Problem 1

Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another.

Step 1 of 6: Set up angle and side notation

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In plain words

Picture stretching or shrinking a triangle while keeping its angles in the ratio $2:1$ at two vertices; as the shape changes continuously, the ratios between its side lengths change too. Consecutive integer side lengths is a very rigid, discrete condition, so it is plausible that only a handful of shapes (perhaps only one) can satisfy both requirements simultaneously.

\angle BAC = 2\alpha, \quad \angle ABC = \alpha, \quad BC=a,\ CA=b,\ AB=c

Detailed analysis

Let triangle $ABC$ have $\angle ABC = \alpha$ and $\angle BAC = 2\alpha$ , and write $a=BC$ , $b=CA$ , $c=AB$ for the side lengths facing $A$ , $B$ , $C$ respectively. The three sides are three consecutive positive integers in some order; the goal is to find every such triangle in which one angle is exactly twice another, and to show there is only one.