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This is a little exercise found in Robin Hartshorne's Euclid: Geometry and Beyond: enter image description here

I believe I have found a solution, which only exists when $AB$ has length at least that of the chord bisected by $P$, and length at most the diameter of the circle. So suppose that is the case. I do the following construction. $\odot cXrYZ$ is an abbreviation for the circle centered at $X$ with radius $YZ$. (I assume the compass is not collapsible.) Also, it's not necessary to know my solution to answer my question, so feel free to skip.

  1. Pick $C$ at random on circle $O$. Describe the circle with center $C$ and radius $AB$. Get intersection $D$.
  2. Describe the circle with center $D$ and radius $DC$. This will intersect with the previous circle at $E$ and $F$, say.
  3. Connect $EF$.
  4. Connect $CD$, get intersection $G$.
  5. Draw circle with center $O$ and radius $OG$. (This gives the circle to which all chords of length $AB$ will be tangent to.)
  6. Connect $PO$. Get $H$, the intersection of $PO$ with $\odot cRrOG$. We know this exists, since $P$ must fall outside $\odot cOrOG$.
  7. $\odot cHrPH$. Get $J$.
  8. $\odot cPrPJ$.
  9. $\odot cJrJP$, which will intersect with $\odot cPrPJ$ at $K$ and $L$.
  10. $\odot cOrPO$.
  11. Extend $LK$ through to $M$, the intersection with $\odot cOrPO$.
  12. Connect $MO$, get $N$, the intersection on $\odot cOrOH$.
  13. Extend $PN$ through to intersect the original circle to get $Q$ and $R$. This line is congruent to $AB$ and passes through $P$.

I apologize that this construction may be hard to follow without the picture. This takes me $13$ steps, but I see that Hartshorne suggests it's possible in $5$, hence the (par=$5$). Can anyone produce a more efficient solution in only $5$ steps? I've been thinking about it, but don't see how to make this any simpler from a compass and straightedge construction.

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What exactly is a "step"? –  Aryabhata Jan 26 '11 at 2:55
    
@Moron, drawing a circle or connecting/extending two points in a line each constitute a step. Simply labeling an intersection as a certain point or picking a point at random do not constitute steps. I think that covers all possible operations. –  yunone Jan 26 '11 at 2:57
    
I've got it in something like 8-10 steps, but not yet in only 5. –  Isaac Jan 26 '11 at 3:01
    
To make that more clear for example, if I wanted to bisect a segment $AB$, I couldn't just say, bisect $AB$ at $C$, and call it one step. I would have to describe the circles of radius $AB$ centered at $A$ and $B$ respectively, which counts as 2 steps, and then connect the intersections on opposite sides of the segment to find the midpoint of $AB$, for a total of 3 steps. –  yunone Jan 26 '11 at 3:01
    
@Isaac, thanks for taking a look at the problem. I'd be curious to follow any construction more efficient than mine. –  yunone Jan 26 '11 at 3:03

3 Answers 3

up vote 4 down vote accepted

1. Pick an arbitrary point C on the circle. Draw an arc centered at C and radius AB, intersecting the original circle at point D.

If there is no intersection, stop: AB is too long for a solution to exist.

2. Draw line segment CD.

We now have a chord which is congruent to AB. We need to find a congruent chord which passes through P, so the first thing we need to do is figure out what point on CD will correspond to P.

3. Draw an arc with center O and radius OP. Let this intersect CD in point E.

If there is no intersection, stop: AB is too short for a solution to exist. Otherwise, E is the point we needed. OE is congruent to OP, and CED is congruent to the desired chord. Now if we find the point Q that C corresponds to, we'll have the chord we want.

4. Draw an arc with center P and radius EC to intersect the original circle at point Q.

The (undrawn) triangles OPQ and OEC are congruent, so PQ and EC make the same angle with the radius, and that determines the length of the chord.

5. Draw line segment PQ, extended beyond P to meet the circle on the other side.

This is the desired chord. It is congruent to CD, which is congruent to AB.

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This also doesn't seem to work in the case where the arc with center $O$ and radius $OP$ is tangent to $CD$. –  mjqxxxx Jan 26 '11 at 3:28
    
@Isaac: fixed that, thanks. –  Rahul Jan 26 '11 at 4:14
    
@mjqxxxx: That's a good point. I think I fixed it with my edit; can you verify? –  Rahul Jan 26 '11 at 4:16
    
Very nice solution, many thanks Rahul. –  yunone Jan 26 '11 at 4:25
    
@Rahul: Yes, I think this fixes it. –  mjqxxxx Jan 26 '11 at 4:52

Shorter, but not down to par:

  1. construct the circle centered at A with the same radius as circle O
  2. construct the circle centered at B with the same radius as circle O
  3. join the intersection points of these two circles; call one of these intersection points C and call the intersection that is the midpoint of segment AB D
  4. construct the circle centered at O with radius CD (note that the desired chord is the chord through P that is tangent to this circle)
  5. construct the circle centered at O with radius OP
  6. construct the circle centered at P with radius OP
  7. join the intersection points of these two circles, which will intersect OP at its midpoint
  8. construct the circle centered at the midpoint of OP and passing through O and P
  9. draw the line through P and one of the two intersection points of the circles in steps 4 and 8; this line contains the desired chord.

edit Having seen Rahul Narain's answer, I was able to apply a similar idea but starting from my basic idea in steps 1-2 above, which constructed the center of a circle congruent to circle O for which AB is a chord, to get down to par.

  1. construct the circle centered at A with the same radius as circle O
  2. construct the circle centered at B with the same radius as circle O
  3. construct the circle centered at an intersection of the circles from steps 1-2 (if AB is too long, this intersection point will not exist) with radius OP; label as C one of its points of intersection with AB (if P is too close to O, this intersection will not exist)
  4. construct the circle centered at P with radius AC; label as Q one of its intersections with circle O
  5. construct the line through P and Q, which contains the desired chord
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Thank you Isaac, I like how this solution shows a different way to find the tangent from a point to a circle than what Euclid did. –  yunone Jan 26 '11 at 4:14
  1. Draw the circle $cOP$ with radius $OP$ and center $O$.
  2. Draw ray $OP$; let it intersect the original circle at $Q$.
  3. Draw an arc $cQrAB$; let it intersect the original circle at $R$. (If it has no intersection with the original circle, stop; $AB$ is longer than any chord of the given circle.)
  4. Draw the chord $QR$. Let it intersect the inner circle $cOP$ at $P'$. (If it does not intersect the inner circle, stop; $AB$ is shorter than any chord of the given circle that passes through $P$.) Now the chord $QP'R$ is congruent to $AB$, and $OP'$ is congruent to $OP$.
  5. Draw ray $OP'$; let it intersect the original circle at $Q'$.
  6. Extend the line $PQ'$ to construct the desired chord.
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Thanks mjqxxxx, steps 3 and 4 are a nice way to check if the situation is even possible. –  yunone Jan 26 '11 at 4:15

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