, 2009

Let $E_1$ be the reflection of $E$ across the line $CI$. Because $CI$ is the angle bisector of $\angle BCA$, $E_1$ lies on ray $CB$. We consider two cases.

In the first case, we assume that $E_1 = D$. Then $CDIE$ is a kite with $CI$ as the symmetry axis. In particular, $\angle IEC = \angle IDC = 90^\circ$; that is, $BE$ is also an altitude of triangle $ABC$, implying that $AB = BC = CA$ or $\angle BAC = 60^\circ$. (See the left-hand side figure shown below.)

In the second case, we assume that $\angle D \neq E_1$. By symmetry, $\angle IE_1K = \angle IEK = 45^\circ = \angle IDK$, from which it follows that $I, D, E_1, K$ lie on a circle. Hence either $\angle IKE = \angle IDE_1 = 90^\circ$ or $\angle IKE_1 = 180^\circ - \angle IDE = 90^\circ$; that is, $\angle IKE_1 = 90^\circ$. Therefore, we deduce the right-hand side figure shown below. In cyclic quadrilateral, we have $\angle KIE_1 = \angle KDE_1 = \angle KDC = 45^\circ$. By symmetry, we have $\angle EIK = \angle KIE_1 = 45^\circ$, from which it follows that $\angle IBC = \angle ICB = 22.5^\circ$ and $\angle BAC = 90^\circ$.

Set $\angle ABC = \angle BCA = 2x$, with $0^\circ < x < 45^\circ$. Then $\angle EIK = 2x$, $\angle ECK = \angle DCK = x$, and $\angle CEK = 135^\circ - x$.

It is clear that $CID$ is a right triangle, and, by the angle-bisector theorem, we have
$$\frac{IK}{KC} = \frac{ID}{DC} \quad \text{or} \quad \frac{IK}{KC} = \frac{\sin x}{\cos x}.$$
Applying the law of sines in triangles $EIK$ and $CEK$ gives
$$\frac{\sin 45^\circ}{IK} = \frac{\sin 2x}{EK} \quad \text{and} \quad \frac{KC}{\sin(135^\circ - 3x)} = \frac{EK}{\sin x}.$$
Multiplying the last three equations yields
$$\frac{\sin 45^\circ}{\sin(135^\circ - 3x)} = \frac{\sin 2x}{\cos x} = 2 \sin x \quad \text{or} \quad \sin 45^\circ = 2 \sin x \sin(135^\circ - 3x).$$
By the addition-subtraction formulas, we obtain
$$\sin 45^\circ = 2 \sin x \sin(135^\circ - 3x) = 2 \sin x(\sin 135^\circ \cos 3x - \cos 135^\circ \sin 3x),$$
hence $1 = 2 \sin x(\cos 3x + \sin 3x)$. By the product-to-sum formulas and the double-angle formulas, we have
$$\begin{aligned} 1 &= 2 \sin x(\cos 3x + \sin 3x) = 2 \sin x \cos 3x + 2 \sin x \sin 3x \\ &= \sin 4x - \sin 2x + \cos 2x - \cos 4x \\ &= 2 \sin 2x \cos 2x - \sin 2x + \cos 2x - 2 \cos^2 2x + 1 \\ &= 1 + (2 \cos x - 1)(\sin 2x - \cos 2x); \end{aligned}$$
that is, we obtain $(2 \cos x - 1)(\sin 2x - \cos 2x) = 0$. Because $0^\circ < x < 45^\circ$, the possible values of $x$ are $x = 30^\circ$ (corresponding to $2 \cos x - 1 = 0$) and $x = 22.5^\circ$ (corresponding to $\sin 2x - \cos 2x = 0$), leading to $\angle CAB = 60^\circ$ and $\angle CAB = 90^\circ$, respectively.