Quadrilateral from a Segment

Here is Problem 22 from the 1968 Annual High School Mathematics Examination:

A segment of length 1 is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is

  1. equal to $\frac{1}{4}$
  2. equal to or greater than $\frac{1}{8}$ and less than $\frac{1}{4}$
  3. greater than $\frac{1}{8}$ and less than $\frac{1}{2}$
  4. greater than $\frac{1}{8}$ and less than $\frac{1}{4}$
  5. less than $\frac{1}{2}$

A solution is also included in the book [The Contest Problem Book, pp. 102-103]:

Fundamental in the proof is the fact that a quadrilateral with four given segments as sides exists if and only if the length of each segment is less than the sum of the lengths of the other three. (The "only if" part is obvious since the lengths of a polynomial path is at least equal to the distance between its endpoints. On the other hand, if the length of each segment is less than the sum of the lengths of the other three, label the segments so that $s_1 + s_2 \ge s_3 + s_4$. Then there is a triangle with sides $s_1$, $s_2$, and $s_3 +s_4$. This triangle may be viewed as a quadrilateral with sides $s_1$, $s_2$, $s_3$ and $s_4$.) Now let $s_1$, $s_2$, $s_3$ and $s_4$ denote the lengths of the segments. If a quadrilateral exists, then by the fact mentioned above,

$s_1 \lt s_2 + s_3 + s_4$.

By hypothesis,

$s_1 + s_2 + s_3 + s_4 = 1$.

If we replace the sum $s_2 + s_3 + s_4$ by a smaller number $s_1$, we obtain the inequality

$1 = s_1 + s_2 + s_3 + s_4 \gt s_1 + s_1 = 2s_1$,

so that $s_1 \lt \frac{1}{2}$.

Since none of the four segments is any way special, we can deduce, by the same argument, that

$s_2 \lt \frac{1}{2}$, $s_3 \lt \frac{1}{2}$, and $s_4 \lt \frac{1}{2}$.

Conversely, if , $s_I \lt \frac{1}{2}$ $(I = 1, 2, 3, 4)$ and $s_1 + s_2 + s_3 + s_4 = 1$, then

$s_2 + s_3 + s_4 = 1 - s_1 > 1-\frac{1}{2} = \frac{1}{2} \gt s_1$,

so that $s_1 \lt s_2 + s_3 + s_4$. Corresponding inequalities hold for the other segments. Choice (E) is therefore correct. All the other choices fail; for example, a rectangle with adjacent sides of lengths $\frac{1}{16}$ and $\frac{7}{16}$ has perimeter $1$, yet is excluded by all other choices. Choice (D) is clearly excluded, since there is no such division into four segments.

(This is just a sample problem from a collection that has been reviewed elsewhere.)


Speaking candidly, I have misread this proof and thought to categorize it as invalid. Putting it down on a page made clear that I was mistaken. However, the proof is still incomplete. The "if" part in the parenthesized portion of the proof misses the case where all four segments are equal, in which case there is no triangle with sides $s_1$, $s_2$, and $s_3 +s_4$. So that the argument, as published, fails. However, in that case the four sides may form a square or rhombi, showing that even then the four segments form a quadrilateral.


  1. C. T. Salkind, J. M. Earl, The Contest Problem Book III, MAA, 1973

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