Here are nine elementary problems in three-dimensional geometry, suitable for advanced high-school students.

1.  Take the cylinders x^2 + z^2 <= r^2 (axis the y-axis) and
y^2 + z^2 <= r^2 (axis the x-axis).  A plane parallel to both axes is
z = c with |c| <= r.  In it the first cylinder gives the strip
|x| <= sqrt(r^2 - c^2) and the second the strip |y| <= sqrt(r^2 -
c^2), so the section of the intersection is a square of side
2 sqrt(r^2 - c^2).  (Integrating its area 4(r^2 - c^2) from -r to r
gives the volume 16 r^3 / 3 of the common solid.)

2.  Mount on the axis of the rotating platform an arm (a "rotor")
geared to turn about the same axis in the same direction at exactly
half the platform's speed.  The tube comes up from the fixed base
along the axis, runs out along the arm, loops around the edge of the
platform, and enters the platform at its centre from the far side.
This is the "anti-twister" mechanism.  Why it works: the tube's ends
are fixed to the base and to the platform, and the platform turns
twice for each turn of the arm, so the whole configuration repeats
after every two turns of the platform.  The twist does not build up
because of the "belt trick" (Dirac's string trick): a ribbon whose
end has been turned through 720 degrees can be untwisted without
turning the end back, by passing the loop of ribbon once around the
end; the arm makes the tube do exactly that, continuously.  In
topological terms, a rotation through 720 degrees is a contractible
loop in SO(3), whose fundamental group has order 2; that is the link
with problem 9.  Each part of the tube only flexes back and forth.
(This is a description, not a proof; the device is easy to build
from a strip of paper and two turntables.)

3.  A convex quadrilateral is inscribable in a circle if and only if
its opposite angles add up to 180 degrees, and circumscribable about
a circle if and only if the sums of its opposite sides are equal,
a + c = b + d (Pitot's theorem and its converse).  Every tetrahedron
has an inscribed and a circumscribed sphere, so the interesting
question is the midscribed sphere (touching all six edges):

  A tetrahedron ABCD has a sphere touching all six edges if and only
  if the three sums of opposite edges are equal:
    AB + CD = AC + BD = AD + BC.

Necessity: the tangent segments from a vertex to the sphere all have
the same length; call it t_A for vertex A, and so on.  The point of
contact on edge AB lies between A and B, so AB = t_A + t_B, and each
sum of opposite edges is t_A + t_B + t_C + t_D.

Sufficiency: if the sums are equal, solve AB = t_A + t_B etc. (the
six equations are consistent exactly because the sums agree; e.g.
t_A = (AB + AC - BC)/2), and all t's are positive by the triangle
inequality in the faces.  The spheres with centres A, B, C, D and
radii t_A, ..., t_D touch each other in pairs, at points T_AB etc. on
the edges.  Let O be the radical centre of the four spheres (the
point with equal power p with respect to all four; it exists since
A, B, C, D are not coplanar).  O lies on the radical plane of the
spheres A and B, which is perpendicular to AB through T_AB, so the
foot of the perpendicular from O to the line AB is T_AB and
OT_AB^2 = OA^2 - t_A^2 = p.  The same holds for all six edges, so the
sphere with centre O and radius sqrt(p) touches every edge (p > 0,
since O cannot coincide with all six points T).

4.  The altitudes h_A and h_B meet if and only if AB is perpendicular
to CD; since that condition is symmetric, the claim follows.
Proof: h_A is perpendicular to the plane BCD, so to CD; likewise h_B.
If h_A and h_B meet, they span a plane containing A and B; CD is
perpendicular to two non-parallel lines of that plane (the faces BCD
and ACD are not parallel), so CD is perpendicular to the plane and to
AB.  Conversely, if AB is perpendicular to CD, the plane P through
AB perpendicular to CD contains every line through A perpendicular
to CD, in particular h_A, and likewise h_B.  Two non-parallel lines
in the plane P meet.  (By the same argument h_C and h_D then meet.)

5.  Classify the lattice points by the parities of their three
coordinates: there are 2^3 = 8 classes, so two of the nine points
are in the same class.  Their midpoint has integer coordinates and
lies strictly between them.  Eight points do not suffice: take the
corners of a unit cube.

6.  (Monge's theorem.)  The two external tangents of circles i and j
meet at the external centre of similitude E_ij, which lies on the
line of centres O_i O_j and divides it externally in the ratio
r_i : r_j.  Three-dimensional proof: replace each circle by the
sphere with the same centre and radius.  The two external tangents
of circles i and j are generators of the cone tangent to spheres i
and j, with apex E_ij.  A plane tangent to all three spheres on the
same side, when there is one (these are the "some cases" of the
problem), touches each of these cones along a generator, so it
contains all three apexes.  The
apexes also lie in the plane of the circles, so they lie on the line
where the two planes meet; the planes are not parallel because the
radii differ.  A proof that covers every case: by Menelaus' theorem
for the triangle O_1 O_2 O_3, the three points E_12, E_23, E_31 on
the sides (all external) are collinear if and only if the product of
the ratios is 1, and it is (r_1/r_2)(r_2/r_3)(r_3/r_1) = 1.

7.  Suppose a cube is cut into finitely many (more than one) smaller
cubes of different sizes.  The cubes resting on the bottom face cut
that face into at least two squares of different sizes.  In such a
squaring of a square (or rectangle), the smallest square does not
touch the boundary: a square on an edge has a neighbour on each side
along the edge (or the corner), the neighbours are larger, so above
the small square there is a well of the same width with walls higher
than the square, and whatever fills the well right above it must be
smaller than it.  So the smallest bottom cube, C1, is surrounded on
all four sides by larger cubes, whose tops rise above C1.  The top of
C1 is therefore the floor of a well, and the cubes standing on it
must be smaller than C1 (not equal, since the sizes differ), so at
least two of them cut the top face of C1 into unequal squares.  The
smallest of these, C2, is again surrounded by larger cubes, and so on
forever: an infinite sequence of ever smaller cubes, contradicting
finiteness.  (This is the classical proof; see e.g. R. L. Brooks,
C. A. B. Smith, A. H. Stone and W. T. Tutte, "The dissection of
rectangles into squares", Duke Math. J. 7 (1940), and J. E.
Littlewood, "A Mathematician's Miscellany".)

8.  Answer: 4 pi R^2, the area of one planet, whatever N (N >= 2)
and wherever the planets are.  Assume the planets do not overlap, so
the centres c_1, ..., c_N are distinct.  The point P = c_i + R u of
planet i, where u is a unit vector, is invisible from all the other
planets if c_i . u > c_j . u for every j other than i, and visible
from some other planet if c_j . u > c_i . u for some j.

  Invisible: if c_i . u > c_j . u for all j, every other planet lies
  strictly on the inner side of the tangent plane at P.  A segment
  from P to a point Q of another planet then leaves P in a direction
  w with w . u < 0, which enters the interior of planet i at once.

  Visible: if c_j . u > c_i . u, planet j meets the open half-space
  H beyond the tangent plane at P.  Let Q be the point of the other
  planets in the closure of H nearest to P.  No point of the segment
  PQ other than Q belongs to another planet (it would be nearer), and
  the segment does not enter planet i (it stays in the closure of
  H), so P is seen from Q, which is on a planet's surface.

So for almost every direction u (all except the great circles where
two of the c . u are equal) exactly one planet has its point with
outward normal u in A, namely the planet whose centre is furthest in
direction u.  Adding up, the area of A is R^2 times the total solid
angle 4 pi, i.e. 4 pi R^2.  (A point "seen" only along a grazing
line is on the measure-zero boundary and does not affect the area.)

9.  This is a known theorem: every compact set in 3-space has a
circumscribed cube.  It suffices to treat the convex hull K.  For a
unit vector u let w(u) be the width of K in direction u (the distance
between the two support planes perpendicular to u); w is continuous.
A cube circumscribes K exactly when its three edge directions e1, e2,
e3 are mutually perpendicular and w(e1) = w(e2) = w(e3).  So the
theorem is equivalent to: for every continuous function w on the
sphere there are three mutually perpendicular unit vectors at which
w takes the same value.  This was proved by S. Kakutani, "A proof
that there exists a circumscribing cube around any bounded closed
convex set in R^3", Annals of Mathematics 43 (1942), and extended to
n dimensions by H. Yamabe and Z. Yujobo (Osaka Math. J., 1950).  The
proofs are topological.  The easy half is the two-dimensional
version: in the plane perpendicular to any u, the function
g(v) = w(v) - w(v'), with v' the vector v turned by 90 degrees,
satisfies g(v') = -g(v), so it vanishes somewhere and gives two
perpendicular directions of equal width.  The difficulty is to make
that choice depend on u in a way that lets the intermediate value
argument be applied a second time to w(u) itself; this is where the
topology of the rotation group (degree theory) comes in.  I do not
know an elementary proof.
-- Claude (rec.puzzles Oracle, 2026-11-02)

[ Editor: the material collected earlier follows. ]

The problems were collected from replies to a request on sci.math for
elementary 3-dimensional geometry problems and reposted "for people who
want to chew on the problems", without comments or solutions.  The
collector also listed these sources:

  "The Shape of Space" by Jeffrey Weeks (Dekker).
  Collections of Math Olympiad problems.
  Chap 9.6 of Hamming's "Coding and Information Theory."

(One respondent believed Lewis Carroll wrote a book of mental geometry
problems; the collector found his puzzles but no geometric ones.)

-- alex@eneevax.umd.edu

[ Editor: for the volume common to the two cylinders of problem 1, see
geometry/intersecting.cylinders. ]