5. Two triangular pyramids $M A B C$ and $N A B C$ have a common base and no other common points. All vertices of both pyramids lie on the same sphere. Find the lengths of the edges $M A$ and $M B$, given that they are equal to each other, and the lengths of all other edges of these pyramids are $\sqrt{3}$.
4.II.5. First, let's find the diameter \( d \) of the sphere circumscribed around the regular tetrahedron \( NABC \). Since \( CO = \frac{AB}{\sqrt{3}} = 1 \), then \( NO = \sqrt{3-1} = \sqrt{2} \). In a regular tetrahedron, the center of the circumscribed sphere coincides with its centroid, which lies on the height \( NO \) of the tetrahedron and divides it in the ratio \( 3:1 \), counting from the vertex. Therefore, \( d = 2 \cdot \frac{3}{4} \cdot NO = \frac{3}{\sqrt{2}} \) (figure).

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![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-085.jpg?height=551&width=579&top_left_y=410&top_left_x=727)

Now consider the pyramid \( MABC \) inscribed in the same sphere. The center of this sphere lies outside this pyramid. Since \( CA = CB = CM \), this center lies on the line containing the height \( CO_1 \) of the pyramid \( MABC \). Extend this height to the point \( D \) of its intersection with the sphere (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-085.jpg?height=497&width=468&top_left_y=1348&top_left_x=794)

The triangle \( CDM \) is a right triangle, so \( CM^2 = CO_1 \cdot d \), from which \( CO_1 = \sqrt{2} \). The triangle \( CMO_1 \) is also a right triangle, so \( MO_1 = \sqrt{CM^2 - CO_1^2} = 1 \). Now consider the triangle \( ABM \). Since the point \( O_1 \) is the center of the circumscribed circle around it and lies outside this triangle, \( \angle AMB > 90^\circ \). The radius \( R \) of this circle is \( MO_1 = 1 \). By the sine rule, \( AB = 2R \sin AMB \), \( \sqrt{3} = 2 \sin AMB \), so \( \sin AMB = \frac{\sqrt{3}}{2} \), and \( \cos AMB = -\frac{1}{2} \). Finally, by the cosine rule,

\[
AB^2 = AM^2 + BM^2 - 2 AM \cdot BM \cos AMB = 3 AM^2
\]

so \( AM = BM = 1 \).