The TV series *Numb3rs* has highlighted how mathematics can play a role in solving crimes. Even though the episodes are sometimes rather fanciful, they still illustrate ways in which various types of math can help illuminate mysteries, confirm conjectures, and point to villains.

In real life, math can also be relevant in the courtroom or come up in legal disputes.

Last year, the Pythagorean theorem was a deciding factor in a case before the New York State Court of Appeals. A man named James Robbins was convicted of selling drugs within 1,000 feet of a school. In the appeal, his lawyers argued that Robbins wasn’t actually within the required distance when caught and so should not get the stiffer penalty that school proximity calls for.

The arrest occurred on the corner of Eighth Avenue and 40th Street in Manhattan. The nearest school, Holy Cross, is on 43rd Street between Eighth and Ninth Avenues.

Law enforcement officials applied the Pythagorean theorem to calculate the straight-line distance between the two points. They measured the distance up Eighth Avenue (764 feet) and the distance to the church along 43rd Street (490 feet), using the data to find the length of the hypotenuse, 907.63 feet.

Robbins’ lawyers contended that the school is more than 1,000 feet away from the arrest site because the shortest (as the crow flies) route is blocked by buildings. They said the distance should be measured as a person would walk the route.

However, the seven-member Court of Appeals unanimously upheld the conviction, asserting that the distance in such cases should be measured “as the crow flies.”

“Plainly, guilt under the statute cannot depend on whether a particular building in a person’s path to a school happens to be open to the public or locked at the time of a drug sale,” Chief Justice Judith S. Kaye wrote in the opinion, as reported in the Nov. 23, 2005, *New York Times*.

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Another application of mathematics in a legal context is described by mathematician William H. Kazez of the University of Georgia in the November *Math Horizons*.

This conundrum involved a man living in subsidized housing, his landlord, and the state of New York. The man received $1,000 per month from the state. This figure was based on a formula that included 30 percent of the amount he was paying in rent, and his rent was $300 per month. However, his rental contract specified that, if his income went up, his rent would increase by 20 percent of his raise.

So, when the state increased the man’s stipend by $100 per month, the landlord raised the rent by 20 percent of $100. The man then went back to state officials, who agreed to raise his stipend by 30 percent of $20, which went from $1,000 to $1,100 to $1,106. This was just the beginning of what, in principle, would be an infinite cycle of stipend and rent increases.

You end up with two geometric series:

Stipend: $1000 + 100 + (.06)100 + (.06)^{2}100 + . . .

Rent: $300 + (.2)100 + (.2)(.06)100 + .2(.06)^{2}100 + . . .

Of course, there are formulas for summing geometric series. Kazez, however, worked out a simpler solution.

Here’s his argument. Suppose that the man ends up with a stipend of 1100 + *x* and a rent of 300 + *y*. The landlord would insist that *y* = .2(100 + *x*), and the man would be entitled to a stipend increase of *x* = .3*y*. So, *y* = .2(100 + *x*) = .2(100 + .3*y*) = 20 + .06*y*.

But then .94*y* = 20, and the landlord will have to do with *y* = 2000/94, or roughly 21.28. Similarly, *x* = .3*y* = .3(.2(100 + *x*)) = 6 + .06*x*. So, .94*x* = 6, or *x* = 600/94, or roughly 6.38.

In the end, the man receives a state stipend of $1106.38 and the landlord gets a rent of $321.28.

Case resolved.

If you wish to comment on this article, see the MathTrek blog version. For more math fun, go to http://blog.sciencenews.org/mathtrek/.