Problem:
Positive integers a,b, and c are chosen so that a<b<c, and the system of equations
2x+y=2003 and y=∣x−a∣+∣x−b∣+∣x−c∣
has exactly one solution. What is the minimum value of c ?
Answer Choices:
A. 668
B. 669
C. 1002
D. 2003
E. 2004
Solution:
Since the system has exactly one solution, the graphs of the two equations must intersect at exactly one point. If x<a, the equation y=∣x−a∣+∣x−b∣+ ∣x−c∣ is equivalent to y=−3x+(a+b+c). By similar calculations we obtain
y=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧−3x+(a+b+c), if x<a−x+(−a+b+c), if a≤x<bx+(−a−b+c), if b≤x<c3x+(−a−b−c), if c≤x
Thus the graph consists of four lines with slopes −3,−1,1, and 3 , and it has corners at (a,b+c−2a),(b,c−a), and (c,2c−a−b).
On the other hand, the graph of 2x+y=2003 is a line whose slope is -2 . If the graphs intersect at exactly one point, that point must be (a,b+c−2a). Therefore
2003=2a+(b+c−2a)=b+c
Since b<c, the minimum value of c is 1002 .
The problems on this page are the property of the MAA's American Mathematics Competitions