6. The first circle with center at point $O$ is inscribed in triangle $A B C$. Points $A$ and $B$ lie on the second circle with the same center $O$. Line $A C$ intersects the second circle at point $D$ $(D \neq A)$, and line $B C$ intersects the second circle at point $E(B \neq E)$. It is known that angle $A B C$ is equal to angle $C A E$. Find the cosine of angle $B A C$.
Solution. Points $D$ and $E$ lie outside segments $AC$ and $BC$ respectively (Fig. 1). Otherwise (Fig. 2), arc $DE$ is a part of arc $AE$ (and does not coincide with it), which contradicts the condition of the equality of angles $ABC$ and $CAE$.

Let $\angle ABC = \beta$. Then $\angle CAE = \beta$, $\angle CAE = \angle CBD = \beta$. Point $O$ is the center of the inscribed circle in triangle $ABD$. $\angle ABO = \beta / 2$. Since $O$ is the center of the circumscribed circle around triangle $ABC$, triangle $AOB$ is isosceles, and $\angle BAO = \angle ABO = \beta / 2$. $AO$ is the angle bisector of $\angle A$, so $\angle OAC = \beta / 2$. Triangles $AOD$ and $BOD$ are isosceles, and $\angle BDO = 3\beta / 2$, $\angle ADO = \beta / 2$. Since $\angle A + \angle B + \angle D = 180^\circ$, we have $3\beta + 3\beta / 2 + \beta / 2 = 180^\circ$, and $\beta = 36^\circ$. Triangle $ABD$ is isosceles, $\angle B = \angle D = 72^\circ$, $\angle A = 36^\circ$.

Let $AC = x$, $CD = y$. Triangles $ABD$ and $CBD$ are similar, $\frac{x}{y} = \frac{x + y}{x}$, $x^2 - xy - y^2 = 0$, $\frac{x}{y} = \frac{\sqrt{5} + 1}{2}$. Since $\cos(\angle BAC) = \cos \beta$, by the cosine rule for triangle $CBD$, we have $\cos \beta = \frac{2x^2 - y^2}{2x^2} = 1 - \frac{1}{2} \cdot \left(\frac{y}{x}\right)^2 = 1 - \frac{1}{2} \cdot \left(\frac{\sqrt{5} - 1}{2}\right)^2 = 1 - \frac{1}{4} \cdot (3 - \sqrt{5}) = \frac{1 + \sqrt{5}}{4}$.

Answer: $\frac{1 + \sqrt{5}}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9a8ac98263d65b486910g-4.jpg?height=571&width=560&top_left_y=1214&top_left_x=288)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_9a8ac98263d65b486910g-4.jpg?height=583&width=666&top_left_y=1216&top_left_x=1186)

Fig. 2

## Solution for Variant 5