# 2018 USAJMO Problems/Problem 4

## Problem 4

Triangle $ABC$ is inscribed in a circle of radius $2$ with $\angle ABC \geq 90^\circ$, and $x$ is a real number satisfying the equation $x^4 + ax^3 + bx^2 + cx + 1 = 0$, where $a=BC,b=CA,c=AB$. Find all possible values of $x$.

## Solution

Notice that $$x^4 + ax^3 + bx^2 + cx + 1 = \left(x^2 + \frac{a}{2}x\right)^2 + \left(\frac{c}{2}x + 1\right)^2 + \left(b - \frac{a^2}{4} - \frac{c^2}{4}\right)x^2.$$ Thus, if $b > \frac{a^2}{4} + \frac{c^2}{4},$ then the expression above is strictly greater than $0$ for all $x,$ meaning that $x$ cannot satisfy the equation $x^4 + ax^3 + bx^2 + cx + 1 = 0.$ It follows that $b\le\frac{a^2}{4} + \frac{c^2}{4}.$

Since $\angle ABC\ge 90^{\circ},$ we have $b^2\ge a^2 + c^2.$ From this and the above we have $4b\le a^2 + c^2\le b^2,$ so $4b\le b^2.$ This is true for positive values of $b$ if and only if $b\ge 4.$ However, since $\triangle ABC$ is inscribed in a circle of radius $2,$ all of its side lengths must be at most the diameter of the circle, so $b\le 4.$ It follows that $b=4.$

We know that $4b\le a^2 + c^2\le b^2.$ Since $4b = b^2 = 16,$ we have $4b = a^2 + c^2 = b^2 = 16.$

The equation $x^4 + ax^3 + bx^2 + cx + 1 = 0$ can be rewritten as $\left(x^2 + \frac{a}{2}x\right)^2 + \left(\frac{c}{2}x + 1\right)^2 = 0,$ since $b = \frac{a^2}{4} + \frac{c^2}{4}.$ This has a real solution if and only if the two separate terms have zeroes in common. The zeroes of $\left(x^2 + \frac{a}{2}x\right)^2$ are $0$ and $-\frac{a}{2},$ and the zero of $\left(\frac{c}{2}x + 1\right)^2 = 0$ is $-\frac{2}{c}.$ Clearly we cannot have $0=-\frac{2}{c},$ so the only other possibility is $-\frac{a}{2} = -\frac{2}{c},$ which means that $ac = 4.$

We have a system of equations: $ac = 4$ and $a^2 + c^2 = 16.$ Solving this system gives $(a, c) = \left(\sqrt{6}+\sqrt{2}, \sqrt{6}-\sqrt{2}\right), \left(\sqrt{6}-\sqrt{2}, \sqrt{6}+\sqrt{2}\right).$ Each of these gives solutions for $x$ as $-\frac{\sqrt{6}+\sqrt{2}}{2}$ and $-\frac{\sqrt{6}-\sqrt{2}}{2},$ respectively. Now that we know that any valid value of $x$ must be one of these two, we will verify that both of these values of $x$ are valid.

First, consider a right triangle $ABC,$ inscribed in a circle of radius $2,$ with side lengths $a = \sqrt{6}+\sqrt{2}, b = 4, c = \sqrt{6}-\sqrt{2}.$ This generates the polynomial equation $$x^4 + \left(\sqrt{6}+\sqrt{2}\right)x^3 + 4x^2 + \left(\sqrt{6}-\sqrt{2}\right)x + 1 = \left(\frac{\sqrt{6}-\sqrt{2}}{2}x + 1\right)^2\left(\left(\frac{\sqrt{6}+\sqrt{2}}{2}x\right)^2+1\right) = 0.$$ This is satisfied by $x=-\frac{\sqrt{6}+\sqrt{2}}{2}.$

Second, consider a right triangle $ABC,$ inscribed in a circle of radius $2,$ with side lengths $a = \sqrt{6}-\sqrt{2}, b = 4, c = \sqrt{6}+\sqrt{2}.$ This generates the polynomial equation $$x^4 + \left(\sqrt{6}-\sqrt{2}\right)x^3 + 4x^2 + \left(\sqrt{6}+\sqrt{2}\right)x + 1 = \left(\frac{\sqrt{6}+\sqrt{2}}{2}x + 1\right)^2\left(\left(\frac{\sqrt{6}-\sqrt{2}}{2}x\right)^2+1\right) = 0.$$ This is satisfied by $x=-\frac{\sqrt{6}-\sqrt{2}}{2}.$

It follows that the possible values of $x$ are $-\frac{\sqrt{6}+\sqrt{2}}{2}$ and $-\frac{\sqrt{6}-\sqrt{2}}{2}.$

Fun fact: these solutions correspond to a $15$- $75$- $90$ triangle.

(sujaykazi)

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 