Find the Fourth Proportional of Three Lengths

Geometric Construction with the Compass Alone

Let the quantities a, b, c be defined as the lengths of three given segments. Find x such that a/b = c/x.


We shall consider three cases:

  1. c<2a
  2. c ≥ 2a, b < 2a
  3. c ≥ 2a, b ≥ 2a

In the case 1, take an arbitrary point O and describe two circles (I and II) with radii a and b, respectively. Pick a point A on the first circle (I) as the center and swing an arc with radius c to find the intersection point B.

Now, with A and B as centers draw two circles of an arbitrary radius d > |a - b| which would intersect the circle II at the points C and D, respectively. The segment CD has the required length x.


Indeed, the triangles ABO and CDO are similar isosceles triangles. AB = c, OA = OB = a, CO = DO = b. Therefore, CD = x. Q.E.D.

In the second case, consider the proportion a/c = b/x instead of a/b = c/x and apply the first case.

In the third case, use Problem #1 to construct a segment of length na such that c < 2(na). Apply case 1 to the proportion (na)/b = c/y. Use Problem #1 again to find x = ny.

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