Difference between revisions of "2005 AIME I Problems/Problem 10"
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=== Solution 3 === | === Solution 3 === | ||
− | Again, the [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is at <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>. | + | Again, the [[midpoint]] <math>M</math> of [[line segment]] <math>\overline{BC}</math> is at <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} = \fbox{047}</math>. |
== See also == | == See also == |
Revision as of 14:52, 28 August 2016
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
![[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)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.