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So the problem is that there is one circle with radius of five and one circle with radius of 1. There centers are 8 units apart and there is a pulley belt that goes around the outside as shown in the image. It is given that the belt touches 2/3 of the edge of the larger circle and 1/3 of the edge of the smaller circle. The goal is to find the total length of the belt. I know that the belt is $(2/3)10\pi + (1/3)2\pi + 2$ (distance between the points of tangency on the circles). However, I am unable to come up with that last component. I thought of using triangles, but I can't assume that there are $90^\circ$ angles when I draw the triangles. Help would be appreciated enter image description here

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    $\begingroup$ You continue the black lines of the pulley until they meet, also draw a line through the two circle centers that meets there as well, you get some similar triangles that way. Note that a line tangent to a circle is perpendicular to the radius of the circle that meets it. $\endgroup$ – Will Jagy Jul 30 '14 at 0:23
  • $\begingroup$ Ahhh perfect, thank you @WillJagy $\endgroup$ – mmm Jul 30 '14 at 0:25
  • $\begingroup$ "Any problem in geometry becomes trivial once the correct triangle is pointed out." $\endgroup$ – John Alexiou Jul 25 '16 at 20:05
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Trick is in constructing a right angled triangle by drawing line parallel to inclined belt through center of smaller circle, and the two radial lines passing through points of tangentcy and centers of circles:

$$ L^2 =d^2 - {\Delta r} ^2 = 8^2 - 4^2 ;$$

Belt Length Constrn

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Draw a line from the center O of circle 5 to the tangent point A on one side, and similarly for circle 1 from center Q to tangent point B. Each such line is perpendicular to the tangent line AB. Now from the center Q of circle 1, draw a line parallel to AB meeting OA at C.

AC = 1 so OC = 4. So in right triangle QOC with QO = 8 and OC = 4, QC = $4\sqrt{3}$, and this is equal to AB as well. So the total belt length is $$ \frac{22}{3} \pi + 8\sqrt{3} $$ By the way, since triangle QOC is a 30-60-90 triangle, it is easy to see that the statement about the belt touching 1/3 of the small circle and 2/3 of the large circle is correct.

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Hint: Let $A$ and $B$ be the centres of the bigger and smaller circles, respectively. Now let $C$ and $D$ be the endpoints of the upper part of the belt (so that $C$ is the point of tangency of the larger circle and $D$ is the point of tangency of the smaller circle). Now draw a line parallel to $CD$ that goes through point $B$ and let $E$ be the point where this line intersects the radius $AC$. Then observe that $ECDB$ is a rectangle and $AEB$ is a right triangle. Hence, if $x = CD$, then we can use Pythagoras to solve for this length as follows: $$ x^2 + 4^2 = 8^2 \iff x = 4\sqrt{3} $$

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