Difference between revisions of "2005 AIME I Problems/Problem 10"
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== Problem == | == Problem == | ||
− | [[Triangle]] <math> ABC </math> lies in the [[ | + | [[Triangle]] <math> ABC </math> lies in the [[cartesian plane]] and has an [[area]] of <math>70</math>. The coordinates of <math> B </math> and <math> C </math> are <math> (12,19) </math> and <math> (23,20), </math> respectively, and the coordinates of <math> A </math> are <math> (p,q). </math> The [[line]] containing the [[median of a triangle | median]] to side <math> BC </math> has [[slope]] <math> -5. </math> Find the largest possible value of <math> p+q. </math> |
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Use [[determinant]]s to find that the [[area]] of <math>\triangle ABC</math> is <math>\frac{1}{2} \begin{vmatrix}p & 12 & 23 \\ q & 19 & 20 \\ 1 & 1 & 1\end{vmatrix} = 70</math> (note that there is a missing [[absolute value]]; we will assume that the other solution for the triangle will give a smaller value of <math>p+q</math>, which is provable by following these steps over again). We can calculate this determinant to become <math>140 = \begin{vmatrix} 12 & 23 \\ 19 & 20 \end{vmatrix} - \begin{vmatrix} p & q \\ 23 & 20 \end{vmatrix} + \begin{vmatrix} p & q \\ 12 & 19 \end{vmatrix}</math> <math>\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q</math> <math>= -197 - p + 11q</math>. Thus, <math>q = \frac{1}{11}p - \frac{337}{11}</math>. | Use [[determinant]]s to find that the [[area]] of <math>\triangle ABC</math> is <math>\frac{1}{2} \begin{vmatrix}p & 12 & 23 \\ q & 19 & 20 \\ 1 & 1 & 1\end{vmatrix} = 70</math> (note that there is a missing [[absolute value]]; we will assume that the other solution for the triangle will give a smaller value of <math>p+q</math>, which is provable by following these steps over again). We can calculate this determinant to become <math>140 = \begin{vmatrix} 12 & 23 \\ 19 & 20 \end{vmatrix} - \begin{vmatrix} p & q \\ 23 & 20 \end{vmatrix} + \begin{vmatrix} p & q \\ 12 & 19 \end{vmatrix}</math> <math>\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q</math> <math>= -197 - p + 11q</math>. Thus, <math>q = \frac{1}{11}p - \frac{337}{11}</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 = 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>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 17:40, 26 April 2008
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
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 |