The derivation process comes from Baidu students, as follows.
As shown in the figure: the side length P-ABCD =a and the side length PA = Pb = A of the bottom surface of a regular pyramid.
Then, the oblique height PM=PN=√3a/2, the height PO' = √ 2a/2, the inscribed circle of △ PMN is the great circle of the ball, O is the center of the ball, and the tangent point T is on the oblique height.
It can be obtained from rt △ PTO ∽ rt △ po' n.
T0/NO'=PO/PN，
Namely. r/(a/2)=(√2a/2-r)/(√3a/2)
By solving the above formula, we can get r=(√6-√2)a/4.
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A regular quadrangular pyramid has eight sides, the length of which is a, the base is square and the sides are regular triangles. ?
If there is an outer sphere, the distances from its center to the five vertices of a regular pyramid must be equal, all of which are r?
As you can imagine, the projection of the center of this sphere at the bottom of a regular pyramid must be at the center of the square (because it must be symmetrical). ?
It can be said that the connecting line between the center of the circle and the vertex is exactly the height h of the regular pyramid, and the so-called spherical center must also be at this height. The distance from the center (the center of the bottom of a square) to the four vertices at the bottom is (√2)a/2, and the side length is a, then the right triangle formed by the center and the height h can be calculated as the height H = √{ A & amp； sup2-[(√2)a/2]& amp； sup2} =√(a & amp； sup2/2)=(√2)a/2 .？
Now, the distance from the center of the sphere to the vertex is R. In the right triangle just analyzed, the center of the sphere divides the right-angled side with a height of H into two parts, the distance from the center of the sphere to the bottom surface is l=h-r=(√2)a/2-r, the triangle formed by the center of the sphere, the length of the hypotenuse of the vertex and the center of the bottom surface of the regular quadrangular pyramid is R (the distance from the center of the sphere to the vertex of the bottom surface of the quadrangular pyramid), and the right-angled sides are respectively
r & ampsup2=[(√2)a/2]& amp； sup2+[(√2)a/2-r]& amp； sup2？
r & ampsup2= a & ampsup2/2+a & amp； sup2/2-(√2)ar+r & amp； sup2？
a & ampsup2-(√2)ar=0？
A≠0, ∴a-(√2)r=0, r=(√2)a/2 (This result shows that the spherical center of the circumscribed circle of a regular quadrangular pyramid is the center of the bottom surface. )?
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