Difference between revisions of "2009 AIME I Problems/Problem 11"

Latest revision as of 14:01, 16 August 2020

Problem

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 (Matrix's, Determinants)

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).

$\det \left(\begin{array}{c} P \\ Q\end{array}\right)=\det \left(\begin{array}{cc}x_1 &y_1\\x_2&y_2\end{array}\right).$

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

The determinant is

$(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))$

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 Solution 1, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ . Let the line $41x+y=2009$ intersect the x-axis at $X$ and the y-axis at $Y$ . $X$ has coordinates $(49,0)$ , and $Y$ has coordinates $(0,2009)$ . As such, there are exactly $50$ lattice points on this line that can be used for $P$ and $Q$ .

WLOG, let the x-coordinate of $P$ be less than the x-coordinate of $Q$ . Note that $[OPQ]=[OYX]-[OYP]-[OXQ]$ . We know that $OY=2009$ and $OX= 49$ ; as such, $[OYX]=\frac{1}{2} \cdot OY \cdot OX = \frac{1}{2} \cdot 2009 \cdot 49$ . In addition, $[OYP]=\frac{1}{2} \cdot 2009 \cdot x_1$ and $[OXQ]=\frac{1}{2} \cdot 49 \cdot y_2$ .

Since $2009 \cdot 49$ is odd, the total area of $OYX$ is not an integer; rather, it is of the form $k + \frac{1}{2}$ where $k$ is an integer. To ensure $[OPQ]$ has an integral value, exactly one of $[OPY]$ and $[OQX]$ must have an integral value as well (the other must be of the form $k + \frac{1}{2}$ where $k$ is an integer).

Returning to $41x+y=2009$ , we notice that integer pairs of $x$ and $y$ that satisfy the equation always have different parities. To satisfy exactly one of $[OPY]$ and $[OQX]$ having an integral area, we must have $x_1$ and $y_2$ having different parities. This is because having an even number for $x_1$ or $y_2$ makes the area of the triangle an integer. We can therefore deduce that $x_1$ and $x_2$ have the same parity.

Out of the $50$ usable lattice points for $P$ and $Q$ , $25$ have even x-coordinates and $25$ 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 $\binom{25}{2}+\binom{25}{2}=300+300=\fbox{600}$ .

Solution 3 (Analytic geometry)

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.

Solution 4

We present a non-analytic solution; consider the lattice points on the line $41x+y=2009$ . The line has intercepts $(0, 2009)$ and $(49, 0)$ , so the lattice points for $x=0, 1, \ldots, 49$ divide the line into $49$ equal segments. Call the area of the large triangle $A$ . Any triangle formed with the origin having a base of one of these segments has area $A/49$ (call this value $B$ ) because the height is the same as that of large triangle, and the bases are in the ratio $1:49$ . A segment comprised of $n$ small segments (all adjacent to each other) will have area $nB$ . Rewriting in terms of the original area, $A=(\frac{1}{2})(49)(2009)$ , $B=\frac{2009}{2}$ , and $nB=n(\frac{2009}{2})$ . It is clear that in order to have a nonnegative integer for $nB$ as desired, $n$ must be even. This is equivalent to finding the number of ways to choose two distinct $x$ -values $x_1$ and $x_2$ ( $0 \leq x_1, x_2 \leq 49$ ) such that their positive difference ( $n$ ) is even. Follow one of the previous methods above to choose these pairs and arrive at the answer of 600.

@@ Line 1: / Line 1: @@
 == 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.
-== Solution ==
+== 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>
 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({\matrix {P \above Q}}\right)=\det \left({\matrix {x_1 \above x_2}\matrix {y_1  \above y_2}\right).</math>
+<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.
@@ Line 16: / Line 15: @@
 <cmath>(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))</cmath>
-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 then <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.
+== 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>.
+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>.
+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).
+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.
+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>.
+== 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>.
+If the coordinates of <math>P</math> and <math>Q</math> have nonnegative integer coordinates, <math>P</math> and <math>Q</math> 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 <math>41x+y=2009</math> to be <math>(0,2009)</math> and the x-intercept to be <math>(49,0)</math>.
+Using the point-to-line distance formula, we can calculate the height of <math>\triangle OPQ</math> from vertex <math>O</math> (the origin) to be:
+<math>\dfrac{|41(0) + 1(0) - 2009|}{\sqrt{41^2 + 1^2}} = \dfrac{2009}{\sqrt{1682}} = \dfrac{2009}{29\sqrt2}</math>
+Let <math>b</math> be the base of the triangle that is part of the line <math>41x+y=2009</math>.
+The area is calculated as:
+<math>\dfrac{1}{2}\times b \times \dfrac{2009}{29\sqrt2} = \dfrac{2009}{58\sqrt2}\times b</math>
+Let the numerical area of the triangle be <math>k</math>.
+So, <math>k = \dfrac{2009}{58\sqrt2}\times b</math>
+We know that <math>k</math> is an integer. So, <math>b = 58\sqrt2 \times z</math>, where <math>z</math> is also an integer.
+We defined the points <math>P</math> and <math>Q</math> as <math>P=(x_1,y_1)</math> and <math>Q=(x_2,y_2)</math>.
+Changing the y-coordinates to be in terms of x, we get:
+<math>P=(x_1,2009-41x_1)</math> and <math>Q=(x_2,2009-41x_2)</math>.
+The distance between them equals <math>b</math>.
+Using the distance formula, we get
+<math>PQ = b = \sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\sqrt2 \times |x_2 - x_1| = 58\sqrt2\times z</math> <math>(*)</math>
+WLOG, we can assume that <math>x_2 > x_1</math>.
+Taking the last two equalities from the <math>(*)</math> string of equalities and putting in our assumption that <math>x_2>x_1</math>, we get
+<math>29\sqrt2\times (x_2-x_1) = 58\sqrt2\times z</math>.
+Dividing both sides by <math>29\sqrt2</math>, we get
+<math>x_2-x_1 = 2z</math>
+As we mentioned, <math>z</math> is an integer, so <math>x_2-x_1</math> is an even integer. Also, <math>x_2</math> and <math>x_1</math> are both positive integers. So, <math>x_2</math> and <math>x_1</math> are between 0 and 49, inclusive. Remember, <math>x_2>x_1</math> as well.
+*There are 48 ordered pairs <math>(x_2,x_1)</math> such that their positive difference is 2.
+*There are 46 ordered pairs <math>(x_2,x_1)</math> such that their positive difference is 4.
+...
+*Finally, there are 2 ordered pairs <math>(x_2,x_1)</math> such that their positive difference is 48.
+Summing them up, we get that there are <math>2+4+\dots + 48 = \boxed{600}</math> triangles.
+== 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 <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 ==
 {{AIME box|year=2009|n=I|num-b=10|num-a=12}}
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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