Method 1: Don't be too complicated, and teach you a simple method! !

Because it is a regular triangular pyramid, SB is perpendicular to AC. MN is parallel to SB, so SB is perpendicular to AM.

So SB has a vertical capsule.

Similarly, according to the symmetry of the regular triangular pyramid, the vertical plane of SA is SBC, and the vertical plane of SC is SAB.

So SA, SB and SC are perpendicular to each other.

Next, S-ABC is simplified as a cube, the radius of its circumscribed circle is half of the diagonal of the cube, that is, R=√3a/2, and the surface area of the circumscribed ball S = 4 π r 2 = 3 π a 2.

Method 2: sa = sb = sc = a.

So we can let AB=BC=CA=b,

Through a&B, the shape of the whole graph can be solved, because the center of the circle can be found by knowing the quantitative relationship of edges.

Where is the quantitative relationship? MN is perpendicular to AM

An 2 = am 2+Mn 2 can be obtained by Pythagorean theorem.

AB = BC = CA = B, so we can know that an = B * (√ 3)/2 has been set on the bottom surface.

The median theorem is Mn =1/2 * sb =1/2 * a.

AM=？ AM is the center line of Δ sac on the side of SC, cos∠SCA=cos∠SAC=b/2a.

In δδAMC, MC=a/2, AC=b, AM=x, cos∠SCA=b/2a.

With cosine theorem, cos ∠ SCA = (MC 2+AC 2-AM 2)/2 * AC * MC.

The solution is (a 2)/4+(b 2)/2 = x 2, x=AM.

Substituting an 2 = am 2+Mn 2, we get b 2 = 2 * (a 2), in other words, b = √ 2 * a.

I finally know, AB = BC = CA = B = √ 2 * A.

I believe you will understand the following words,

If the bottom center is p, Δ Δ SPC is a right triangle.

SC=a，PC=(√6)/3*a，SP=√3/3*a

The radius of the S-ABC circumscribed ball is R=√3/2*a, and the center of the circle is outside the shape.

The surface area of the circumscribed sphere is S = 4 * π * R 2 = 3 π a 2.