2005 AIME I Problems/Problem 10
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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 
Contents |
Solution
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 • Resources) | ||
| Preceded by Problem 9 | Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
