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

Latest revision as of 12:36, 16 July 2024

Problem

$[asy] size(120); real t = 2/sqrt(3); real x = 1 + sqrt(3); pair A = t*dir(90), D = x*A; pair B = t*dir(210), E = x*B; pair C = t*dir(330), F = x*C; draw(D--E--F--cycle); draw(Circle(A, 1)); draw(Circle(B, 1)); draw(Circle(C, 1)); //Credit to MSTang for the diagram [/asy]$

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

$\textbf{(A) }36+9\sqrt{2}\qquad \textbf{(B) }36+6\sqrt{3}\qquad \textbf{(C) }36+9\sqrt{3}\qquad \textbf{(D) }18+18\sqrt{3}\qquad \textbf{(E) }45$

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

Latest revision as of 12:36, 16 July 2024

Problem

Solution

@@ Line 1: / Line 1: @@
-Three circles are drawn, so that each circle is externally tangent to the other two circles. Each circle has a radius of <math>3.</math> 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.
+== Problem ==
-==Diagram==
-Diagram by Aryamans
-[asy]
+<asy>
-unitsize(1 cm);
+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>
-draw(Circle(dir(90),sqrt(3)/2));
+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
-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
+<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
-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 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>.
+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>.