Problem:
The graph of f(x)=β£βxββ£ββ£β1βxββ£ is symmetric about which of the following? (Here βxβ is the greatest integer not exceeding x.)
Answer Choices:
A. the y-axis
B. the line x=1
C. the origin
D. the point (21β,0)
E. the point (1,0)
Solution:
Let x=n+r, where n is an integer and 0β€r<1. First suppose that r=0. Then f(x)=β£nβ£ββ£1βnβ£, which is equal to nβ(nβ1)=1 if nβ₯1 and (βn)β(1βn)=β1 if nβ€0. Hence the graph of f cannot be symmetric about the y-axis, the line x=1, the origin, or the point (1,0). However, note that in all cases f(1βn)=βf(n), so the graph of f is symmetric about the point (21β,0) when x is restricted to integer values. Finally, when r>0, observe that βxβ=n and β1βxβ=β1βnβrβ=βn, so f(x)=β£nβ£ββ£βnβ£=0, and again f(1βx)=βf(x). Hence the graph of f is symmetric about the point (D)(21β,0)β.
OR
Note that f(1βx)=β£β1βxββ£ββ£β1β(1βx)ββ£=βf(x). Thus a point (x,y) is on the graph of f if and only if (1βx,βy) is on the graph. The midpoint of the segment joining these two points is the point (D)(21β,0)β; that is, the graph is symmetric about this point.
To show that f(x) does not have any of the properties described in the four other alternatives, it is useful to note that
f(x)=β©βͺβͺβ¨βͺβͺβ§β10β1β if n is an integer, nβ₯1 if n is not an integer if n is an integer, nβ€0β
The graph of f is not symmetric about the y-axis because f(1)=1 and f(β1)=β1ξ =1. The graph of f is not symmetric about the line x=1 because f(3)=1 and f(β1)=β1ξ =1. The graph of f is not symmetric with respect to the origin because f(0)=1 and the reflected point (0,β1) is not in the graph of f. The graph of f is not symmetric about the point (1,0) because f(1)=1 and the reflected point (1,β1) is not in the graph of f.
The problems on this page are the property of the MAA's American Mathematics Competitions