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> |
− | + | <center><asy>defaultpen(fontsize(8)); | |
− | == Solution == | + | size(170); |
− | 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{ | + | 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> | ||
+ | |||
+ | == Solution 1 == | ||
+ | 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{39}{2}}{p - \frac{35}{2}}</math>. Cross multiply and simplify to yield that <math>-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}</math>, so <math>q = -5p + 107</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) (alternatively, we could use the [[Shoelace Theorem]]). 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 = \boxed{047}</math>. | |
− | |||
− | + | == Solution 2 == | |
+ | Using the equation of the median from above, 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 = \fbox{047}</math>. | ||
− | + | == 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} = \fbox{047}</math>. | |
− | + | == Solution 4 == | |
− | + | Plug points into the [[Shoelace Theorem]]. This will provide you with the equation <math>|-p - 197 + 11q| = 140</math>. The find the midpoint of the line <math>BC</math> which is <math>(17.5,19.5)</math>. Now using this post and the given slope of the median, <math>-5</math>, using basic algebra we can find the equation of the median which is <math>q = -5p + 107</math>. Now that we have been given <math>q</math> in terms of <math>p</math> plug this equation back into <math>|-p - 197 + 11q| = 140</math>. The result is the equation <math>|980 - 56p| = 140</math>. Solve this equation for two possible answers <math>p = 15, 20</math>. Plugging into <math>q = -5p + 107</math> these inputs produce <math>q</math> values <math>32</math> and <math>7</math>. Obviously <math>15 + 32</math> is the greater sum so the answer is <math>47</math> and we are done. | |
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:23, 4 October 2023
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) (alternatively, we could use the Shoelace Theorem). 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
.
Solution 4
Plug points into the Shoelace Theorem. This will provide you with the equation . The find the midpoint of the line
which is
. Now using this post and the given slope of the median,
, using basic algebra we can find the equation of the median which is
. Now that we have been given
in terms of
plug this equation back into
. The result is the equation
. Solve this equation for two possible answers
. Plugging into
these inputs produce
values
and
. Obviously
is the greater sum so the answer is
and we are done.
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 |
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