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Given two circles $S_1$ and $S_2$, a line $l_1$, and a length $a$ that is less than the sum of the diameters of the circles, construct a line $l$, parallel to $l_1$, so that the sum of the chords that $S_1$ and $S_2$ cut from $l$ is $a$. (You may assume that the interiors of circles $S_1$ and $S_2$ are disjoint, and that the circles are positioned so that at least one such line $l$ exists.)

I tried working backwards, and translating the circles such to connect the two chords, but I did not get anything useful from that. Can I have a little hint as to how to start/procede from my current position? Thanks!

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let me see if i can construct the required line geometrically.

(a) draw lines $s_1, s_2$ orthogonal to line $l_1$ through the centers $O_1, O_2$ of $S_1$ and $S_2$ respectively. let $d$ be the distance between the parallel lines constructed.

(b) construct the length $d- {1\over 2} a$ (added later: this is the gap between the two circles in the direction of $l_1$ needed.)

(c) construct a point $O$ so that $O$ is between lines $s_1, s_2$ and $O_1O = d - {1 \over 2}a$

(d) draw a circle $S$ centered at $O$ and with the same radius as $S_1.$

(e) find the points where $S$ and $S_2$ intersect.

(f) draw lines through the points found in (e) parallel to line $l_1$.

the line/s in (f) is the answer to your problem.

ADDED BY RORY DAULTON:

Here is a graphic to better understand the construction. Point $A$ and the dotted circle were not mentioned in abel's description: they are used to find the point $O$.

enter image description here

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  • $\begingroup$ I tried this in Geogebra, and as long as the input sizes are within certain limits, this works! (1) Could you show the justification for this? (2) May I add a graphic of my verification to clarify your construction? +1 and this answer should be accepted! $\endgroup$ – Rory Daulton Dec 23 '14 at 14:27
  • $\begingroup$ @RoryDaulton, thanks for your verification. please add the figures from geogebra. i don't have and don't know how to add figures. adding figures would certainly make it easier to follow my reasoning. anyway, i will add some justification. $\endgroup$ – abel Dec 23 '14 at 14:42
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NOTE: This post has been edited due to @RicardoCruz catching a mistake. I also changed the desired variable from $z$ to $u$, for my own reasons.

Here is an approach that leads to a not-so-easy but possible construction.

Rotate your circles and line onto a Cartesian frame of reference so line $l$ is horizontal and translate so circle $S_1$ is centered at the origin with radius $s$. Its equation is then $x^2+y^2=s^2$. Let's then say that circle $S_2$ is centered at point $(c,d)$ and has radius $r$. (Reflections could be made to make both $c$ and $d$ nonnegative.) Let's finally say that desired line $l_1$ has the equation $y=u$. Given $a$, $c$, $d$, $r$, and $s$, we want to find $u$ so the sum of the chords of line $l$ with circles $S_1$ and $S_2$ is $a$.

Two circles with chords

We can easily find that the length of the chord in $S_1$ is $2\sqrt{s^2-u^2}$ and the length of the chord in $S_2$ is $2\sqrt{r^2-(d-u)^2}$. Therefore, the equation we want to solve for $u$ is

$$a=2\sqrt{s^2-u^2}+2\sqrt{r^2-(d-u)^2}$$

Removing the second square root by getting it alone and squaring leads to

$$\left(\frac{a^2}4+d^2+s^2-r^2 \right)-2du=a\sqrt{s^2-u^2}$$

I am too lazy to continue, but squaring this equation gives a quadratic equation in $u$. The solution (or solutions: there may be two valid ones!) can be constructed with compass and straightedge, which can easily be used to construct line $l_1$. Constructing the solution(s) for $u$ looks messy but there may be some tricks to make it easier.

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  • $\begingroup$ Thank you for your input. I'm just wondering, since I can't prove it, is there any way to find the distance between the two chords in terms of the lengths of the two chords? $\endgroup$ – Pakquebchsoflwty Dec 20 '14 at 18:27
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    $\begingroup$ There is something that doesn't fit in the equation you got. I think the second term of the RHS should be $2 \sqrt{r^2-(d-z)^2}$. If you correct it, you will get a quadratic equation. $\endgroup$ – RicardoCruz Dec 20 '14 at 19:36
  • $\begingroup$ @RicardoCruz: You are absolutely right! (And I was so wrong!) Thanks for the catch. I'll edit my answer immediately. How can I give you some of my (as-yet-non-existent) reputation points for this answer? $\endgroup$ – Rory Daulton Dec 20 '14 at 21:20
  • $\begingroup$ @Pakquebchsoflwty: Note that I made major changes to my answer. You problem is indeed solvable with compass and straightedge. $\endgroup$ – Rory Daulton Dec 20 '14 at 21:36
  • $\begingroup$ @RoryDaulton You're welcome. I'm happy to help. $\endgroup$ – RicardoCruz Dec 20 '14 at 22:07

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