Difference between revisions of "2003 AMC 12B Problems/Problem 24"
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− | From the graph, it is clear that <math>f(x)</math> and <math>2x+y=2003</math> have one intersection point if and only if they intersect at <math>x=a</math>. Since the line where <math>a<x<b</math> has slope <math>-1</math>, the positive difference in <math>y</math>-coordinates from <math>x=a</math> to <math>x=b</math> must be <math>b-a</math>. Together with the fact that <math>(b,c-a)</math> is on <math>f(x)</math>, we see that <math>P=(a,c-a+b-a)</math>. Since this point is on <math>x=a</math>, the only intersection point with <math>2x+y=2003</math>, we have <math>2 \cdot a+(b+c-2a)=2003 \implies | + | From the graph, it is clear that <math>f(x)</math> and <math>2x+y=2003</math> have one intersection point if and only if they intersect at <math>x=a</math>. Since the line where <math>a<x<b</math> has slope <math>-1</math>, the positive difference in <math>y</math>-coordinates from <math>x=a</math> to <math>x=b</math> must be <math>b-a</math>. Together with the fact that <math>(b,c-a)</math> is on <math>f(x)</math>, we see that <math>P=(a,c-a+b-a)</math>. Since this point is on <math>x=a</math>, the only intersection point with <math>2x+y=2003</math>, we have <math>2 \cdot a+(b+c-2a)=2003 \implies b+c=2003</math>. As <math>c>b</math>, the smallest possible value of <math>c</math> occurs when <math>b=1001</math> and <math>c=1002</math>. This is indeed a solution as <math>a=1000</math> puts <math>P</math> on <math>y=2003-2x</math>, and thus the answer is <math>\boxed{\mathrm{(C)}\ 1002}</math>. |
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+ | This indeed works for the two right segments of slope <math>1</math> and <math>3</math>. We already know that the minimum is achieved between slopes <math>-3</math> and <math>-1</math> with <math>b+c=2003</math>: | ||
+ | <cmath>2003-2x=-a-b+c+x\longrightarrow 3x\ne a+b-c+2003 \{b<x<c\}\rightarrow (3b,3c)\ne a+2b\rightarrow b>a\text{ (true)}</cmath> | ||
+ | <cmath>2003-2x=-a-b-c+3x\longrightarrow 5x\ne a+b+c+2003 \{x>c\}\rightarrow (5c,+5\infty)\ne a+2b+2c\rightarrow 3c>a+2b\text{ (true)}</cmath> | ||
+ | Indeed, within the restricted domain of <math>x</math> in each segment, these inequalities prove to be unequal everywhere. So <math>y=2003-2x</math> is strictly below <math>y=|x-a|+|x-b|+|x-c|</math> at these domains. | ||
== Solution 2 == | == Solution 2 == | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 14:23, 2 August 2024
Contents
Problem
Positive integers and
are chosen so that
, and the system of equations
![$2x + y = 2003 \quad$](http://latex.artofproblemsolving.com/7/0/6/706aa31c279b4f4744165bfddf5f8bd451518a57.png)
![$\quad y = |x-a| + |x-b| + |x-c|$](http://latex.artofproblemsolving.com/5/7/c/57c363455cd9e9bf8c2dceadb7298327d953e0bc.png)
has exactly one solution. What is the minimum value of ?
Solution 1
Consider the graph of .
When , the slope is
.
When , the slope is
.
When , the slope is
.
When , the slope is
.
Setting gives
, so
is a point on
. In fact, it is the minimum of
considering the slope of lines to the left and right of
. Thus, graphing this will produce a figure that looks like a cup:
From the graph, it is clear that
and
have one intersection point if and only if they intersect at
. Since the line where
has slope
, the positive difference in
-coordinates from
to
must be
. Together with the fact that
is on
, we see that
. Since this point is on
, the only intersection point with
, we have
. As
, the smallest possible value of
occurs when
and
. This is indeed a solution as
puts
on
, and thus the answer is
.
This indeed works for the two right segments of slope and
. We already know that the minimum is achieved between slopes
and
with
:
Indeed, within the restricted domain of
in each segment, these inequalities prove to be unequal everywhere. So
is strictly below
at these domains.
Solution 2
Step 1: Finding some promising bound
Does the system have a solution where ?
For such a solution we would have , hence
, which solves to
.
If we want to avoid this solution, we need to have
, hence
, hence
.
In other words, if
, there will always be one solution
such that
.
Step 2: Showing one solution
We will now find out whether there is a for which (and some
) the system has only one solution. We already know of one such solution, so we need to make sure that no other solution appears.
Obviously, there are three more theoretically possible solutions: one in
, one in
, and one in
.
The first case solves to
, the second to
, and the third to
.
We need to make sure that the following three conditions hold:
.
Let and
. We then have:
Hence for ,
and any valid
the system has exactly one solution
.
Step 3: Proving the optimality of our solution
We will now show that for the system always has a solution such that
. This will mean that the system has at least two solutions, and thus the solution with
is optimal.
- As we are looking for a
, we have
, hence
. To make sure that the value falls outside
, we need to make it larger than
, thus
, or equivalently
.
- The condition we just derived,
, can be rewritten as
, then as
, which becomes
. Thus to make sure that the second value falls outside
, we need to make it larger than
. The inequality
simplifies to
.
- To avoid the last solution, we must have
, which simplifies to
.
The last two inequalities contradict each other, thus there are no that would satisfy both of them.
Conclusion
We just showed that whenever , the system has at least two different solutions: one with
and one with
.
We also showed that for there are some
for which the system has exactly one solution.
Hence the optimal value of is
.
See also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.