Difference between revisions of "2018 USAJMO Problems/Problem 4"
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Latest revision as of 13:53, 21 April 2018
Problem 4
Triangle is inscribed in a circle of radius
with
, and
is a real number satisfying the equation
, where
. Find all possible values of
.
Solution
Notice that
Thus, if
then the expression above is strictly greater than
for all
meaning that
cannot satisfy the equation
It follows that
Since we have
From this and the above we have
so
This is true for positive values of
if and only if
However, since
is inscribed in a circle of radius
all of its side lengths must be at most the diameter of the circle, so
It follows that
We know that Since
we have
The equation can be rewritten as
since
This has a real solution if and only if the two separate terms have zeroes in common. The zeroes of
are
and
and the zero of
is
Clearly we cannot have
so the only other possibility is
which means that
We have a system of equations: and
Solving this system gives
Each of these gives solutions for
as
and
respectively. Now that we know that any valid value of
must be one of these two, we will verify that both of these values of
are valid.
First, consider a right triangle inscribed in a circle of radius
with side lengths
This generates the polynomial equation
This is satisfied by
Second, consider a right triangle inscribed in a circle of radius
with side lengths
This generates the polynomial equation
This is satisfied by
It follows that the possible values of are
and
Fun fact: these solutions correspond to a -
-
triangle.
(sujaykazi)
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2018 USAJMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |