Given a square of area A, perimeter P. For which values of r between zero and one is it possible to draw a line across the square so that it cuts off a section of area equal to rA and a piece of perimeter equal to rP?

OKAY, go ahead and try it, I'll wait a bit.....

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As you probably discovered, there is only one, r=1/2. But then there are an infinite number of lines that make that so. One of the things I like about this problem is that it calls up so many possible generalizations. Is it possible to slice a Cube with a plane that divides volume and surface area into equal halves?

And what about triangles, Equilateral or in general. As early as 1959 Dov Aveshalom was writing about Cleavers. Unfortunately for me, his first paper was in Hebrew, but in 1963 he wrote "The Perimetric Bisection of Triangles." and used (created?) the term for perimeter bisectors which passed through a midpoint of one side of a triangle.

As far as I am aware, the first use of Splitters was in Ross Honsberger's Episodes in Nineteenth and Twentieth Century Euclidean Geometry in 1995. Shortly after that, in 1997, George Berzsenyi used (created?) the term equalizers in an article in Quantum Magazine. He gave no proofs, but posed three interesting conjectures:

1) Prove that every triangle has at least one equalizer

2) Prove that No triangle has more than three equlizers

3) Prove that there can (or cannot) be a triangle with exactly two equalizers.

The three questions resisted cracking until in 2010, Dimitrios Kodokostas published "Triangle Equalizers" in Mathematics Magazine. He not only proved that every triangle had at lest one equalizer, he described the conditions under which a triangle could have one, two or three equalizers, and that none had more. The breakthrough idea was his discovery that:

any equalizer goes through the incenter and any line through the incenter is an area splitter, if, and only if, it is a perimeter splitter.

By rotating a line through the incenter, a triangle would be cut off between two sides of the triangle and the cutting leg. When the area of that triangle changed for greater than 1/2 to less than, or vice-versa, there must exist an equalizer. He also worked out why there had been no discovery of a triangle with only two equalizers. The only conditions under which that can occur is a triangle with a smallest angle of less than about 49 degrees, and the second angle fell into a very narrow range of values dependent upon the measure of angle A.

So Squares have an infinite number of equalizing lines, and triangles have between one and three, but what about Mr Tanton's challenge to see if some other r was possible, ie, some sort of

**fractionalizer**in general.

I set my sights on an equilateral triangle first, figuring it would be my easiest to work with. Experimenting after a few seconds I realized that a regular unit triangle that had a line cutting smaller equilateral triangle with all sides of 2/3, would be exactly the r=4/9 that was impossible in the square. The perimeter cut off would be 2/3+2/3 = 4/3.

After a moment of playing with the numbers, I came up with two equations that established the conditions under which lengths of a and b cut off from a vertex of the unit triangle would produce a "fractionalizer" \( \frac{a+b}{3} = r\) and \( ab = r \) (since both the original triangle and the cut off triangle could be computed using the formula A= 1/2(side*side)sin(60deg) we could ignore the actual area, and deal with the product of the adjacent sides.

Restructuring the equations we arrive at the condition that a+b = 3ab. So I quickly set up a geogebra sketch to test the range of possible r's that could occur.

At a=b=2/3 I knew that 4/9=r, and as I adjusted the line on the x-axis to more than 2/3, the ratio increased toward 1/2 as the bottom (a side) approached a length of one, and the b side approached 1/2. When I shortened side a, the same thing happened, as the b side became larger. When a was 3/5 for example, the b side worked out to be 3/4, and r was 9/20.

I switched to a 3-4-5 right triangle to make sure that what I found was not limited to regular triangles, and indeed it was not. For example cutting off a right triangle with sides of 2 and 2, popped up right away. a+b= 4 which is 1/3 of P=12, and ab/2 = 2, or 1/3 of 6. In this case since (a+b)/12=r and (ab/2)/6 = r, it worked out to a+b=ab. Cutting off sides of 5/2 and 5/3 for example, gave r=25/72 or abt .347. As the length of side a on the 4 side of the 3-4-5 reached 4, r=4/9, the smallest possible r in the equilateral triangle.

I suspect that there are an infinite range of fractionalizers for every triangle, and that any number in the range \( 0 < r < 1/2 \) is achievable..... but alas the hour has grown late.