# Difference between revisions of "2018 USAJMO Problems/Problem 4"

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Fun fact: these solutions correspond to a <math>15</math>-<math>75</math>-<math>90</math> triangle. | Fun fact: these solutions correspond to a <math>15</math>-<math>75</math>-<math>90</math> triangle. | ||

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## Revision as of 00:57, 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)