0

I've been working on this problem for a while but I can't seem to figure it out, so any explanations regarding how to solve it would be appreciated. Here it is:

Let $AB$, $CD$, and $EF $be three parallel chords that are non-diameters of a circle on the same side of the center. Let $GJ$ be tangent to the circle at $H$ such that $GJ$ is parallel to $AB$. The distance between $AB$ and $CD$ is equal to the distance between $CD$ and $EF$ and is also equal to the distance between $EF$ and $GJ$. If $AB = 24$ and $CD = 20$, what is the distance from the center of the circle to $AB$?

Thanks

BlackAdder
  • 4,029
John
  • 1
  • 1
    Draw it, it is the first and the most useful step(if you are not skilled enough to picture it in your mind :) ). As a sidenote: This is not a descriptive title that will help others to know what do you mean before reading your post, and it is never a good idea to believe a problem is tough. It just biases you and makes you think you are not clever enough to solve it before even trying it in a serious way. – chubakueno Dec 24 '13 at 01:22

2 Answers2

4

Hint:

$\hspace{3.8cm}$enter image description here $$ \begin{align} r^2-(r-2s)^2&=10^2\\ r^2-(r-3s)^2&=12^2 \end{align} $$

robjohn
  • 345,667
1

Remember that, if two chords of a circle intersect, and you think of each of these chords as cut into two pieces by the intersection point, then the product of the two pieces of one chord equals the product of the two pieces of the other chord. Apply this twice, first to the chord AB and the diameter perpendicular to it, and second to the chord CD and the same diameter. Note that the diameter bisects both chords. You get two equations for two unknowns, namely the radius of the circle and the distance that the question asks for.

Andreas Blass
  • 71,833