{(C) }36+9\sqrt{3}\qquad
-\textbf{(D) }18+18\sqrt{3}\qquad
-\textbf{(E) }45   </math>
 ==Solution==
 Solution by e_power_pi_times_i

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

Revision as of 00:20, 18 May 2020

Solution