Solution for January 2013

The Problem:
	You have five line segments such that every choice of three of them can be used as the sides of a triangle. (This means that for any three of the segments, the sum of the lengths of the smaller two segments must exceed the length of the third.) Prove that at least one of these ten possible triangles must be acute.

The solution:

Let's begin with a review of the relevant geometry. It is customary to use the same capital letter to represent both the vertex of a triangle and its measure, and to use the corresponding lower case letter for both the opposite side and its length. Thus triangle $ABC$ has sides $a, b$, and $c$. The angle $A$ is acute, right, or obtuse according as $a^2,$ the square of the opposite side, is less than, equal to, or greater than $b^2 + c^2$. An easy way to see this is by the cosine law,
$$a^2 = b^2+c^2 - 2bc\cos A,$$
and to observe that $\cos A$ is positive when $A < 90^\circ$ (which makes $a^2$ smaller than $b^2 + c^2$), zero when $A = 90^\circ$, and negative when $A > 90^\circ$ (which makes $a^2$ bigger than $b^2 + c^2$).

For our problem we are given five line segments $a,b,c,d,e$. We may as well label them in increasing order:
$$a \le b \le c \le d \le e.$$
With this convention it is both necessary and sufficient to assume that
$$a + b > e$$
for all 10 sets of triangle inequalities to be satisfied. We now assume to the contrary that all ten triangles are obtuse or right angled. That would imply, among other things, that
$$e^2 \ge c^2 + d^2, \quad c^2 \ge a^2 + b^2, \quad \mbox{ and } \quad d^2 \ge a^2 + b^2.$$
Then, using the inequalities in the order we have listed them, we find that
$$(a+b)^2 > e^2 \ge c^2 + d^2 \ge a^2 + b^2 + a^2 + b^2.$$
That is,
$$ a^2 + 2ab + b^2 > 2a^2 + 2b^2,$$
or
$$0 > a^2 - 2ab + b^2 = (a-b)^2.$$
But zero cannot exceed the square of a real number. This contradiction forces us to conclude that at least one of the triangles must have three acute angles.

Comments. Our January problem was Challenge no. 6 of Pi in the Sky, issue 12 (fall 2008); their solution was published in the following issue 13 (fall 2009), page 29. The on-line edition of the journal can be found at

http://media.pims.math.ca/pi_in_sky/pi13.pdf.

Several readers observed that with four segments (instead of five), it is possible for all the triangles to be obtuse. The four segments in LoPresti's example are 6, 6, 9, and 11: Note that $6+6>11$, so that we can be certain that these segments can be used to form the triangles {6, 6, 11}, {6, 6, 9}, and {6, 9, 11}; furthermore, these triangles are all obtuse (because $6^2+6^2<81$ and $6^2+9^2<121)$.

Collignon and Desprez both looked briefly at the question of whether with five given segments we are guaranteed to have at least two acute triangles. Although the answer might be yes, they (independently) showed that the answer is no if we allow degenerate obtuse triangles; their example used the segments $1,1,\sqrt 2, \sqrt 3, 2$. Unfortunately the segments 1, 1, and 2 do not form a proper triangle.