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Difference between revisions of "1994 AIME Problems/Problem 7"

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== Solution ==
 
== Solution ==
−
bro whoever came up with this problem must've been drunk tbh, there are literally infinite solutions...
+
bro whoever came up with this problem must've been drunk tbh, there are literally infinite solutions...  
 +
The vertices are chosen from the points in a <math>4\times4</math> grid as shown below. Since there are <math>4\cdot 4=16</math> points in the grid and triangles consist of <math>3</math> distinct vertices, we get an initial count of<cmath>\binom{16}{3}=560.</cmath>
 +
[asy]
 +
size(100);
 +
dot((1,1));
 +
dot((2,1));
 +
dot((3,1));
 +
dot((4,1));
 +
 
 +
dot((1,2));
 +
dot((2,2));
 +
dot((3,2));
 +
dot((4,2));
 +
 
 +
dot((1,3));
 +
dot((2,3));
 +
dot((3,3));
 +
dot((4,3));
 +
 
 +
dot((1,4));
 +
dot((2,4));
 +
dot((3,4));
 +
dot((4,4));
 +
 
 +
[/asy]
 +
 
 +
However, degenerate triangles (those with <math>0</math> area) are precisely those formed by <math>3</math> collinear points, so we have to remove these from the count. In each column or row, there are <math>\binom{4}{3}=4</math> sets of <math>3</math> collinear points. There are <math>8</math> total columns and rows, so this gives us <math>4\cdot 8=32</math> invalid triangles.
 +
 
 +
There are also <math>4</math> such sets of <math>3</math> collinear points in each of the two main diagonals and <math>1</math> for each of the <math>4</math> shorter diagonals. This means that we counted a total of<cmath>32+8+4=44</cmath>invalid triangles in our initial computation of <math>560</math> triangles, so the answer is <math>560-44=\boxed{516}</math>.
 +
 
 +
Source: 1993 AHSME #28
 +
Hint(s):
 +
There are probably more degenerate triangles than you think.
 +
Your Response(s):
 +
<math> 516
 +
Problem 8 – Give Up – Score: 0 / 7 (25910)
 +
Problem: Report Error
 +
For certain ordered pairs </math>(a,b)<math> of real numbers, the system of equations\begin{eqnarray*} && ax+by =1\\ &&x^2+y^2=50\end{eqnarray*}has at least one solution, and each solution is an ordered pair </math>(x,y)<math> of integers. How many such ordered pairs </math>(a,b)<math> are there?
 +
Solution:
 +
The equation </math>x^2+y^2=50<math> is that of a circle of radius </math>\sqrt{50}<math>, centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are </math>(\pm1,\pm7)<math>, </math>(\pm5,\pm5)<math>, and </math>(\pm7,\pm1)<math> where the signs are all independent of each other, for a total of </math>3\cdot 2\cdot 2=12<math> lattice points. They are indicated by the blue dots below.
 +
 
 +
[asy]
 +
size(150);
 +
draw(circle((0,0),sqrt(50)));
 +
 
 +
draw((1,7)--(-1,-7),red);
 +
draw((7,1)--(5,-5), green);
 +
 
 +
dot((0,0));
 +
 
 +
dot((1,7),blue);
 +
dot((1,-7),blue);
 +
dot((-1,7),blue);
 +
dot((-1,-7),blue);
 +
 
 +
dot((5,5),blue);
 +
dot((5,-5),blue);
 +
dot((-5,5),blue);
 +
dot((-5,-5),blue);
 +
 
 +
dot((7,1),blue);
 +
dot((7,-1),blue);
 +
dot((-7,1),blue);
 +
dot((-7,-1),blue);
 +
[/asy]
 +
 
 +
Since </math>(x,y)=(0,0)<math> yields </math>a\cdot 0+b\cdot 0=0 \neq 1<math>, we know that </math>ax+by=1<math> is the equation of a line that does not pass through the origin. So, we are looking for the number of lines which pass through at least one of the </math>12<math> lattice points on the circle, but do not pass through the origin or through any non-lattice point on the circle. An example is the green line above. It is straightforward to show that a line passes through the origin precisely when there exist two opposite points </math>(p,q)<math> and </math>(-p,-q)<math> through which it passes. And example is the red line above.
 +
 
 +
There are </math>\binom{12}{2}=66<math> ways to pick two distinct lattice points, and subsequently </math>66<math> distinct lines which pass through two distinct lattice points on the circle. Then we subtract the lines which pass through the origin by noting that the lattice points on the circle can be grouped into opposite pairs </math>(p,q)<math> and </math>(-p,-q)<math>, for a total of </math>\frac{12}{2}=6<math> lines. Finally, we add the </math>12$ unique tangent lines to the circle at each of the lattice points.
 +
 
 +
Therefore, our final count of distinct lines which pass through one or two of the lattice points on the circle, but do not pass through the origin, is<cmath>66-6+12=\boxed{72}.</cmath>
  
 
== See also ==
 
== See also ==

Revision as of 21:47, 15 September 2023

Problem

For certain ordered pairs $(a,b)\,$ of real numbers, the system of equations

$ax+by=1\,$
$x^2+y^2=50\,$

has at least one solution, and each solution is an ordered pair $(x,y)\,$ of integers. How many such ordered pairs $(a,b)\,$ are there?

Solution

bro whoever came up with this problem must've been drunk tbh, there are literally infinite solutions... The vertices are chosen from the points in a $4\times4$ grid as shown below. Since there are $4\cdot 4=16$ points in the grid and triangles consist of $3$ distinct vertices, we get an initial count of\[\binom{16}{3}=560.\] [asy] size(100); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1));

dot((1,2)); dot((2,2)); dot((3,2)); dot((4,2));

dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3));

dot((1,4)); dot((2,4)); dot((3,4)); dot((4,4));

[/asy]

However, degenerate triangles (those with $0$ area) are precisely those formed by $3$ collinear points, so we have to remove these from the count. In each column or row, there are $\binom{4}{3}=4$ sets of $3$ collinear points. There are $8$ total columns and rows, so this gives us $4\cdot 8=32$ invalid triangles.

There are also $4$ such sets of $3$ collinear points in each of the two main diagonals and $1$ for each of the $4$ shorter diagonals. This means that we counted a total of\[32+8+4=44\]invalid triangles in our initial computation of $560$ triangles, so the answer is $560-44=\boxed{516}$.

Source: 1993 AHSME #28 Hint(s): There are probably more degenerate triangles than you think. Your Response(s): $516 Problem 8 – Give Up – Score: 0 / 7 (25910) Problem: Report Error For certain ordered pairs$(a,b)$of real numbers, the system of equations\begin{eqnarray*} && ax+by =1\\ &&x^2+y^2=50\end{eqnarray*}has at least one solution, and each solution is an ordered pair$ (Error compiling LaTeX. Unknown error_msg)(x,y)$of integers. How many such ordered pairs$(a,b)$are there? Solution: The equation$x^2+y^2=50$is that of a circle of radius$\sqrt{50}$, centered at the origin. By testing integers until the left side becomes too big, we see that the lattice points on this circle are$(\pm1,\pm7)$,$(\pm5,\pm5)$, and$(\pm7,\pm1)$where the signs are all independent of each other, for a total of$3\cdot 2\cdot 2=12$lattice points. They are indicated by the blue dots below.

[asy] size(150); draw(circle((0,0),sqrt(50)));

draw((1,7)--(-1,-7),red); draw((7,1)--(5,-5), green);

dot((0,0));

dot((1,7),blue); dot((1,-7),blue); dot((-1,7),blue); dot((-1,-7),blue);

dot((5,5),blue); dot((5,-5),blue); dot((-5,5),blue); dot((-5,-5),blue);

dot((7,1),blue); dot((7,-1),blue); dot((-7,1),blue); dot((-7,-1),blue); [/asy]

Since$ (Error compiling LaTeX. Unknown error_msg)(x,y)=(0,0)$yields$a\cdot 0+b\cdot 0=0 \neq 1$, we know that$ax+by=1$is the equation of a line that does not pass through the origin. So, we are looking for the number of lines which pass through at least one of the$12$lattice points on the circle, but do not pass through the origin or through any non-lattice point on the circle. An example is the green line above. It is straightforward to show that a line passes through the origin precisely when there exist two opposite points$(p,q)$and$(-p,-q)$through which it passes. And example is the red line above.

There are$ (Error compiling LaTeX. Unknown error_msg)\binom{12}{2}=66$ways to pick two distinct lattice points, and subsequently$66$distinct lines which pass through two distinct lattice points on the circle. Then we subtract the lines which pass through the origin by noting that the lattice points on the circle can be grouped into opposite pairs$(p,q)$and$(-p,-q)$, for a total of$\frac{12}{2}=6$lines. Finally, we add the$12$ unique tangent lines to the circle at each of the lattice points.

Therefore, our final count of distinct lines which pass through one or two of the lattice points on the circle, but do not pass through the origin, is\[66-6+12=\boxed{72}.\]

See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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