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.
Solution
We shall consider three cases:
- c<2a
- c ≥ 2a, b < 2a
- 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.
Proof
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.

Problems (Use a compass only)
- Multiply a Line Segment by a Whole Number
- Reflect a Point in a Line Sgement
- Cross a Circle by a Line Segment
- Drop a Perpendicular to a Line from a Point
- Detect Collinearity
- Complete a Parallelogram
- Bisect an Arc
- Find the Points of Intersection of a Circle with a Line Segment
- Build a Square on a Given Side
- Find the Fourth Proportional of Three Lengths
- Find the Intersection of Two Straight Lines
- Divide a Line Segment into a Whole Number of Parts
- Find the Center of a Given Circle
- Bisect a given line
- Mascheroni Construction of a Regular Pentagon
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Copyright © 1996-2012 Alexander Bogomolny
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