Difference between revisions of "2001 AIME I Problems/Problem 10"
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There are <math>12</math> EEEs (the first position denotes the parity of <math>x,</math> the second <math>y,</math> and the third <math>z.</math>), <math>8</math> EEOs, <math>12</math> EOEs, <math>6</math> OEEs, <math>8</math> EOOs, <math>4</math> OEOs, <math>6</math> OOEs, and <math>4</math> OOOs. Doing a sanity check, <math>12 + 8 + 12 + 6 + 8 + 4 + 6 + 4 = 60,</math> which is the total number of points. | There are <math>12</math> EEEs (the first position denotes the parity of <math>x,</math> the second <math>y,</math> and the third <math>z.</math>), <math>8</math> EEOs, <math>12</math> EOEs, <math>6</math> OEEs, <math>8</math> EOOs, <math>4</math> OEOs, <math>6</math> OOEs, and <math>4</math> OOOs. Doing a sanity check, <math>12 + 8 + 12 + 6 + 8 + 4 + 6 + 4 = 60,</math> which is the total number of points. | ||
− | Now, we can see that there are <math>12 \cdot 11</math> to choose two EEEs (respective to order), <math>8 \cdot 7</math> ways to choose two EEOs, and so on. Therefore, we get <cmath>12 | + | Now, we can see that there are <math>12 \cdot 11</math> to choose two EEEs (respective to order), <math>8 \cdot 7</math> ways to choose two EEOs, and so on. Therefore, we get <cmath>12\cdot11 + 8\cdot7 + 12\cdot11 + 6\cdot5 + 8\cdot7 + 4\cdot3 + 6\cdot5 + 4\cdot3 = 460</cmath> ways to choose two points where order matters. There are <math>60 \cdot 59</math> total ways to do this, so we get a final answer of <cmath>\dfrac{460}{60 \cdot 59} = \dfrac{23}{3 \cdot 59} = \dfrac{23}{177},</cmath>for our answer of <math>23 + 177 = \boxed{200}.</math> |
Solution by Ilikeapos | Solution by Ilikeapos |
Revision as of 22:58, 19 June 2019
Contents
[hide]Problem
Let be the set of points whose coordinates and are integers that satisfy and Two distinct points are randomly chosen from The probability that the midpoint of the segment they determine also belongs to is where and are relatively prime positive integers. Find
Solutions
Solution 1
The distance between the , , and coordinates must be even so that the midpoint can have integer coordinates. Therefore,
- For , we have the possibilities , , , , and , possibilities.
- For , we have the possibilities , , , , , , , and , possibilities.
- For , we have the possibilities , , , , , , , , , , , , and , possibilities.
However, we have cases where we have simply taken the same point twice, so we subtract those. Therefore, our answer is .
Solution 2
There are points in total. We group the points by parity of each individual coordinate -- that is, if is even or odd, is even or odd, and is even or odd. Note that to have something that works, the two points must have this same type of classification (otherwise, if one doesn't match, the resulting sum for the coordinates will be odd at that particular spot).
There are EEEs (the first position denotes the parity of the second and the third ), EEOs, EOEs, OEEs, EOOs, OEOs, OOEs, and OOOs. Doing a sanity check, which is the total number of points.
Now, we can see that there are to choose two EEEs (respective to order), ways to choose two EEOs, and so on. Therefore, we get ways to choose two points where order matters. There are total ways to do this, so we get a final answer of for our answer of
Solution by Ilikeapos
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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