Difference between revisions of "2005 AIME I Problems/Problem 10"
Hashtagmath (talk | contribs) |
Fuzimiao2013 (talk | contribs) (→Solution 1) |
||
Line 15: | Line 15: | ||
The [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is <math>\left(\frac{35}{2}, \frac{39}{2}\right)</math>. The equation of the median can be found by <math>-5 = \frac{q - \frac{39}{2}}{p - \frac{35}{2}}</math>. Cross multiply and simplify to yield that <math>-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}</math>, so <math>q = -5p + 107</math>. | The [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is <math>\left(\frac{35}{2}, \frac{39}{2}\right)</math>. The equation of the median can be found by <math>-5 = \frac{q - \frac{39}{2}}{p - \frac{35}{2}}</math>. Cross multiply and simplify to yield that <math>-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}</math>, so <math>q = -5p + 107</math>. | ||
− | Use [[determinant]]s to find that the [[area]] of <math>\triangle ABC</math> is <math>\frac{1}{2} | + | Use [[determinant]]s to find that the [[area]] of <math>\triangle ABC</math> is <math>\frac{1}{2} |
Setting this equation equal to the equation of the median, we get that <math>\frac{1}{11}p - \frac{337}{11} = -5p + 107</math>, so <math>\frac{56}{11}p = \frac{107 \cdot 11 + 337}{11}</math>. Solving produces that <math>p = 15</math>. [[Substitution|Substituting]] backwards yields that <math>q = 32</math>; the solution is <math>p + q = \boxed{047}</math>. | Setting this equation equal to the equation of the median, we get that <math>\frac{1}{11}p - \frac{337}{11} = -5p + 107</math>, so <math>\frac{56}{11}p = \frac{107 \cdot 11 + 337}{11}</math>. Solving produces that <math>p = 15</math>. [[Substitution|Substituting]] backwards yields that <math>q = 32</math>; the solution is <math>p + q = \boxed{047}</math>. |
Revision as of 19:08, 2 January 2022
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) (alternatively, we could use the Shoelace Theorem). 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 . Solve 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.