## February 2015 solution

This month, readers definitely rose to the challenge, sending in proofs in a wide variety of disciplines, ranging from Euclidean geometry to trigonometry, to algebraic (Cartesian) geometry to complex-plane calculations to calculus of variations.

Here is a solution based on trigonometry (though, in a somewhat less elegant wording, it can be translated step-for-step to a Euclidean argument).

First, let's prove the following lemma.

Lemma: For any triangle ABC it is true that the triangle A'B'C' satisfying that the angle ABC equals the angle A'B'C' and that |AB|+|BC|=|A'B'|+|B'C'| and that |A'B'|=|B'C'| satisfies ACA'C'.

Proof of the lemma:
The general edge-length relation in triangles is that

|AB|2+|BC|2-2|AB||BC|cos(ABC) = |AC|2.

To minimise |AC| we would need to minimise the right-hand side of this equation, and therefore also the left-hand side. Because of the condition |AB|+|BC|=|A'B'|+|B'C'|, we can treat

(|AB|+|BC|)2 = |AB|2 + 2|AB||BC| + |BC|2

as a constant. Subtracting from this constant the left-hand side of the first displayed equation we get that we need to maximise

2|AB||BC|(cos(ABC)+1)

Because angle ABC equals angle A'B'C', we can treat it also as a constant. Furthermore, cos(ABC) is bounded from below by -1, so we have that we need to maximise |AB||BC| while not changing their sum. The solution is therefore to make them equal to each other. This is the definition of A'B'C'.

Q.E.D.

(An alternate and perhaps more elegant proof of the same lemma is to see that (|AB|+|BC|)sin(ABC/2) is the length of the projection of AC onto the perpendicular of ABC's angle bisector, which is a lower bound on the length of AC itself. A'C' equals this bound.)

With this lemma, it is possible to prove the main theorem.

Proof:
First, note that we can assume that ABCD is convex. (Suppose, on the contrary that, without loss of generality, angle ABC is greater than 180deg. Then by taking AB'CD, where B' is the mirror image of B on the other side of AC, we can construct a quadrilateral with the same side lengths but longer diagonal sum. By repeating this at most twice, we get a convex quadrilateral.)

With this assumption, the two diagonals of ABCD intersect at a point inside the quadrilateral. Let us call this point E, let angle AEB be called θ and let θ/2 be ψ.

By the lemma, |AB| can be bounded from below by (|AE|+|EB|)*sin(ψ). Similarly, |CD| can be bounded from below by (|CE|+|ED|)*sin(ψ). Altogether, |AB|+|CD| can be bounded from below by (|AC|+|BD|)*sin(ψ).

For exactly the same reason, |BC|+|DA| can be bounded from below by (|AC|+|BD|)*sin((180-θ)/2) = (|AC|+|BD|)*cos(ψ).

The largest of |AB| and |CD| can be bounded from below by (|AC|+|BD|)*sin(ψ)/2, and the largest of |BC| and |DA| can be bounded from below by (|AC|+|BD|)*cos(ψ)/2.

So the sum of the largest three of the edges can be bounded from below by (|AC|+|BD|)*max(sin(ψ)+cos(ψ)/2, sin(ψ)/2 + cos(ψ)).

To prove the theorem, we need to show that the minimum of max(sin(ψ)+cos(ψ)/2, sin(ψ)/2 + cos(ψ)) is greater than 1.

The minimum of this expression is attained at the extremes: ψ=0 or ψ=90deg. This minimum is 1 (although, because ψ can never actually attain either extreme value, neither can an exact ratio of 1 be attained by any quadrilateral).

Q.E.D.

Note that a ratio arbitrarily close to 1 can be attained when ABCD is a very long and narrow rectangle.

Last note:
For all the geometry purists in the audience, I decided to also publish this month one solution that is purely classic Euclidean geometry. For this, I am using the solution sent to me by Daniel Bitin. It appears here exactly as it was sent to me. I haven't altered a word. Thanks, Daniel!