Difference between revisions of "2009 AIME I Problems/Problem 11"
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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. | ||
− | == Solution 1 == | + | == Solution 1 (Matrix's, Determinants) == |
Let the two points <math>P</math> and <math>Q</math> be defined with coordinates; <math>P=(x_1,y_1)</math> and <math>Q=(x_2,y_2)</math> | Let the two points <math>P</math> and <math>Q</math> be defined with coordinates; <math>P=(x_1,y_1)</math> and <math>Q=(x_2,y_2)</math> | ||
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem). | We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem). | ||
− | <math>\det \left({ | + | <math>\det \left(\begin{array}{c} P \\ Q\end{array}\right)=\det \left(\begin{array}{cc}x_1 &y_1\\x_2&y_2\end{array}\right).</math> |
Since the triangle has half the area of the parallelogram, we just need the determinant to be even. | Since the triangle has half the area of the parallelogram, we just need the determinant to be even. | ||
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Since <math>2009</math> is not even, <math>((x_1)-(x_2))</math> must be even, thus the two <math>x</math>'s must be of the same parity. Also note that the maximum value for <math>x</math> is <math>49</math> and the minimum is <math>0</math>. There are <math>25</math> even and <math>25</math> odd numbers available for use as coordinates and thus there are <math>(_{25}C_2)+(_{25}C_2)=\boxed{600}</math> such triangles. | Since <math>2009</math> is not even, <math>((x_1)-(x_2))</math> must be even, thus the two <math>x</math>'s must be of the same parity. Also note that the maximum value for <math>x</math> is <math>49</math> and the minimum is <math>0</math>. There are <math>25</math> even and <math>25</math> odd numbers available for use as coordinates and thus there are <math>(_{25}C_2)+(_{25}C_2)=\boxed{600}</math> such triangles. | ||
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== Solution 2 == | == Solution 2 == | ||
+ | |||
+ | As in Solution 1, let the two points <math>P</math> and <math>Q</math> be defined with coordinates; <math>P=(x_1,y_1)</math> and <math>Q=(x_2,y_2)</math>. Let the line <math>41x+y=2009</math> intersect the x-axis at <math>X</math> and the y-axis at <math>Y</math>. <math>X</math> has coordinates <math>(49,0)</math>, and <math>Y</math> has coordinates <math>(0,2009)</math>. As such, there are exactly <math>50</math> lattice points on this line that can be used for <math>P</math> and <math>Q</math>. | ||
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+ | WLOG, let the x-coordinate of <math>P</math> be less than the x-coordinate of <math>Q</math>. Note that <math>[OPQ]=[OYX]-[OYP]-[OXQ]</math>. We know that <math>OY=2009</math> and <math>OX= 49</math>; as such, <math>[OYX]=\frac{1}{2} \cdot OY \cdot OX = \frac{1}{2} \cdot 2009 \cdot 49</math>. In addition, <math>[OYP]=\frac{1}{2} \cdot 2009 \cdot x_1</math> and <math>[OXQ]=\frac{1}{2} \cdot 49 \cdot y_2</math>. | ||
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+ | Since <math>2009 \cdot 49</math> is odd, the total area of <math>OYX</math> is not an integer; rather, it is of the form <math>k + \frac{1}{2}</math> where <math>k</math> is an integer. To ensure <math>[OPQ]</math> has an integral value, exactly one of <math>[OPY]</math> and <math>[OQX]</math> must have an integral value as well (the other must be of the form <math>k + \frac{1}{2}</math> where <math>k</math> is an integer). | ||
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+ | Returning to <math>41x+y=2009</math>, we notice that integer pairs of <math>x</math> and <math>y</math> that satisfy the equation always have different parities. To satisfy exactly one of <math>[OPY]</math> and <math>[OQX]</math> having an integral area, we must have <math>x_1</math> and <math>y_2</math> having different parities. This is because having an even number for <math>x_1</math> or <math>y_2</math> makes the area of the triangle an integer. We can therefore deduce that <math>x_1</math> and <math>x_2</math> have the same parity. | ||
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+ | Out of the <math>50</math> usable lattice points for <math>P</math> and <math>Q</math>, <math>25</math> have even x-coordinates and <math>25</math> have odd x-coordinates. Since we must pick two points with even x-coordinates or two points with odd x-coordinates, our desired answer is <math>\binom{25}{2}+\binom{25}{2}=300+300=\fbox{600}</math>. | ||
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+ | == Solution 3 (Analytic geometry) == | ||
As in the solution above, let the two points <math>P</math> and <math>Q</math> be defined with coordinates; <math>P=(x_1,y_1)</math> and <math>Q=(x_2,y_2)</math>. | As in the solution above, let the two points <math>P</math> and <math>Q</math> be defined with coordinates; <math>P=(x_1,y_1)</math> and <math>Q=(x_2,y_2)</math>. | ||
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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 | + | == Solution 4 == |
− | 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> small segments (all adjacent to each other) 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 | + | 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> small segments (all adjacent to each other) 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 <math>x</math>-values <math>x_1</math> and <math>x_2</math> (<math>0 \leq x_1, x_2 \leq 49</math>) such that their positive difference (<math>n</math>) 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}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:01, 16 August 2020
Contents
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 (Matrix's, Determinants)
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).
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 Solution 1, let the two points and be defined with coordinates; and . Let the line intersect the x-axis at and the y-axis at . has coordinates , and has coordinates . As such, there are exactly lattice points on this line that can be used for and .
WLOG, let the x-coordinate of be less than the x-coordinate of . Note that . We know that and ; as such, . In addition, and .
Since is odd, the total area of is not an integer; rather, it is of the form where is an integer. To ensure has an integral value, exactly one of and must have an integral value as well (the other must be of the form where is an integer).
Returning to , we notice that integer pairs of and that satisfy the equation always have different parities. To satisfy exactly one of and having an integral area, we must have and having different parities. This is because having an even number for or makes the area of the triangle an integer. We can therefore deduce that and have the same parity.
Out of the usable lattice points for and , have even x-coordinates and have odd x-coordinates. Since we must pick two points with even x-coordinates or two points with odd x-coordinates, our desired answer is .
Solution 3 (Analytic geometry)
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 4
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 small segments (all adjacent to each other) 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 -values and () such that their positive 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.