Difference between revisions of "2003 AMC 12B Problems/Problem 24"

Revision as of 09:27, 4 July 2013

Problem

Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$ , and the system of equations

$2x + y = 2003 \quad$ and $\quad y = |x-a| + |x-b| + |x-c|$

has exactly one solution. What is the minimum value of $c$ ?

$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$

Solution

Step 1: Finding some promising bound

Does the system have a solution where $x\leq a$ ?

For such a solution we would have $y=(a-x)+(b-x)+(c-x)$ , hence $2x+(a+b+c-3x)=2003$ , which solves to $x=a+b+c-2003$ . If we want to avoid this solution, we need to have $a+b+c-2003>a$ , hence $b+c>2003$ , hence $c\geq 1003$ . In other words, if $c<1003$ , there will always be one solution $(x,y)$ such that $x\leq a$ .

Step 2: Showing one solution

We will now find out whether there is a $c<1003$ for which (and some $a,b$ ) 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 $x$ in $\left(a,b\right]$ , one in $\left(b,c\right]$ , and one in $\left(c,\infty\right)$ . The first case solves to $x=2003+a-b-c$ , the second to $3x=2003+a+b-c$ , and the third to $5x=2003+a+b+c$ . We need to make sure that the following three conditions hold:

$2003+a-b-c\not\in\left(a,b\right]$
$\frac{2003+a+b-c}3\not\in \left(b,c\right]$
$\frac{2003+a+b+c}5\not\in\left(c,\infty\right)$ .

Let $c=1002$ and $b=1001$ . We then have:

$2003+a-b-c=a$
$\frac{2003+a+b-c}3 = \frac{2002+a}3 \leq \frac{2002+1000}3 < 1001 = b$
$\frac{2003+a+b+c}5 = \frac{4006+a}5 \leq \frac{4006+1006}5 < 1002 = a$

Hence for $c=1002$ , $b=1001$ and any valid $a$ the system has exactly one solution $(x,y)=(a,2003-2a)$ .

Step 3: Proving the optimality of our solution

We will now show that for $c<1002$ the system always has a solution such that $x>a$ . This will mean that the system has at least two solutions, and thus the solution with $c=1002$ is optimal.

As we are looking for a $c<1002$ , we have $b+c\leq 2001$ , hence $2003+a-b-c > a$ . To make sure that the value falls outside $\left(a,b\right]$ , we need to make it larger than $b$ , thus $2003+a-b-c > b$ , or equivalently $2003+a > 2b+c$ .
The condition we just derived, $2003+a > 2b+c$ , can be rewritten as $2003+a+b > 3b+c$ , then as $2003+a+b-c > 3b$ , which becomes $\frac{2003+a+b-c}3 > b$ . Thus to make sure that the second value falls outside $\left(b,c\right]$ , we need to make it larger than $c$ . The inequality $\frac{2003+a+b-c}3 > c$ simplifies to $2003+a+b > 4c$ .
To avoid the last solution, we must have $\frac{2003+a+b+c}5\leq c$ , which simplifies to $2003+a+b \leq 4c$ .

The last two inequalities contradict each other, thus there are no $a,b,c$ that would satisfy both of them.

Conclusion

We just showed that whenever $c<1002$ , the system has at least two different solutions: one with $x\leq a$ and one with $x>a$ .

We also showed that for $c=1002$ there are some $a,b$ for which the system has exactly one solution.

Hence the optimal value of $c$ is $\boxed{\mathrm{(C)}\ 1002}$ .

@@ Line 57: / Line 57: @@
 == See also ==
 {{AMC12 box|year=2003|ab=B|num-b=23|num-a=25}}
+{{MAA Notice}}

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

Page

Toolbox

Search