Difference between revisions of "2005 AIME I Problems/Problem 10"
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[[Triangle]] <math> ABC </math> lies in the [[Cartesian Plane]] and has an [[area]] of 70. 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> | [[Triangle]] <math> ABC </math> lies in the [[Cartesian Plane]] and has an [[area]] of 70. 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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== Solution == | == Solution == | ||
| + | === Solution 1 === | ||
| + | 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} \Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q = -197 - p + 11q</math>. Thus, <math>q = \frac{1}{11}p - \frac{337}{11}</math>. | ||
| + | The equation of the median can be found by <math>-5 = \frac{q - \frac{12 + 23}{2}}{p - \frac{19 + 20}{2}}</math>, which yields that <math>q = -5p + 107</math>. Setting the two equations equal to each other, 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>. | ||
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| + | === Solution 2 === | ||
The [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is <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>. | The [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is <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>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=2005|n=I|num-b=9|num-a=11}} | |
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 15:19, 4 March 2007
Problem
Triangle 
lies in the Cartesian Plane and has an area of 70. 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
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, 
.
The equation of the median can be found by 
, which yields that 
. Setting the two equations equal to each other, we get that 
, so 
. Solving produces that 
. Substituting backwards yields that 
; the solution is 
.
Solution 2
The midpoint 
of line segment 
is 
. 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 | ||
