3. An isosceles triangle can be cut into two isosceles triangles (not necessarily equal). Find the values that the smallest angle of such a triangle can take. In the answer, specify the smallest of these values in degrees, multiplied by 6006.

ANSWER: 154440.
Solution: Consider triangle $ABC$ with angles $\angle A=\angle C=\alpha, \angle B=180^{\circ}-2 \alpha$. For the line to divide the triangle into two, it must pass through one of the vertices. Let's consider the case where it passes through vertex A and divides the triangle into two: $ADB$ and $ADC$ (see fig.).

![](https://cdn.mathpix.com/cropped/2024_05_06_ae561e403711a867145fg-2.jpg?height=686&width=551&top_left_y=248&top_left_x=638)

Triangle ADC is isosceles in two cases:
I) $\angle ADC=\alpha$. By equating $\angle BAD=\angle ABD$ (since angle $\angle ADB$ is obtuse), we get the equation $180^{\circ}-2 \alpha=3 \alpha-180^{\circ}$, from which $\alpha=72^{\circ}$. The smallest angle is then $\angle B=36^{\circ}$.

II) $\angle ADC=\angle DAC=\frac{180^{\circ}-\alpha}{2}$. Then $\frac{3 \alpha}{2}-90^{\circ}=180^{\circ}-2 \alpha$, from which $\alpha=\frac{540^{\circ}}{7}$. In this case, the smallest angle is $\angle B=\frac{180^{\circ}}{7}$.

If the line passes through vertex B, the cases are considered similarly, but the smallest angle is $36^{\circ}$ or $45^{\circ}$, which is greater than $\frac{180}{7}$.

Multiplying $\frac{180}{7}$ by 6006 and simplifying by 7, we get the answer.