1965 IMO Problems/Problem 1
Problem
Determine all values in the interval
which satisfy the inequality
Solution 1
We shall deal with the left side of the inequality first () and the right side after that.
It is clear that the left inequality is true when is non-positive, and that is when
is in the interval
. We shall now consider when
is positive. We can square the given inequality, and the resulting inequality will be true whenever the original left inequality is true.
. This inequality is equivalent to
. I shall now divide this problem into cases.
Case 1: is non-negative. This means that
is in one of the intervals
or
. We must find all
in these two intervals such that
. This inequality is equivalent to
, which is only true when
or
.
Case 2: is negative. This means that
is in one of the interavals
or
. We must find all
in these two intervals such that
, which is equivalent to
, which is true for all
in these intervals.
Therefore the left inequality is true when is in the union of the intervals
,
, and
, which is the interval
. We shall now deal with the right inequality.
As above, we can square it and have it be true whenever the original right inequality is true, so we do that. , which is always true. Therefore the original right inequality is always satisfied, and all
such that the original inequality is satisfied are in the interval
.
Solution 2
Manipulate the inequality so that it becomes:
Inside the absolute value, we identify the half-angle formulas. However, since we do not know the sign of the resultant expression, we have to use absolute value signs since the principal square root is always positive; then the inequality becomes
which, considering absolute value, is the same as
Since both inner absolute values range between
and
, their positive difference also ranges from
to
, so the right inequality is always true. Thus we take a look at the left inequality. For
, the left-hand side is never positive; however, the right-hand side is always nonnegative due to absolute value, so the inequality holds for these
-values. As a result, we consider the remaining two sections of the interval.
For and
, when considering the
, inside the absolute value, cosine is positive but sine is negative; thus our inequality becomes
, then due to the interval,
and
are both never negative; thus,
If , then due to the interval,
is negative and
is positive; thus,
For , inside the absolute value, when considering the
, both cosine and sine are positive; thus our inequality becomes
For , inside the absolute value, when considering the
, both cosine and sine are negative; thus our inequality becomes
, so the inequality holds.
Combining our findings, we find that the solutions are .
~eevee9406
1965 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
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