Suppose that the side lengths of a triangle are three consecutive integers and one of the angles is twice another. The number of such triangles is are
Step-by-Step Solution
Key Concept: Use the angle condition and the law of sines to establish relationships between consecutive integer sides.
Let $B = 2A$ and $BD$ be the bisector of angle $B$, then $CD = \frac{a}{1+c}$ and $AD = \frac{bc}{a+c}$. Since $\triangle ABC$ and $\triangle BDC$ are similar, we have $\frac{bc}{CD} = \frac{b^2}{a(a+c)}$. This gives $b^2 = a(a+c)$. Since $b > a$, either $b = a+1$ or $b = a+2$. If $b = a+1$ or $b = a+2$, then from $b^2 = a(a+c)$, we get $c = 2+\frac{1}{a}$. For $c$ to be an integer, $a = 1$, but then $b^2 = (a+1)^2 = 2a+1$, which gives $2 = 1$, a contradiction.
Correct Answer: 1