2008 AMC 10B Problem 15

P11-15~3 minPrint

Diophantine Equations Number Properties Triangles: Area & Pythagorean Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Bound the quantity above/below or estimate to pin it down

How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$ , where $b<100$ ?

Show solution without answering

Key insight

The Pythagorean theorem simplifies to a squared equals 2b plus 1. So a is odd, and b stays below 100 only up to a = 13.

Solution

The Pythagorean theorem ties $a$ and $b$ together. Expanding it shows that $a$ alone decides $b$ .

1. The hypotenuse is $b + 1$ , so
$a^2 + b^2 = (b + 1)^2 = b^2 + 2b + 1.$
2. Subtract $b^2$ from both sides:
$a^2 = 2b + 1, \qquad\text{so}\qquad b = \frac{a^2 - 1}{2}.$
3. $2b + 1$ is odd, so $a^2$ is odd, so $a$ is odd. And for every odd $a$ , $a^2 - 1$ is even, so $b$ comes out a whole number.
4. List the odd values of $a$ and the $b$ each one gives:

$a$	$1$	$3$	$5$	$7$	$9$	$11$	$13$	$15$
$b$	$0$	$4$	$12$	$24$	$40$	$60$	$84$	$112$

$a = 1$ gives $b = 0$ , which is not a triangle, so drop it.
$a = 15$ gives $b = 112$ , which is too big. Larger $a$ make $b$ even bigger.
That leaves $a = 3, 5, 7, 9, 11, 13$ : six triangles, such as $3$ - $4$ - $5$ and $5$ - $12$ - $13$ .

The answer is $\boxed{\textbf{(A)}\ 6}$ .

Why this works

When two sides differ by a fixed amount, write one in terms of the other and expand. The squares cancel and leave a simple equation. Then an odd/even check tells you which values work, and the size limit tells you when to stop. The triangles found here, $3$ - $4$ - $5$ , $5$ - $12$ - $13$ , $7$ - $24$ - $25$ , are a well-known family where the hypotenuse is one more than a leg.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The Pythagorean theorem gives $a^2 + b^2 = (b + 1)^2 = b^2 + 2b + 1$ , so
$a^2 = 2b + 1.$
Thus $a^2$ is odd, so $a$ is odd, and $b = \dfrac{a^2 - 1}{2}$ is automatically an integer for odd $a$ . Each odd $a$ therefore yields exactly one triangle, provided $b$ is a positive integer below $100$ .

Positivity: $a = 1$ gives $b = 0$ , which is not a triangle, so $a \ge 3$ .

Bound: $b < 100$ means $a^2 - 1 < 200$ , so $a^2 < 201$ , giving $a \le 14$ . The odd values are $a = 3, 5, 7, 9, 11, 13$ , with $b = 4, 12, 24, 40, 60, 84$ .

That is $6$ triangles. The answer is $\boxed{\textbf{(A)}\ 6}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Algebra, Chapter 11: Special Factorizations