Exercise 10. Let $ABC$ be a triangle and $I$ the center of its inscribed circle. Suppose that $AI = BC$ and that $\widehat{ICA} = 2 \widehat{IAC}$.

What is the value of $\widehat{\overline{ABC}}$?
Solution to Exercise 10 Let $\alpha$ be the angle $\widehat{\mathrm{IAB}}$ and $\beta$ be the angle $\widehat{\mathrm{IBA}}$. By hypothesis, $\widehat{\mathrm{ICA}}=2 \alpha$.

![](https://cdn.mathpix.com/cropped/2024_06_04_e23d725cd2c5ceb81650g-5.jpg?height=469&width=783&top_left_y=1625&top_left_x=671)

It is easy to verify that $\beta=90^{\circ}-3 \alpha$ and $\widehat{\mathrm{IAC}}=90^{\circ}+\beta$. We will use the law of sines in triangles AIC and ABC, which gives respectively

$$
\frac{A I}{\sin (2 \alpha)}=\frac{A C}{\sin \left(90^{\circ}-\beta\right)} \text { and } \frac{A C}{\sin (2 \beta)}=\frac{B C}{\sin (2 \alpha)}
$$

By hypothesis $A \mathrm{I}=\mathrm{BC}$. Therefore, we have,

$$
\frac{A I}{A C} \cdot \sin (2 \alpha)=\sin \left(90^{\circ}-\beta\right)=\sin (2 \beta)
$$

According to the properties of the sine function, if two angles $x$ and $y$ less than $180^{\circ}$ satisfy $\sin (x)=\sin (y)$, then $x=y$ or $x=180^{\circ}-y$. Let's consider the two cases:

- if $90^{\circ}-\beta=2 \beta$, then $\beta=90^{\circ}$ and $\widehat{A B C}=180^{\circ}$, and the triangle is flat, which is not a valid solution
- if $90^{\circ}-\beta=180^{\circ}-2 \beta$, then $\beta=30^{\circ}$ and $\widehat{A B C}=60^{\circ}$, which gives the figure presented above.

The solution is $\widehat{A B C}=60^{\circ}$.