1981 USAMO Problems/Problem 3
Problem
Show that for any triangle, .
When does the equality hold?
Solution
Given three angles that add to , one can construct a triangle from them. However, its angles must all be nonnegative; thus the constraints on the angles of a triangle are
and
.
In fact, at this point, we only care about and
. Let us call them
and
. We have:
and we must prove that
Without loss of generality, assume . It follows that
and
(otherwise,
would be strictly greater than
or strictly less than
, respectively).
Since is nonnegative over the interval
and
for all real
, we can immediately use the
result to show that
This proves the lower bound; equality occurs when and
, and this is reachable only when
and
, which translates into
and
.
Now for the upper bound. It is true that is non-positive over the interval
. Also,
(by a simple squaring argument). Then
. If
, then
, and then
Therefore, we need to handle the case where
.
, so we can subtract
from
due to periodicity. Then the constraints become
and
.
Let us suppose that is the selection of
given the above constraints with the largest value of
. We shall see that, given these constraints, this value increases when the distance between two of these variables decreases, and it is maximized when
.
If
is well-behaved and
is strictly decreasing over an open interval
, then, for any
selected from that interval with
, then
It is true that, for any well-behaved function
,
, and likewise,
. Without loss of generality, suppose
; let
. Since
is given to be strictly decreasing over the relevant interval, we have
for all
; therefore
Likewise,
for all
; therefore
Therefore, adding our inequalities together,
But
. In that case, the
and
terms cancel, and we get our desired result:
.
Since is well-behaved, and
, which is strictly decreasing over the open interval
, then we can apply our lemma. Suppose for the sake of contradiction that
are not all the same value, and this produces a maximum. Then, without loss of generality, suppose
. Let
. Then the selection
also satisfies the constraints
and
. Furthermore, by the above lemma,
, so this new selection
has a larger corresponding value
than the selection
. This contradicts our original assumption that
was chosen to have the maximum. Therefore,
.
Then, since , we conclude
. Then
So
is indeed the maximum value of
, proving the bound. Translating back into the previous problem, the maximum occurs when
and
; translating into the original problem, we have proven the upper bound, and equality with the upper bound occurs when
and
.
~ by eevee9046
See Also
1981 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.