The Tetrahedron
Coordinate-based proof:
B
21
11
R
Q
C
A
P
20
Proof that the four triangles of 4ABC are congruent:
• We know that CQ and QB, BR and RA, AP and P C, are equal in length to each other by definition of a
midpoint.
• We also know that each of 4QCP , 4P RA, and 4QRB are similar to 4ABC through SAS.
• Since 6 BQR and 6 BCA, 6 QP R and 6 CAB, and 6 P RQ and 6 ABC are identical through similar triangle
properties, we know that QR and CA, QP and AB, and P R and BC are parallel.
• Knowing that the abovementioned sets of lines are parallel, we can use ASA to prove that all triangles are
congruent.
Finding the area of the base:
The general formula for the volume of a tetrahedron, or pyramid, is 1/3ah, where a is the altitude, or point-plane distance
from the vertex to the base, and b is the area of the base. In order to find the area of the base, we must find its height, which
is equal to its altitude. By projecting the base onto a coordinate system and assigning coordinates (0,0) and (10,0) to A and
C, respectively, we find that the altitude is equal to the y-value of the coordinates of point B. Since we know the distances
between every vertex, we can apply the distance formula to solve for the coordinates of point B, (x,y) or:
2
11
(x − 0) + (y − 0) =
2
2
21
2
2
(x − 10) + (y − 0) =
2
2
2
√
Since y must be positive, we have (x, y) = (1, 3
13
2 ),
meaning that the height of the base is
1
√
3 13
2 .
y
B
11/2
√
3 13
2
21/2
C
A
10
x
Finding the altitude of the tetrahedron:
Now that we know the altitude and coordinates of the vertices of a base, we can apply the 3-dimensional distance formula
(the proof of which is trivial) to solve for the altitude of the entire tetrahedron. Since the altitude of the tetrahedron is
perpendicular to the base, it is equal to the y-value of the vertex
it intersects. By projecting the base onto a 3-D coordinate
√
3 13
system with the vertices occupying (0,0,0), (10,0,0), and (1, 2 ,0) (yes, the figures are not to scale, but at least they make
sense), we can apply the distance formula for the coordinates of point D, (x,y,z) based on the known lengths of the edges:
2
21
(x − 0) + (y − 0) + (z − 0) =
2
2
11
2
2
2
(x − 10) + (y − 0) + (z − 0) =
2
√ !2
3 13
2
2
= 102
(x − 1) + (y − 0) + z −
2
2
2
2
15
). By plugging this into the formula for volume, we have:
Solving, we have a positive (x, y, z) solution of (9, √1813 , 2√
13
√
1
3 13
18
v = ∗ 10 ∗
∗√
3
2
13
v = 45cm3
y
z
D
10
B
11/2
11/2
C
21/2
10
21/2
2
A
x
Orthocenter-based proof:
Orthocenter of the net is contained in altitude:
If we reflect the tetrahedron over the plane containing its base and connect the topmost and lowest vertices,
we find that this line is perpendicular to the shared base and contains both tetrahedral altitudes.
D
10
B
11/2
11/2
21/2
10
C
21/2
A
E
Next, if we construct the vertical altitudes of each side meeting the topmost vertex and unfold the net, we
have the figure below.
B
21
R
11 Q
C
P
A
20
Notice that each of the altitudes are perpendicular to the bases of the smaller triangle they are contained
in, and since the bases of these small triangles are parallel to the larger 4ABC, they are the three altitudes
of 4ABC, and meet at its orthocenter.
Since these altitudes of the smaller triangles all meet at the top vertex of the tetrahedron (therefore
sharing an endpoint with the tetrahedral altitude), and are rotatable only across the axis formed by the
base of their corresponding face, the altitude of the tetrahedron is coplanar with each altitude of the three
faces that has an endpoint at the topmost vertex. Additionally, we can show through three points per face,
the tetrahedral vertex, the orthocenter of the net, and the point at which each altitude intersects the base
3
of its face, that the three planes formed from each face intersect at the line containing the altitude. Thus,
the line connecting the orthocenter of the net to the tetrahedral vertex is the altitude of the tetrahedron,
meaning the orthocenter has the same x and z coordinates as the vertex.
Since we already know the coordinates of all points A, B, and C, we can find the orthocenter through
solving x = 9 and y = 3√1813 x − 2√9313 by taking and solving for perpendicular slopes. Solving, we get
coordinates (x, y) = (9, 2√1513 ). Since we already know the coordinates for the other three vertices of the
tetrahedron (see coordinate-based proof), we can solve a simpler set of equations, namely
!2
15
(9 − 10) + (y − 0) + √ − 0
2 13
2
2
!2
15
(9 − 0) + (y − 0) + √ − 0
2 13
2
for positive y =
√18 .
13
=
2
11
2
21
=
2
2
2
Again using the formula for volume, we have:
√
1
3 13
18
v = ∗ 10 ∗
∗√
3
2
13
v = 45cm3
y
B
11/2
21/2
C
A
x
10
Cuboid proof:
Tetrahedron can be inscribed in cuboid:
E
F
G
a
H
A
B
c
C
D
b
4
If we have a cuboid (a rectangular prism) of dimensions a, b, and c, we find that by connecting four vertices
through diagonals as shown below, the resultant shape enclosed has the same edge proportions and arrangement as the tetrahedron. Specifically, opposite edges, such
√as CE and BH are of equal lengths, which
can be shown through the fact that both are equal in length to a2 + c2 through the Pythagorean theorem.
Since we know the lengths of all sides of the tetrahedron, and therefore all lengths of the diagonals,
we can set up a system of equations to solve for a, b, and c, namely:
11
a +b =
2
2
2
2
2
21
b +c =
2
2
2
c + a = 102
2
2
again
using the Pythagorean
theorem. This system is easy enough to be solved by hand, for which we have
√
√
10, 92 , and 3 ∗ 10 for a, b, and c. Since we know the dimensions of the cuboid, we can find its volume
through v = lwh for v = 135cm3√
, subtracting
the volumes of the four tetrahedrons CGEH, DCHB,
√
CEBA, and BHF E, or ( 12 )( 92 )(3 ∗ 10)( 10)( 13 )(4) = 90cm3 . Thus, the volume of CEBH is equal to the
volume of the cuboid minus the four volumes of the other tetrahedrons, or
135 − 90 = 45cm3
General solution:
Volume of cuboid:
We can generalize from the system of equations presented in the cuboid proof to solve for the volume of
any tetrahedron of congruent faces. Specifically, for any such tetrahedron of side lengths d, e, and f , we
have:
a2 + b 2 = d 2
b2 + c2 = e2
c 2 + a2 = f 2
which can easily be solved for a, b, and c in terms of d, e, and f , namely
√ 2
d − e2 + f 2
√
a=
2
√ 2
d + e2 − f 2
√
b=
2
√ 2
−d + e2 + f 2
√
c=
2
The volume of the cuboid is abc, or
q
(d2 − e2 + f 2 )(d2 + e2 − f 2 )(−d2 + e2 + f 2 )
√
8
The volume of the tetrahedron is 1/3 that of the cuboid, or
q
(d2 − e2 + f 2 )(d2 + e2 − f 2 )(−d2 + e2 + f 2 )
√
8∗3
q
=
(d2 − e2 + f 2 )(d2 + e2 − f 2 )(−d2 + e2 + f 2 )
√
72
5