Question 148, In a regular tetrahedron \(P-ABC\) with all edge lengths equal to 1, \(L\), \(M\), and \(N\) are the midpoints of edges \(PA\), \(PB\), and \(PC\) respectively. The area of the section formed by the plane \(LMN\) cutting through the circumsphere of the regular tetrahedron is \(\qquad\) -
Question 148, Solution: As shown in the figure, let the circumcenter of $\triangle ABC$ be $O$, the circumcenter of $\triangle ABC$ be $Q$, and the circumcenter of $\triangle LMN$ be $S$. Then $S$ is the midpoint of $PQ$, and $\frac{PO}{OQ}=3$, so $O$ is the midpoint of $SQ$. Since $AB=1$, we have $AQ=AQ=\frac{1}{\sqrt{3}}$, hence $PQ=\frac{\sqrt{2}}{\sqrt{3}}$, and thus $OP=\frac{3\sqrt{2}}{4\sqrt{3}}, OS=\frac{\sqrt{2}}{4\sqrt{3}}$. According to the power of a point theorem, the radius of the section of the circumcircle of the regular triangular pyramid cut by $\triangle LMN$ is $\sqrt{OP^2 - OS^2} = \frac{1}{\sqrt{3}}$, so the area of the section is $\frac{\pi}{3}$.