Difference between revisions of "1998 AIME Problems/Problem 2"
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=== Solution 2 === | === Solution 2 === | ||
| − | First, note that all pairs of the form <math> | + | First, note that all pairs of the form <math>(a,a)</math>, <math>1\le a\le30</math> work. |
| − | Now, considered the ordered pairs with <math>x < y</math>, so that <math>x < 2y</math> is automatically satisfied. Since <math>x < y\le 2x</math>, there are <math>2x - x = x | + | Now, considered the ordered pairs with <math>x < y</math>, so that <math>x < 2y</math> is automatically satisfied. Since <math>x < y\le 2x</math>, there are <math>2x - x = x</math> possible values of <math>y</math>. Hence, given <math>x</math>, there are <math>x</math> values of possible <math>y</math> for which <math>x < y</math> and the above conditions are satisfied. But <math>2y\le60</math>, so this only works for <math>x\le15</math>. Thus, there are |
| − | <math>\sum_{i=1}^{15} i=\frac{( | + | <math>\sum_{i=1}^{15} i=\frac{(15)(16)}{2}</math> |
| − | ordered pairs. For <math>x > 15</math>, <math>y</math> must follow <math>x < y\le 30 | + | ordered pairs. For <math>x > 15</math>, <math>y</math> must follow <math>x < y\le 30</math>. Hence, there are <math>30 - x</math> possibilities for <math>y</math>, and there are |
<math>\sum_{i=16}^{30}(30-i)=\sum_{i=0}^{14}i=\frac{(14)(15)}{2}</math> | <math>\sum_{i=16}^{30}(30-i)=\sum_{i=0}^{14}i=\frac{(14)(15)}{2}</math> | ||
| Line 29: | Line 29: | ||
ordered pairs. | ordered pairs. | ||
| − | By symmetry, there are also <math> | + | By symmetry, there are also <math>\frac {(15)(16)}{2} + \frac {(14)(15)}{2}</math> ordered pairs with <math>x > y</math> and the above criteria satisfied. |
Hence, the total is | Hence, the total is | ||
Revision as of 00:00, 6 March 2011
Problem
Find the number of ordered pairs 
of positive integers that satisfy 
and 
.
Solution
Solution 1
Pick's theorem states that:
The conditions give us four inequalities: 
, 
, 
, 
. These create a quadrilateral, whose area is 
of the 30 by 30 square it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of the area of the 30 by 30 square.
So 
. 
we can calculate by just counting. Ignoring the vertices, the top and right sides have 14 lattice points, and the two diagonals each have 14 lattice points (for the top diagonal, every value of 
corresponds with an integer value of 
as 
and vice versa for the bottom, so and there are 14 values for x not counting vertices). Adding the four vertices, there are 60 points on the borders.
Since the inequalities also include the equals case, we include the boundaries, which gives us 
ordered pairs. However, the question asks us for positive integers, so 
doesn't count; hence, the answer is 
.
Solution 2
First, note that all pairs of the form 
, 
work.
Now, considered the ordered pairs with 
, so that 
is automatically satisfied. Since 
, there are 
possible values of . Hence, given , there are values of possible for which and the above conditions are satisfied. But 
, so this only works for 
. Thus, there are
ordered pairs. For 
, must follow 
. Hence, there are 
possibilities for , and there are
ordered pairs.
By symmetry, there are also 
ordered pairs with 
and the above criteria satisfied.
Hence, the total is
See also
| 1998 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
