Difference between revisions of "2009 AIME I Problems/Problem 11"
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== Problem == | == Problem == | ||
Consider the set of all triangles <math>OPQ</math> where <math>O</math> is the origin and <math>P</math> and <math>Q</math> are distinct points in the plane with nonnegative integer coordinates <math>(x,y)</math> such that <math>41x + y = 2009</math>. Find the number of such distinct triangles whose area is a positive integer. | Consider the set of all triangles <math>OPQ</math> where <math>O</math> is the origin and <math>P</math> and <math>Q</math> are distinct points in the plane with nonnegative integer coordinates <math>(x,y)</math> such that <math>41x + y = 2009</math>. Find the number of such distinct triangles whose area is a positive integer. | ||
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Summing them up, we get that there are <math>2+4+\dots + 48 = \boxed{600}</math> triangles. | Summing them up, we get that there are <math>2+4+\dots + 48 = \boxed{600}</math> triangles. | ||
+ | === Solution 3 === | ||
+ | |||
+ | We present a non-analytic solution; consider the lattice points on the line <math>41x+y=2009</math>. The line has intercepts <math>(0, 2009)</math> and <math>(49, 0)</math>, so the lattice points for <math>x=0, 1, \ldots, 49</math> divide the line into <math>49</math> equal segments. Call the area of the large triangle <math>A</math>. Any triangle formed with the origin having a base of one of these segments has area <math>A/49</math> (call this value <math>B</math>) because the height is the same as that of large triangle, and the bases are in the ratio <math>1:49</math>. A segment comprised of <math>n</math> of these segments consecutively will have area <math>nB</math>. Rewriting in terms of the original area, <math>A=(\frac{1}{2})(49)(2009)</math>, <math>B=\frac{2009}{2}</math>, and <math>nB=n(\frac{2009}{2})</math>. It is clear that in order to have a nonnegative integer for <math>nB</math> as desired, <math>n</math> must be even. This is equivalent to finding the number of ways to choose two distinct nonnegative <math>x</math>-values <math>x_1</math> and <math>x_2</math> ($1 \leq x_1, x_2 \leq 49) such that their difference is even. Follow one of the previous methods above to choose these pairs and arrive at the answer of 600. | ||
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=10|num-a=12}} | {{AIME box|year=2009|n=I|num-b=10|num-a=12}} |
Revision as of 12:29, 1 April 2013
Problem
Consider the set of all triangles where is the origin and and are distinct points in the plane with nonnegative integer coordinates such that . Find the number of such distinct triangles whose area is a positive integer.
Solution 1
Let the two points and be defined with coordinates; and
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 is not even, must be even, thus the two 's must be of the same parity. Also note that the maximum value for is and the minimum is . There are even and odd numbers available for use as coordinates and thus there are such triangles.
Solution 2
As in the solution above, let the two points and be defined with coordinates; and .
If the coordinates of and have nonnegative integer coordinates, and 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 to be and the x-intercept to be .
Using the point-to-line distance formula, we can calculate the height of from vertex (the origin) to be:
Let be the base of the triangle that is part of the line .
The area is calculated as:
Let the numerical area of the triangle be .
So,
We know that is an integer. So, , where is also an integer.
We defined the points and as and .
Changing the y-coordinates to be in terms of x, we get:
and .
The distance between them equals .
Using the distance formula, we get
WLOG, we can assume that .
Taking the last two equalities from the string of equalities and putting in our assumption that , we get
.
Dividing both sides by , we get
As we mentioned, is an integer, so is an even integer. Also, and are both positive integers. So, and are between 0 and 49, inclusive. Remember, as well.
- There are 48 ordered pairs such that their positive difference is 2.
- There are 46 ordered pairs such that their positive difference is 4.
...
- Finally, there are 2 ordered pairs such that their positive difference is 48.
Summing them up, we get that there are triangles.
Solution 3
We present a non-analytic solution; consider the lattice points on the line . The line has intercepts and , so the lattice points for divide the line into equal segments. Call the area of the large triangle . Any triangle formed with the origin having a base of one of these segments has area (call this value ) because the height is the same as that of large triangle, and the bases are in the ratio . A segment comprised of of these segments consecutively will have area . Rewriting in terms of the original area, , , and . It is clear that in order to have a nonnegative integer for as desired, must be even. This is equivalent to finding the number of ways to choose two distinct nonnegative -values and ($1 \leq x_1, x_2 \leq 49) such that their difference is even. Follow one of the previous methods above to choose these pairs and arrive at the answer of 600.
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |