When I am working with my child, I am stuck in this geometry problem.

"We have two different points $M, N$ in the plane. Using only compass to construct the midpoint $I$ of the segment $MN$."

Thank you for all helping and comments.

  • $\begingroup$ Do you mean to find the midpoint of $MN$? $\endgroup$ – Neal Nov 2 '12 at 3:48
  • $\begingroup$ @Jasper Loy: Yeah. Only compass. $\endgroup$ – blindman Nov 2 '12 at 3:48
  • $\begingroup$ @Neal: Dear Sir. Thank you for your construction. $\endgroup$ – blindman Nov 2 '12 at 3:49

Some Googling revealed the following comments to this answer:

  • I know it is possible, but is there an easy way to divide a segment in half with only a compass? – robjohn♦ May 20 at 3:46
  • I don't know if that's "easy", but here's one method:
    1. Find the point $C$ on the ray from $A$ through $B$ such that $|AC|=2|AB|$ using my previous comment [The relevant part: "To double the distance along a ray, use the construction of a regular hexagon with vertex $A$ and center $B$".]
    2. Intersect the circle with center $C$ through $A$ with the circle with center $A$ through $B$ to find $D_1,D_2$.
    3. The midpoint of $AB$ is the second point of intersection of the two circles with center $D_i$ through $A$. – t.b. May 20 at 9:28
  • Here is a picture of what I have in mind: - t.b. May 20 at 12:38

Division in half

The dotted line is not used in the construction.


The triangles $\Delta ACD_1$ and $\Delta AMD_1$ are isosceles by construction and they share a common angle, hence they are similar. Therefore $AM : AB = AM : AD_1 = AD_1 : AC = AB : AC = 1 : 2$.

  • $\begingroup$ If anyone wants to expand on the details, fell free to do so. $\endgroup$ – commenter Nov 2 '12 at 4:04
  • 1
    $\begingroup$ @commenter: Could you give the proof of your construction? Thank you for your construction. $\endgroup$ – blindman Nov 2 '12 at 4:08
  • $\begingroup$ @blindman: I added a proof. $\endgroup$ – commenter Nov 2 '12 at 4:29
  • 1
    $\begingroup$ The fact that $D_1$ and $D_2$ are symmetric about the line $AB$ means that $M$, being equidistant from both, must be on $AB$. The fact that the triangles $AD_1M$ and $ACD_1$ are both isosceles and share a base angle angle makes them similar. And the ratio of $AC$ to $AD_1$ is 2:1, so the ratio of $AD_1$ to $AM$ is 2:1. Length $AB$ = Length $AD_1$ so $M$ is the midpoint. $\endgroup$ – Logan Stokols Nov 2 '12 at 4:30
  • $\begingroup$ @Logan: exactly, thanks! (I added the same argument just a minute ago). $\endgroup$ – commenter Nov 2 '12 at 4:31

Open the compass to any length more than half the distance between $\,A,B\,$ but less than their total distance. Put the compass's point on A and trace part of the circle over the line $\,AB\,$ , and after this do the same putting the point on $\,B\,$, (without changing the compass's openning!) and mark the interesection point of the two circles as $\,P\,$.

Now repeat the above with circles under the line segment and mark the intersection point of the two circles as $\,S\,$ (BTW, no need the compass has the very same openning as in the first part!). Since both points $\,P,S\,$ are at the same distance from $\,A\,$ and from $\,B\,$ (why?) , joining them gives you the perpendicular bisector of $\,AB\,$.

Finally, just take the intersection of the Perp. bisector with the segment $\,AB\,$

  • 1
    $\begingroup$ And how do you find the intersection point of the perpendicular bisector with the segment $AB$ without straightedge? $\endgroup$ – commenter Nov 2 '12 at 3:45
  • $\begingroup$ As the OP talks of the "segment $\,AB\,$" I assumed this segment is given, of course. $\endgroup$ – DonAntonio Nov 2 '12 at 3:46
  • $\begingroup$ @DonAntonio: Dear Sir. We have to use only compass to construct midpoint. Thank you for your answer. $\endgroup$ – blindman Nov 2 '12 at 3:46
  • $\begingroup$ Even if you have the segment $AB$ you still need a straightedge to draw the perpendicular bisector and find the intersection point. $\endgroup$ – commenter Nov 2 '12 at 3:47
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    $\begingroup$ @DonAntonio: This problem is not easy as we think. $\endgroup$ – blindman Nov 2 '12 at 3:55

Finally, by google we find the solution on thess sites: http://gogeometry.com/circle/mascheroni_compass_1.htm http://mathafou.free.fr/themes_en/compas.html

and useful lecture on this problem http://www.math.ualberta.ca/~tlewis/343_10/04sec.pdf


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