12.1 Prisms, Pyramids, Cylinders, Cones, and Spheres

Section 12.1: Prisms, Pyramids, Cylinders, Cones, and Spheres

Solid geometry in the National College Entrance Examination (Gaokao) tests spatial visualization, geometric reasoning, and quantitative metric calculations. Three-dimensional figures are broadly classified into two major categories: Polyhedra (多面体)—bounded by plane polygonal faces—and Solids of Revolution (旋转体)—generated by revolving plane closed regions about a fixed axis.

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1. Structural Definitions and Geometric Classifications

1.1 Polyhedra: Prisms, Pyramids, and Frustums

1. Prism (棱柱):

   • Definition: A polyhedron formed by two congruent parallel polygonal bases and lateral faces that are parallelograms.
   • Right Prism (直棱柱): A prism whose lateral edges are perpendicular to the base planes. All lateral faces are rectangles.
   • Regular Prism (正棱柱): A right prism whose base is a regular polygon. All lateral faces are congruent rectangles.
   • Key Metric Properties: For a right prism with base area $S_{\text{base}}$, base perimeter $C_{\text{base}}$, and height $h$ (equal to lateral edge length $l$): $$\text{Lateral Area: } S_{\text{lat}} = C_{\text{base}} \cdot h$$ $$\text{Volume: } V = S_{\text{base}} \cdot h$$

2. Pyramid (棱锥):

   • Definition: A polyhedron with one polygonal base and lateral faces that are triangles meeting at a single vertex called the apex (顶点).
   • Regular Pyramid (正棱锥): A pyramid whose base is a regular polygon and whose apex projects orthogonally onto the center (circumcenter/incenter) of the base.
   • Properties of Regular Pyramids:
     • All lateral edges are equal in length: $l = \sqrt{h^2 + R_{\text{base}}^2}$, where $R_{\text{base}}$ is the circumradius of the base.
     • All lateral faces are congruent isosceles triangles.
     • The altitude of each lateral face from the apex to the base edge is called the slant height (斜高, denoted $h'$): $$(h')^2 = h^2 + r_{\text{base}}^2$$ where $r_{\text{base}}$ is the inradius of the regular base polygon.
     • Lateral Area: $S_{\text{lat}} = \frac{1}{2} C_{\text{base}} \cdot h'$.
     • Volume: $V = \frac{1}{3} S_{\text{base}} \cdot h$.

3. Frustum of a Pyramid (棱台):

   • Formed by cutting a pyramid with a plane parallel to its base. The section and base are similar polygons. Given top base area $S_1$, bottom base area $S_2$, and height $h$: $$V_{\text{frustum}} = \frac{1}{3} h \left(S_1 + S_2 + \sqrt{S_1 S_2}\right)$$

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1.2 Solids of Revolution: Cylinders, Cones, Frustums, and Spheres

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2. Inscribed and Circumscribed Sphere Models (球的切接问题)

In Gaokao Mathematics, problems combining polyhedra with spheres are extremely frequent. Master the following three canonical models:

Model A: Rectangular Cuboid and Wall/Corner Embedding ("补形法")

• Any triangular pyramid with three mutually perpendicular edges meeting at a vertex (such as a corner of a cube $O-ABC$ where $OA \perp OB \perp OC$) can be completed into a rectangular cuboid with dimensions $a = OA, b = OB, c = OC$.
• The circumscribed sphere of the pyramid is identical to the circumscribed sphere of the completed cuboid.
• Circumradius Formula: $$2R = \sqrt{a^2 + b^2 + c^2} \implies R = \frac{1}{2}\sqrt{a^2 + b^2 + c^2}$$ $$\text{Surface Area of Sphere: } S = 4\pi R^2 = \pi (a^2 + b^2 + c^2)$$

Model B: Regular Polyhedron Metric Constants (Regular Tetrahedron)

For a regular tetrahedron $A-BCD$ with edge length $a$:

• Height: $h = \frac{\sqrt{6}}{3} a$
• Base Circumradius: $R_{\text{base}} = \frac{\sqrt{3}}{3} a$
• Volume: $V = \frac{\sqrt{2}}{12} a^3$
• Circumscribed Sphere Radius: $R = \frac{\sqrt{6}}{4} a$
• Inscribed Sphere Radius: $r = \frac{\sqrt{6}}{12} a$
• Key Ratio: $R : r = 3 : 1$, and $R + r = h = \frac{\sqrt{6}}{3} a$.

Model C: Volume-Area Method for Inscribed Spheres ("等体积法")

For any solid convex polyhedron with total volume $V$ and total surface area $S$ containing an inscribed sphere tangent to all faces:

• Partition the polyhedron into $n$ smaller pyramids, each with a face of area $S_i$ as base and the sphere center $O$ as apex (height $r$): $$V = \sum_{i=1}^n \frac{1}{3} S_i \cdot r = \frac{1}{3} r \left(\sum_{i=1}^n S_i\right) = \frac{1}{3} S \cdot r$$
• Inscribed Sphere Radius Formula: $$r = \frac{3V}{S}$$

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3. Worked Gaokao Exam Examples

Example 1: Regular Tetrahedron Inscribed and Circumscribed Spheres

Problem: A regular tetrahedron $A-BCD$ has edge length $a = 2\sqrt{3}$.

1. Calculate the volume of its circumscribed sphere $V_{\text{sphere}}$.
2. Find the ratio of the surface area of its circumscribed sphere $S_{\text{circum}}$ to its inscribed sphere $S_{\text{in}}$.

Detailed Solution:

1. For a regular tetrahedron of edge length $a = 2\sqrt{3}$, we first determine its circumradius $R$: $$R = \frac{\sqrt{6}}{4} a = \frac{\sqrt{6}}{4} \cdot (2\sqrt{3}) = \frac{\sqrt{18}}{2} = \frac{3\sqrt{2}}{2}$$ The volume of the circumscribed sphere is: $$V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \left(\frac{3\sqrt{2}}{2}\right)^3 = \frac{4}{3}\pi \cdot \frac{54\sqrt{2}}{8} = 9\sqrt{2}\pi$$
2. The radius of the inscribed sphere is: $$r = \frac{\sqrt{6}}{12} a = \frac{\sqrt{6}}{12} \cdot (2\sqrt{3}) = \frac{\sqrt{18}}{6} = \frac{\sqrt{2}}{2}$$ The ratio of surface areas is: $$\frac{S_{\text{circum}}}{S_{\text{in}}} = \frac{4\pi R^2}{4\pi r^2} = \left(\frac{R}{r}\right)^2 = \left(\frac{3\sqrt{2}/2}{\sqrt{2}/2}\right)^2 = 3^2 = 9$$

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Example 2: Wall/Corner Embedding Method for Right Triangular Prism

Problem: In a right triangular prism $ABC-A_1B_1C_1$, $\angle ACB = 90^\circ$, $AC = 3$, $BC = 4$, and the lateral edge length is $AA_1 = 4$. Find the total surface area of the circumscribed sphere of this prism.

Detailed Solution:

1. Analyze the geometric arrangement: Since $ABC-A_1B_1C_1$ is a right prism, the lateral edge $CC_1$ is perpendicular to the base plane $ABC$. Combined with $\angle ACB = 90^\circ$, the three line segments $CA$, $CB$, and $CC_1$ are mutually perpendicular at point $C$.
2. Apply the Cuboid Embedding Method (补形法): Embed the right triangular prism into a rectangular cuboid with length $a = AC = 3$, width $b = BC = 4$, and height $c = CC_1 = 4$.
3. The vertices of the triangular prism are a subset of the 8 vertices of this cuboid. Thus, the circumscribed sphere of the prism is identical to the circumscribed sphere of the rectangular cuboid.
4. Calculate the body diagonal $D$ and circumradius $R$ of the cuboid: $$D = \sqrt{a^2 + b^2 + c^2} = \sqrt{3^2 + 4^2 + 4^2} = \sqrt{9 + 16 + 16} = \sqrt{41}$$ $$R = \frac{D}{2} = \frac{\sqrt{41}}{2}$$
5. Compute the surface area of the circumscribed sphere: $$S = 4\pi R^2 = 4\pi \left(\frac{\sqrt{41}}{2}\right)^2 = 4\pi \cdot \frac{41}{4} = 41\pi$$

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4. Geometric Classification Visualization