|Ivars Peterson's MathTrek|
March 22, 1999
This problem underlies a math project that earned fifth place in the 1999 Intel Science Talent Search for Rio G. Bennin, age 17, of Berkeley, Calif. Titled "Triangular Equalizers, Pythagorean Quadrilaterals, Equalizers of N-Dimensional Convex Sets, and Perfect Equalizers," his paper dealt with dividing two-dimensional figures, such as triangles, into two pieces of equal area and perimeter, then extended those ideas and results to higher-dimensional shapes.
It all started with a math contest question. The problem asked for the length of a line (called an equalizer) that would cut a given triangle into a smaller triangle and a quadrilateral such that the two shapes have equal areas and equal perimeters.
In solving the problem, Bennin noticed that, in this particular case, the base of the quadrilateral, c, squared was equal to the sum of the adjacent sides, (r + s), squared, plus the equalizer, z, squared. He called the shape a Pythagorean quadrilateral and wondered whether that relationship, c2 = (r + s)2 + z2 , is true for all triangle equalizers.
Bennin wrote to the problem's creator, George Berzsenyi of the Rose-Hulman Institute of Technology in Terre Haute, Ind. Berzsenyi replied that he had not heard of the relationship, encouraged Bennin to try to prove it for all triangle equalizers, and referred him to an article in Quantum magazine.
Bennin initially proved that every triangle has precisely one or three equalizers. Each such equalizer always passes through a point known as the triangle's incenter. Located where the bisectors of a triangle's three angles intersect, the incenter is the center of the largest circle that can be inscribed inside that triangle.
Bennin then showed that the quadrilateral formed by a triangle's equalizer is, indeed, always a Pythagorean quadrilateral. From this proof, he also obtained a simple formula for the length of an equalizer, z, based on the lengths of the original triangle's three sides, a, b, c:
z2 = c2 - (a + b - c)2 /4.
That was just the beginning. Bennin also discovered that it's possible to find an equalizer for certain class of three- or higher-dimensional shapes, known as simplexes. In three dimensions, such an equalizer would be a plane that cuts the figure into two blocks of equal volume and surface area.
Finding an equalizer for a tetrahedron is relatively straightforward. For other three-dimensional figures and in higher dimensions, the task is much tougher. Nonetheless, Bennin's proof shows that such equalizers exist, even if they can't be readily found or visualized.
There are a number of ways to extend Bennin's results. Equalizers can be found in two dimensions not only for triangles but also for any convex shape. Convex means that the figure has no indentations. Thus, a square or a circle is convex, but a five-pointed star is not.
Bennin has also discovered certain three-dimensional forms for which it is possible to find what he calls a "perfect" equalizer--one that bisects volume, surface area, and edge length.
Copyright 1999 by Ivars Peterson
Berzsenyi, G. 1997. The equalizer of a triangle. Quantum 7(March-April):51.
Weiss, P. 1999. Neutrinos to buckyballs: 10 talents tower. Science News 155(March 13):163. (Available at http://www.sciencenews.org/sn_arc99/3_13_99/fob3.htm).
Information about the Intel Science Talent Search is available at http://www.sciserv.org/sts/.
Comments are welcome. Please send messages to Ivars Peterson at firstname.lastname@example.org.