2005 AIME I Problems/Problem 10
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
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 .
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 .
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 .
Plug points into the shou lace 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 . Obviossly is the greater sum so the answer is and we are done.
|2005 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|