2009 AIME I Problems/Problem 11

Revision as of 20:29, 25 March 2012 by Williamhu888 (talk | contribs) (Solution 2)


Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$. Find the number of such distinct triangles whose area is a positive integer.

Solution 1

Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$

We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).

$\det \left({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1 \above y_2}\right).$ (Error compiling LaTeX. ! Package amsmath Error: Old form `\matrix' should be \begin{matrix}.)

Since the triangle has half the area of the parallelogram, we just need the determinant to be even.

The determinant is


Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$'s must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$. There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=\boxed{600}$ such triangles.

Solution 2

As in the solution above, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$.

If the coordinates of $P$ and $Q$ have nonnegative integer coordinates, $P$ and $Q$ must be lattice points either

  • on the nonnegative x-axis
  • on the nonnegative y-axis
  • in the first quadrant

We can calculate the y-intercept of the line $41x+y=2009$ to be $(0,2009)$ and the x-intercept to be $(49,0)$.

Using the point-to-line distance formula, we can calculate the height of $\triangle OPQ$ from vertex $O$ (the origin) to be:

$\dfrac{|41(0) + 1(0) - 2009|}{\sqrt{41^2 + 1^2}} = \dfrac{2009}{\sqrt{1682}} = \dfrac{2009}{29\sqrt2}$

Let $b$ be the base of the triangle that is part of the line $41x+y=2009$.

The area is calculated as: $\dfrac{1}{2}\times b \times \dfrac{2009}{29\sqrt2} = \dfrac{2009}{58\sqrt2}\times b$

Let the numerical area of the triangle be $k$.

So, $k = \dfrac{2009}{58\sqrt2}\times b$

We know that $k$ is an integer. So, $b = 58\sqrt2 \times z$, where $z$ is also an integer.

We defined the points $P$ and $Q$ as $P=(x_1,y_1)$ and $Q=(x_2,y_2)$.

Changing the y-coordinates to be in terms of x, we get:

$P=(x_1,2009-41x_1)$ and $Q=(x_2,2009-41x_2)$.

The distance between them equals $b$.

Using the distance formula, we get

$PQ = b = \sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\sqrt2 \times |x_2 - x_1| = 58\sqrt2\times z$ $(*)$

WLOG, we can assume that $x_2 > x_1$.

Taking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$, we get

$29\sqrt2\times (x_2-x_1) = 58\sqrt2\times z$.

Dividing both sides by $29\sqrt2$, we get

$x_2-x_1 = 2z$

As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well.

  • There are 48 ordered pairs $(x_2,x_1)$ such that their positive difference is 2.
  • There are 46 ordered pairs $(x_2,x_1)$ such that their positive difference is 4.


  • Finally, there are 2 ordered pairs $(x_2,x_1)$ such that their positive difference is 48.

Summing them up, we get that there are $2+4+\dots + 48 = \boxed{600}$ triangles.

See also

2009 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions
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