A regular quadrilateral pyramid has a circumscribed sphere whose center coincides with the center of the inscribed sphere. What is the angle between two adjacent side edges?
I. Solution. In the diagram, $O$ is the common center of the circumscribed and inscribed spheres, $R$ and $r$ are the radii, and $AB=1$. The projection of vertex $E$ onto the base plane is $T$, $F$ is the midpoint of $BC$, and $U$ is the point of tangency of the inscribed sphere on the side $ABC$. The right triangles $OTF$ and $OUF$ are congruent because they are equal in two sides and the angle opposite the larger side. Therefore, $FU=FT=\frac{1}{2}$. Similarly, the triangles $OUE$ and $OTB$ are congruent, which means $EU=BT=\frac{\sqrt{2}}{2}$. Therefore, $EF=EU+FU=\frac{1+\sqrt{2}}{2}$. For the half-angle of the dihedral angle: $\operatorname{tg} \frac{\alpha}{2}=\frac{\frac{1}{2}}{\frac{1+\sqrt{2}}{2}}=\sqrt{2}-1$, from which, using a table or a pocket calculator, $\frac{\alpha}{2}=22.5^{\circ}$ and thus $\alpha=45^{\circ}$. However, we might think that the obtained value is not precise. To find the exact value of $\alpha$, let's calculate the slant height of the pyramid:

$$
BE=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1+\sqrt{2}}{2}\right)^{2}}=\sqrt{\frac{2+\sqrt{2}}{2}}
$$

Then, using the cosine rule in triangle $EBC$:

$$
\cos \alpha=\frac{2 \cdot \frac{2+\sqrt{2}}{2}-1}{2 \cdot \frac{2+\sqrt{2}}{2}}=\frac{1+\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}}{2}
$$

from which $\alpha=45^{\circ}$.

David Papp (Budapest, Szent István Gymnasium 10.o.t.)

II. Solution. David Papp provided two solutions to this problem. Based on the diagram, his calculations can be outlined as follows. According to the theorem of parallel chords, $TV=\frac{r(R+r)}{R}$. By the Pythagorean theorem: $EF=$ $\sqrt{(R+r)^{2}+\frac{1}{4}}$. Then, express the square of twice the area of triangle $ETF$ in two ways:

$$
\frac{r^{2}(R+r)^{2}}{R^{2}} \cdot\left[(R+r)^{2}+\frac{1}{4}\right]=\frac{(R+r)^{2}}{4}
$$

From triangle $BOT$:

$$
R^{2}=r^{2}+\frac{1}{2}
$$

From the system of equations (1) and (2), $r=\frac{\sqrt{\sqrt{2}-1}}{2}$, with which

$$
BE=\sqrt{(R+r)^{2}+\frac{1}{2}}=\sqrt{\left(\sqrt{r^{2}+\frac{1}{2}}+r\right)^{2}+\frac{1}{2}}=\sqrt{\frac{2+\sqrt{2}}{2}}
$$

Finally, just as in the first solution, $\alpha=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_116ac1440fe5f6bf2df2g-1.jpg?height=650&width=781&top_left_y=1960&top_left_x=672)