Difference between revisions of "2005 AIME I Problems/Problem 10"
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== Solution == | == Solution == | ||
− | + | <center><asy>defaultpen(fontsize(8)); | |
− | 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{35}{2}}{p - \frac{39}{2}}</math>. Cross multiply and simplify to yield that <math> | + | size(170); |
+ | pair A=(15,32), B=(12,19), C=(23,20), M=B/2+C/2, P=(17,22); | ||
+ | draw(A--B--C--A);draw(A--M);draw(B--P--C); | ||
+ | label("A (p,q)",A,(1,1));label("B (12,19)",B,(-1,-1));label("C (23,20)",C,(1,-1));label("M",M,(0.2,-1)); | ||
+ | label("(17,22)",P,(1,1)); | ||
+ | dot(A^^B^^C^^M^^P);</asy></center> | ||
+ | |||
+ | 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{35}{2}}{p - \frac{39}{2}}</math>. Cross multiply and simplify to yield that <math>-5p + \frac{39 \cdot 5}{2} = q - \frac{35}{2}</math>, so <math>q = -5p + 107</math>. | ||
=== Solution 1 === | === Solution 1 === | ||
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=== Solution 2 === | === Solution 2 === | ||
− | Using the equation of the median, we can write the [[coordinate]]s of <math>A</math> as <math>(p,\ -5p + 107)</math>. The equation of <math>\overline{BC}</math> is <math>\frac{20 - 19}{23 - 12} = \frac{y - 19}{x - 12}</math>, so <math>x - 12 = 11y - 209</math>. In [[general form]], the line is <math>x - 11y + 197 = 0</math>. Use the equation for the distance between a line and point to find the distance between <math>A</math> and <math>BC</math> (which is the height of <math>\triangle ABC</math>): <math>\frac{|1(p) - 11(-5p + 107) + 197|}{1^2 + 11^2} = \frac{|56p - 980|}{\sqrt{122}}</math>. Now we need the length of <math>BC</math>, which is <math>\sqrt{(23 - 12)^2 + (20 - 19)^2} = \sqrt{122}</math>. The area of <math>\triangle ABC</math> is <math>70 = \frac{1}{2}bh = \frac{1}{2}(\frac{|56p - 980|}{\sqrt{122}}) \cdot \sqrt{122}</math>. Thus, <math>|28p - 490| = 70</math>, and <math>p = 15,\ 20</math>. We are looking for <math>p + q = -4p + 107 = 47,\ 27</math>. The maximum possible value of <math>p + q = 47</math>. | + | Using the equation of the median, we can write the [[coordinate]]s of <math>A</math> as <math>(p,\ -5p + 107)</math>. The equation of <math>\overline{BC}</math> is <math>\frac{20 - 19}{23 - 12} = \frac{y - 19}{x - 12}</math>, so <math>x - 12 = 11y - 209</math>. In [[general form]], the line is <math>x - 11y + 197 = 0</math>. Use the equation for the distance between a line and point to find the distance between <math>A</math> and <math>BC</math> (which is the height of <math>\triangle ABC</math>): <math>\frac{|1(p) - 11(-5p + 107) + 197|}{1^2 + 11^2} = \frac{|56p - 980|}{\sqrt{122}}</math>. Now we need the length of <math>BC</math>, which is <math>\sqrt{(23 - 12)^2 + (20 - 19)^2} = \sqrt{122}</math>. The area of <math>\triangle ABC</math> is <math>70 = \frac{1}{2}bh = \frac{1}{2}\left(\frac{|56p - 980|}{\sqrt{122}}\right) \cdot \sqrt{122}</math>. Thus, <math>|28p - 490| = 70</math>, and <math>p = 15,\ 20</math>. We are looking for <math>p + q = -4p + 107 = 47,\ 27</math>. The maximum possible value of <math>p + q = 47</math>. |
=== Solution 3 === | === Solution 3 === |
Revision as of 17:59, 19 November 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
![[asy]defaultpen(fontsize(8)); size(170); pair A=(15,32), B=(12,19), C=(23,20), M=B/2+C/2, P=(17,22); draw(A--B--C--A);draw(A--M);draw(B--P--C); label("A (p,q)",A,(1,1));label("B (12,19)",B,(-1,-1));label("C (23,20)",C,(1,-1));label("M",M,(0.2,-1)); label("(17,22)",P,(1,1)); dot(A^^B^^C^^M^^P);[/asy]](http://latex.artofproblemsolving.com/3/e/0/3e0c65881bed6213e0789bf735b240b830e07554.png)
The midpoint of line segment
is
. The equation of the median can be found by
. Cross multiply and simplify to yield that
, so
.
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,
.
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, 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
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 |