VMHS Math Circle
XV. The Grand Geometry Review (Part V: 3-D Geometry)
This final part of our geometry review will cover three-dimensional geometry, as if this wasn’t
obvious already in our title. To me, 3-D geometry in a problem-solving context is very tough because of
how hard it is to visualize. It’s easy to visualize a simple cube or prism, but when other cubes/prisms are
being drawn in them or when planes slice through various points of them, the process just becomes
confusing.
Here, we’ll cover some formulas pertaining to 3-D geometry, but we’ll also combine what we’ve
learned in other sections to deal with what we’re up against.
Surface Area/Volume Formulas
When dealing with surface area, think of it as the sum of the areas of all faces!
For a cube, the surface area is 6s2, where “s” is the side length. The volume is s3.
For a rectangular prism, the surface area is 2lw + 2lh + 2wh, with “l,” “w,” and “h” being the
dimensions of the prism. The volume is lwh.
The surface area of a pyramid is much less specific. However, simply find the area of the base, find the
areas of the other triangular faces (often, one might have to use the Pythagorean Theorem to find the
height), and compute the sum of each area.
For a sphere, the surface area is 4πr2, where “r” is the radius. The volume is 4πr3/3.
For a cone, the surface area is πr2 + πrl, where “r” is the radius, and “l” is the slant height. The volume is
4πr2h/3, where “h” is the vertical height.
Other Solids
A tetrahedron is basically four triangles glued together into a 3-d shape. Think of a regular tetrahedron
as four equilateral triangles glued together into a 3-d shape.
An octahedron is 8 triangles glued together into a 3-d shape. Similarly, a regular octahedron is eight
equilateral triangles glued together into a 3-d shape.
A polyhedron is basically just a multitude of faces.
3-D Analytic Geometry
There’s just a tiny bit to know here, since 3-D analytic geometry is a bit obscure.
3-D Distance Formula: √((x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2)
Equation of a plane: ax + by + cz = d, where “d” is a constant and “a,” “b,” and “c” are real coefficients.
Equation of a line given two points: P + t(Q – P), where Q and P are marked by different coordinates,
and “t” is any real number. For example, let us have points P (1, 2, 3) and Q (4, 3, 2).
The vector for this line could be marked as <x, y, z> = <1, 2, 3> + t<3, 1, -1>. Algebraically refining this,
we have:
x = 1 + 3t
y=2+t
z = 3 – t.
Don’t worry a lot about the latter two. Like I said, 3-D analytical geometry is very obscure. Equations of
planes and lines are calculus 3 material!
That’s basically all there is to know. Usually, AMC problems dealing with 3-D geometry will be
seen at the beginning or end. Don’t expect more than two in your “qualifying range” of problems (again,
this is problems 1-15 on the AMC 12 and problems 1-20 on the AMC 10). Let’s get to some examples.
A student is working with a graduated cylinder with a radius of 4 cm. The cylinder is filled with enough
water so that the water level is 15 cm. If a tablet of volume 30π cm3 was dropped into the cylinder, by
how many cm will the water level rise?
We start by finding the volume of the water: π x (16) x (15) = 240π cm3. Adding the volume of the tablet
gets us 270π cm3. Dividing this by 16π gets us our second height: 16.875 cm. Subtracting by 15 cm, we
get 1.875 cm.
A plane cuts through the midpoints of three edges of a regular tetrahedron with side length 6, all of
which have a common vertex, as shown. Find the area of the plane’s intersection with the tetrahedron.
Let’s take this to one single face.
We notice that a cut in the form of the midline creates two similar triangles, with the smaller one being
length 3. Since all of these cuts are similar, the intersection is simply an equilateral triangle with length
3. Hence, the answer is (√3/4)(9) = 9√3/4.
Two congruent square pyramids are inscribed in a cylinder with each vertex of their bases tangent to
their respective circular bases on the cylinder. Also, the apexes of the two pyramids are tangent, as
shown. If the distance from the apex to one vertex of the pyramid is 8, and the radius of the cylinder is
3, what is the length of the longest line segment to be contained in this cylinder?
3
8
Let’s look at one of the circular bases.
Playing with our radius, we see that we can use half the diagonal of the square and the edge of the
pyramid with side-length 8 to find the height of one pyramid. Using the Pythagorean Theorem, we get:
h2 + 32 = 82  h = √(55). The entire height of the cylinder is twice this height: 2√(55).
Clearly, the longest line segment within a cylinder is its diagonal. We use Pythagoras again with the
diameter of one of the circular bases and the height of the cylinder. Doing so, we get that the answer is:
36 + 220 = d2  d = 16.
TIPS:
 Really notice similarity ratios when you slice off the top portion of a cone. Apply this reasoning
analogously with pyramids. Also note that, when the top part of a cone gets cut off by a plane
parallel to the base, the radius of the base of that smaller portion of the cone is smaller than the
original radius. Again, apply similarity ratios here to figure out what that radius is.
 Try drawing effective diagrams right off the bat when dealing with 3-D geometry. Again, it’s
harder to visualize than 2-D geometry.
 One tip I forgot to mention that applies not only to 3-D, but 2-D as well: when circles or other
things are tangent to one another, try drawing new lines that connect useful points to such
tangents.
I should have taken the time to make more sample problems… Oh, well. What I will do, however, is post
a link to LOTS of AMC problems. I’ll go over ones that are important or out of the box. With 3-D
geometry, I’ll try to go over as many as possible.