Difference between revisions of "2005 AIME I Problems/Problem 10"
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=== Solution 4 === | === Solution 4 === | ||
− | Plug points into the shoelace theorem. This will provide you with the equation <math>|-p - 197 + 11q| = 140</math>. The find the midpoint of the line <math>BC</math> which is <math>(17.5,19.5)</math>. Now using this post and the given slope of the median, <math>-5</math>, using basic algebra we can find the equation of the median which is <math>q = -5p + 107</math>. Now that we have been given <math>q</math> in terms of <math>p</math> plug this equation back into <math>|-p - 197 + 11q| = 140</math>. The result is the equation <math>|980 - 56p| = 140</math>. Sove this equation for two possible answers <math>p = 15, 20</math>. Plugging into <math>q = -5p + 107</math> these inputs produce <math>q</math> values <math>32</math> and <math>7</math>. Obviously <math>15 + 32</math> is the greater sum so the answer is <math>47</math> and we are done. | + | Plug points into the [[shoelace theorem]]. This will provide you with the equation <math>|-p - 197 + 11q| = 140</math>. The find the midpoint of the line <math>BC</math> which is <math>(17.5,19.5)</math>. Now using this post and the given slope of the median, <math>-5</math>, using basic algebra we can find the equation of the median which is <math>q = -5p + 107</math>. Now that we have been given <math>q</math> in terms of <math>p</math> plug this equation back into <math>|-p - 197 + 11q| = 140</math>. The result is the equation <math>|980 - 56p| = 140</math>. Sove this equation for two possible answers <math>p = 15, 20</math>. Plugging into <math>q = -5p + 107</math> these inputs produce <math>q</math> values <math>32</math> and <math>7</math>. Obviously <math>15 + 32</math> is the greater sum so the answer is <math>47</math> and we are done. |
== See also == | == See also == |
Revision as of 14:44, 26 December 2017
Problem
Triangle lies in the cartesian plane and has an area of . The coordinates of and are and respectively, and the coordinates of are The line containing the median to side has slope Find the largest possible value of
Solution 1
The midpoint of line segment is . The equation of the median can be found by . Cross multiply and simplify to yield that , so .
Use determinants to find that the area of is (note that there is a missing absolute value; we will assume that the other solution for the triangle will give a smaller value of , which is provable by following these steps over again). We can calculate this determinant to become . Thus, .
Setting this equation equal to the equation of the median, we get that , so . Solving produces that . Substituting backwards yields that ; the solution is .
Solution 2
Using the equation of the median from above, we can write the coordinates of as . The equation of is , so . In general form, the line is . Use the equation for the distance between a line and point to find the distance between and (which is the height of ): . Now we need the length of , which is . The area of is . Thus, , and . We are looking for . The maximum possible value of .
Solution 3
Again, the midpoint of line segment is at . Let be the point , which lies along the line through of slope . The area of triangle can be computed in a number of ways (one possibility: extend until it hits the line , and subtract one triangle from another), and each such calculation gives an area of 14. This is of our needed area, so we simply need the point to be 5 times as far from as is. Thus , and the sum of coordinates will be larger if we take the positive value, so and the answer is .
Solution 4
Plug points into the shoelace theorem. This will provide you with the equation . The find the midpoint of the line which is . Now using this post and the given slope of the median, , using basic algebra we can find the equation of the median which is . Now that we have been given in terms of plug this equation back into . The result is the equation . Sove this equation for two possible answers . Plugging into these inputs produce values and . Obviously is the greater sum so the answer is and we are done.
See also
2005 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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