Difference between revisions of "2005 AIME I Problems/Problem 10"
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=== Solution 3 === | === Solution 3 === | ||
− | Again, the [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is at <math>\left(\frac{35}{2}, \frac{39}{2}\right)</math>. Let <math>A'</math> be the point <math>(17, 22)</math>, which lies along the line through <math>M</math> of slope <math>-5</math>. The area of triangle <math>A'BC</math> can be computed in a number of ways (one possibility: extend <math>A'B</math> until it hits the line <math>y = 19</math>, and subtract one triangle from another), and each such calculation gives an area of 14. This is <math>\frac{1}{5}</math> of our needed area, so we simply need the point <math>A</math> to be 5 times as far from <math>M</math> as <math>A'</math> is. Thus <math>A = \left(\frac{35}{2}, \frac{39}{2}\right) \pm 5\left(-\frac{1}{2}, \frac{5}{2}\right)</math>, and the sum of coordinates will be larger if we take the positive value, so <math>A = \left(\frac{35}{2} - \frac{5}2, \frac{39}{2} + \frac{25}{2}\right)</math> and the answer is <math>\frac{35}{2} - \frac{5}2 + \frac{39}{2} + \frac{25}{2} = 047</math>. | + | Again, the [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is at <math>\left(\frac{35}{2}, \frac{39}{2}\right)</math>. Let <math>A'</math> be the point <math>(17, 22)</math>, which lies along the line through <math>M</math> of slope <math>-5</math>. The area of triangle <math>A'BC</math> can be computed in a number of ways (one possibility: extend <math>A'B</math> until it hits the line <math>y = 19</math>, and subtract one triangle from another), and each such calculation gives an area of 14. This is <math>\frac{1}{5}</math> of our needed area, so we simply need the point <math>A</math> to be 5 times as far from <math>M</math> as <math>A'</math> is. Thus <math>A = \left(\frac{35}{2}, \frac{39}{2}\right) \pm 5\left(-\frac{1}{2}, \frac{5}{2}\right)</math>, and the sum of coordinates will be larger if we take the positive value, so <math>A = \left(\frac{35}{2} - \frac{5}2, \frac{39}{2} + \frac{25}{2}\right)</math> and the answer is <math>\frac{35}{2} - \frac{5}2 + \frac{39}{2} + \frac{25}{2} = \fbox{047}</math>. |
== See also == | == See also == |
Revision as of 14:52, 28 August 2016
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 .
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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