value?

In triangle $ABC$, with the usual notation, $\alpha=60^{\circ}, \beta=20^{\circ}$, and $AB=1$. What is the exact value of $\frac{1}{AC}-BC$?
I. solution. For the triangle $ABC$, the sine theorem is written twice:

$$
\frac{1}{AC}=\frac{\sin 100^{\circ}}{\sin 20^{\circ}} \quad \text { and } \quad BC=\frac{\sin 60^{\circ}}{\sin 100^{\circ}}
$$

Since $\sin 100^{\circ}=\cos 10^{\circ}$ and $\sin 20^{\circ}=\sin \left(10^{\circ}+10^{\circ}\right)=2 \sin 10^{\circ} \cos 10^{\circ}$, we have

$$
\frac{1}{AC}-BC=\frac{\sin 100^{\circ}}{\sin 20^{\circ}}-\frac{\sin 60^{\circ}}{\sin 100^{\circ}}=\frac{\cos 10^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}}-\frac{\sin 60^{\circ}}{\cos 10^{\circ}}=\frac{\cos 10^{\circ}-\sqrt{3} \cdot \sin 10^{\circ}}{2 \sin 10^{\circ} \cos 10^{\circ}}=\frac{2 \cdot \sin \left(30^{\circ}-10^{\circ}\right)}{\sin 20^{\circ}}=\frac{2 \cdot \sin 20^{\circ}}{\sin 20^{\circ}}=2
$$

Based on the work of Péter Bakos (Eger, Szilágyi Erzsébet Gymn., IV. grade) and Eszter Dragó (Szeged, Radnóti M. Gymn., III. grade)

II. solution. The triangle $ABC$ in the problem is cut off from the equilateral triangle $ABD$ by one of the angle trisectors at $B$. Let the intersection of $AD$ and the other angle trisector be $E$ (see the figure). Then

$$
BC=BE \quad \text { and }
$$

$CE=1-2AC$.

Applying the angle bisector theorem to triangle $ABE$:

$$
\frac{BE}{1}=\frac{CE}{AC}, \quad \text { from which, based on (1) and (2): } \quad BC=\frac{1-2AC}{AC} \text {. }
$$

From this, $BC=\frac{1}{AC}-2$, and thus $\frac{1}{AC}-BC=2$.

Imre Pólik (Pannonhalma, Bencés Gymn., IV. grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_2fb0fa7a61de1cb1ff0ag-1.jpg?height=374&width=377&top_left_y=1190&top_left_x=863)