美国数学竞赛答案

Solution

Answer: (C)

Lemma) if , then

The tedious algebra is left to the reader. (it is not bad at all)

Well, let us consider the cases where each of those step is definite (is never evaluate).

So, we have ,

--- (exception -> case 2)

--- (exception -> case 3)

--- (exception -> case 4)

--- (exception -> case 5)

If it is not any of the above 5 cases, then

if (--- exception -> case 6), then , ,

Hence, it is possible maximum of and minimum is 1.

2 possible combination of are and . Verification is left upto the reader. Right now, (C) is the most possible answer out of those 5.

Case 2) , then

Case 3) = 1, which is in the range.

Case 5) , hence

Case 6) Since , ,

Case 4) , this is quite an annoying special case. In this case, ,

is not define.

In this case,

and

Hence, and . Once,

you work out this system, you will get no solution with .

Thus, answer is (C).

SOLUTION 2:

After a bit of tedius algebra (that isn't too bad, but a little lengthy) we obtain

where ,

, , and

. In order for , we must have , ,

and . The first implies or , the second implies ,

, or , and the third implies or .

Since , in order to satisfy all 3 conditions we must have either or

. In the first case . For the latter case note that

so that and hence

. On the other hand so that

. Thus . Hence in any case the maximum

value for is while the minimum is (which can be achieved for instance when

or respectively). Hence the answer is .\\ \

Solution

From , we know that .

From the first inequality, we get . Subtracting

from this gives us , and thus .

Since must be an integer, it follows that .

Similarly, from the second inequality, we get . Again

subtracting from this gives us , or

. It follows from this that .

We now have a system of three equations: , , and

. Solving gives us and from this we find that

Since , we find that

.