In the triangle $ABC$, the angle bisectors intersect the opposite sides at points $D, E, F$ respectively. What are the limits within which the angle $BAC$ must lie so that $AB \neq AC$ while $DE = DF$?
We start from the consequence of $D F E \angle = D E F \angle$. Let the projection of $D$ onto $A C$ be $D^{\prime}$ and $D E C \angle = \varepsilon_{2}$. Based on the angle bisector division ratio, with the usual notation,

\[
\begin{aligned}
& \operatorname{tg} \varepsilon_{2} = \frac{D D^{\prime}}{E D^{\prime}} = \frac{C D \sin \gamma}{C E - C D \cos \gamma} = \frac{\frac{a b}{b+c} \sin \gamma}{\frac{a b}{a+c} - \frac{a b}{b+c} \cos \gamma} = \\
& = \frac{a \sin \gamma + c \sin \gamma}{c(1 - \cos \gamma) + (b - a \cos \gamma)} = \frac{c(\sin \alpha + \sin \gamma)}{c(1 - \cos \gamma + \cos \alpha)}
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_ef8c7a5d26c901534d2cg-1.jpg?height=290&width=675&top_left_y=598&top_left_x=714)

By re-labeling for the angle $F E A \angle = \varepsilon_{1}$,

\[
\operatorname{tg} \varepsilon_{1} = \frac{\sin \alpha + \sin \gamma}{1 - \cos \alpha + \cos \gamma}
\]

Further transformations yield

\[
\operatorname{tg} D E F \angle = -\operatorname{tg}(\varepsilon_{1} + \varepsilon_{2}) = \frac{\sin \alpha + \sin \gamma}{\cos \beta + \frac{1}{2}},
\]

and by re-labeling,

\[
\operatorname{tg} D E F \angle = \frac{\sin \alpha + \sin \beta}{\cos \gamma + \frac{1}{2}}.
\]

From this, the statement of problem 2239 also follows: when $\gamma = 120^\circ$, $F D \perp F E$. The vanishing of any denominator, which could lead to a degenerate triangle, does not cause confusion if we require the equality of these two expressions:

\[
\frac{\sin \alpha + \sin \gamma}{\cos \beta + \frac{1}{2}} = \frac{\sin \alpha + \sin \beta}{\cos \gamma + \frac{1}{2}}
\]

If this holds, then based on the identity

\[
\frac{x}{y} = \frac{z}{u} \Leftrightarrow \frac{x - z}{y - u} = \frac{x}{y},
\]

we have

\[
\operatorname{tg} D E F \angle = \frac{\sin \gamma - \sin \beta}{\cos \beta - \cos \gamma} = \operatorname{ctg} \frac{\beta + \gamma}{2} = \operatorname{tg} \frac{\alpha}{2}.
\]

Thus, in the triangles under consideration,

\[
D E F \angle = D F E \angle = \frac{\alpha}{2} = 90^\circ - \frac{E D F \angle}{2}, \quad \operatorname{tg} E D F \angle = -\operatorname{tg} \alpha,
\]

(meaning the circumcircle of $\triangle D E F$ passes through $A$), and writing the expression for $\operatorname{tg} E D F$ according to (1),

\[
\frac{\sin \beta + \sin \gamma}{\cos \alpha + \frac{1}{2}} = -\frac{\sin \alpha}{\cos \alpha}, \quad \left(\alpha \neq \frac{\pi}{2}, \frac{2 \pi}{3}\right)
\]

and again applying (2),

\[
\sin \beta + \sin \gamma + \sin \alpha = -\frac{\sin \alpha}{2 \cos \alpha}
\]

From this, we can read off: $\alpha > \pi / 2$. On the left side, $a > 0$ and on the right side, $b < 0$. Introducing the notation

\[
\sin \frac{\alpha}{2} = z
\]

we get

\[
4 z (2 z^2 - 1) < 0 \quad \text{for} \quad 0 < z < 1
\]

and

\[
f(z) = 4 z (2 z^2 - 1) + \frac{1}{2} \sin \alpha \cos \alpha
\]

Since $f(0) < 0$ and $f(1) > 0$, $f$ has a unique positive zero in the interval $(0, 1)$. Using a small calculator, $f$ becomes positive between $z = 0.7807764$ and $z = 0.78077641$ (values are $-7 \cdot 10^{-8}$ and $+4 \cdot 10^{-8}$, respectively). Thus,

\[
0.78077641 < \sin \frac{\alpha}{2} < 0.7905694
\]

or equivalently,

\[
\begin{aligned}
102.66343^\circ & < \alpha < 104.4775^\circ \\
102^\circ 39' 48.4'' & < \alpha < 104^\circ 28' 39''
\end{aligned}
\]

is a sufficient condition for $\triangle D E F$ to be isosceles.

For a given $\alpha$, the other angles can be calculated from the system $\beta + \gamma = 180^\circ - \alpha$ and (3).

We note that the value $\alpha = 4 \pi / 7 = 102^\circ 51' 25.7''$ examined in F. 2264 falls within the narrow interval (5).

The problem was solved by G. Parry in The Mathematical Gazette, issue 420 (June 1978).

B. T.

[^0]:    ${ }^{1}$ K. M. L. 60 (1980) 206. (May 1980).

[^1]:    ${ }^{1}$ See the solution in K. M. L. 61 (1980), p. 137.