2018 USAJMO Problems/Problem 4
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 |