In an isosceles triangle, the base is one unit, and the length of the legs is $b$. What is the base of an isosceles triangle whose vertex angle is equal to the base angle of the previous triangle and whose legs are one unit long?
![](https://cdn.mathpix.com/cropped/2024_05_02_860bf5709a9af52e9b6ag-1.jpg?height=521&width=350&top_left_y=127&top_left_x=882)

Solution. The base angles of the isosceles triangle $ABC$ are $\alpha$. In the isosceles triangle $ACD$, $AC = AD = 1$, $\angle CAD = \alpha$, and $CD = x$. We apply the Law of Cosines in triangle $ACD$:

$$
x^{2} = 1^{2} + 1^{2} - 2 \cdot 1 \cdot 1 \cdot \cos \alpha
$$

From the right triangle $BEA$: $\cos \alpha = \frac{1}{2b}$. Substituting this into the Law of Cosines and simplifying, we get:

$$
x = \sqrt{2 - \frac{1}{b}}
$$

This is valid if $2 > \frac{1}{b}$, which means $b > \frac{1}{2}$. This is always true because the sides of triangle $ABC$ satisfy the triangle inequality.