Tetrahedral Volume Sum

First consider a side with length 3. All but one edge is adjacent to this side, so we know that both edges with length 3 are adjacent to one of the edges with length 5. We then claim that one of the sides is a 3-4-5 right triangle. We shall prove this by attempting to construct a tetrahedron that does not have such a face. Consider a face with one side of length 3 and one other side of length 5. From the Triangle Inequality, we cannot have the remaining side have length 2, thus it must have length 3, 4, or 5.

Case 1: The remaining side has a length of 3. We now have a triangle with edge lengths 3, 3, 5. The remaining three edges of the tetrahedron must have lengths 2, 4, and 5. We apply the Triangle Inequality again to get that the edges with length 2 and 5 cannot be on the same face as the edges with length 3. It follows that they must be on the face containing an isosceles 2-5-5 triangle, and the edge with length 4 is adjacent to both edges with length 3. Then the most recently created edge with length 5 completes a 3-4-5 triangle.

Case 2: The remaining side has a length of 4. We're done.

Case 3: The remaining side has a length of 5. We now have a triangle with edge lengths 3, 5, 5. The remaining three edges of the tetrahedron must have lengths 2, 3, and 4. We apply the Triangle Inequality again to get that the edges with length 2 and 3 cannot be on the same face as the edges with length 5, thus they must be on the face containing an isosceles 2-3-3 triangle, and the edge with length 4 is adjacent to both edges with length 5. Then the most recently created edge with length 3 completes a 3-4-5 triangle.

This shows that there must be a 3-4-5 triangular face on our tetrahedron. Label the vertices of this face A, B, and C, where BC=3, CA=4, and AB=5. Let D be the fourth vertex of the tetrahedron. Since the volume of the tetrahedron is 1/3 times the base times its respective height, we must maximize the height from vertex D. We will now restrict the number of possible configurations by using the Triangle Inequality. Assume that CD is 5. Then BD and AD are 2 and 3, but then triangle ABD is a degenerate 2-3-5 triangle. Thus CD cannot be 5 - it must be 2 or 3. Assume that AD is 3. Then BD and CD are 2 and 5, but then triangle BCD is a degenerate 2-3-5 triangle. Thus AD cannot be 3 - it must be 2 or 5. This reduces the number of possible configurations of the tetrahedron to 3 (listed below):

Case X: $AD=5, BD=2, CD=3$ . Then triangle ACD is another 3-4-5 right triangle, and so angle DCA is a right angle. Hence the plane containing triangle BCD is perpendicular to that containing ABC. It follows that the height of the tetrahedron from vertex D is simply the height of triangle BCD from vertex D. Heron's Formula can be used to find the area of the triangle; it comes out to be $2\sqrt{2}$ . Thus the height of BCD from vertex D is $4\sqrt{2}/3$ , which is the height of this tetrahedron. The area of triangle ABC is 6, so the volume of this tetrahedron is $4\sqrt{2}/3 \times 6 \times 1/3=8\sqrt{2}/3$ .

Case Y: $AD=5, BD=3, CD=2$ . It's easy to see that the height of the tetrahedron from vertex D is at most the height of triangle BCD from vertex D. However, the height of this triangle was found previously, it's $4\sqrt(2)/3$ . Thus the height in this case is at most the height in Case X.

Case Z: $AD=2, BD=5, CD=3$ . Here we change faces: We consider the height of the tetrahedron from vertex C this time. This is at most the height of triangle BCD from vertex C. Triangle BCD is a 3-3-5 isosceles triangle, so the height from vertex C is $\sqrt{11}/2$ . The area of face ABC (which is a 5-5-2 isosceles triangle) can easily be calculated to be $2\sqrt{6}$ , so the volume of this tetrahedron is at most $2\sqrt{6}\times \sqrt{11}/2 \times 1 /3=\sqrt{66}/3$ . However, $\sqrt{66}<8\sqrt{2}$ , so this volume is strictly less than the volume computed in Case X.

Therefore Case X is the optimal case. The volume of the tetrahedron in this case is $8\sqrt{2}/3$ , so $a+b+c=13$ .

Lemma 1: There must be a face of the tetrahedron that is a 3-4-5 right triangle.
Proof: Consider the edge of length 4. If the opposite edge has length 2, then the two faces containing the edge of length 4 must be 3-3-4 and 4-5-5 triangles, or we are done. However, in this case, there exists a face with sides 2-3-5, which does not satisfy the triangle inequality. If the opposite edge has length 3, then the face containing the edge of length 4 and the other edge of length 3 must be a 2-3-4 triangle, or we are done. However, in this scenario, we again have a face with sides 2-3-5. Finally, if the opposite edge has length 5, then the face containing the edge of length 4 and the other edge of length 5 must be 2-4-5, or we are done. Again, there exists a face with sides 2-3-5. We conclude that there must be a face that is a 3-4-5 right triangle. $\square$

Now consider the 3-4-5 triangle as the base of the tetrahedron. Let this triangle be $\triangle ABC$ with $AB=4, BC=3, AC=5.$ Let the fourth vertice of the tetrahedron be $D.$ We see that there are three cases.

Case 1: $AD=5, BD=2, CD=3.$ The projection $D'$ of $D$ onto plane $ABC$ clearly lies outside $\triangle ABC$ . In $\triangle DBC$ , which is 2-3-3, let the foot of the altitude from $D$ to $BC$ be $D''$ . $DD''$ can be found to be $\frac{4\sqrt{2}}{3}$ . Let the height of the tetrahedron be $DD'=x.$ Thus we have that the distance from $D'$ to $BC$ is $DD''=\sqrt{\frac{32}{9}-x^2}.$ Furthermore, $BD''=\frac{2}{3}$ also from the Pythagorean Theorem. It follows that $AD^2=(AB+D'D'')^2+BD''^2+DD'^2=\left(4+\sqrt{\frac{32}{9}-x^2}\right)^2+\frac{4}{9}+x^2=25.$ We solve this equation to obtain $x=\frac{\sqrt{1823}}{24}.$

Case 2: $AD=5, BD=3, CD=2.$ With reasoning analogous to that of Case 1, we find that the height of the tetrahedron is $x=\frac{4\sqrt{2}}{3}.$ -

Case 3: $AD=2, BD=3, CD=5.$ Here, the maximum value for the height of the tetrahedron $x$ is equal to the length of the altitude from $D$ to $AB$ in $\triangle ABD$ . This triangle has sides 2-3-4, and we find the length of the altitude to be $\frac{3\sqrt{15}}{4}$ , which is smaller than each of the values from the previous cases, so this case will not yield the maximum volume. A quick check with the triangle inequality tells us that are no other cases. The value of $x$ is largest in Case 2, so our desired volume is $\frac{8\sqrt{2}}{3}$ which gives us an answer of $8+2+3=13.\ \blacksquare$

Define $s_x$ by the edge of the tetrahedron with side length $x$ . So, there is one $s_2$ and $s_4$ , and two $s_3$ s and $s_5$ s.

First, we claim that one of the face of the tetrahedron is a triangle where two of its sides are $s_2$ and $s_3$ . There is only one edge which does not touch $s_2$ . But, since there are two $s_3$ s, one of them must have a common vertex with the $s_2$ . So, our claim must be correct.

Consider the triangle mentioned in our claim as the base of the tetrahedron. The possible third side for this triangle are $s_3$ and $s_4$ , since $s_2, s_3, s_5$ do not form a triangle. Since the hypotenuse of any right triangle must be longer than its other sides, the height of our tetrahedron, $h$ , cannot exceed the shortest unused edge i.e. the edges other than three edges in the base.

If the base are $s_2, s_3, s_3$ , then $h \le 4$ and $A = \sqrt{4\times2\times1\times1}$ by the Heron's formula. Here, $h$ and $A$ denotes the height and the area of the base, respectively. So, $V \le \frac{1}{3}Ah = \frac{8\sqrt{2}}{3}$ , where $V$ is the volume of the tetrahedron.

The other case is similar. If the base are $s_2, s_3, s_4$ , then $h \le 3$ and $A = \sqrt{4.5\times2.5\times1.5\times0.5}$ , so $V \le \frac{1}{3}Ah = \frac{3\sqrt{15}}{4}$ .

Since $\frac{8\sqrt{2}}{3}\ > \frac{3\sqrt{15}}{4}$ , the maximum possible volume is $\frac{8\sqrt{2}}{3}$ . We are left to show that this maximum value can be achieved. Put $s_4$ such that it is perpendicular to the base and is connected to the both $s_3$ s. Then, the other two edges must have side length $5$ by the Phytagorean theorem. So, such a tetrahedron can be formed, thus proving the result.

So, the answer to our problem is $8 + 2 + 3 = 13$ .

If it is 2,3,3 then the volume<=1/3 base height<=1/3 base 4=8sqrt2/3 and it can be attained If it is 2,5,5 then the other one shares 2 as side must be 2,3,4 now the volume<=1/3 S(2,3,4) 3<8sqrt2/3

Label the tetrahedron $WXYZ$ . Consider the edge of side length 2. Let the vertices be $W, X$ . Since $WXY$ and $WXZ$ are triangles, they must satisfy the triangle inequality. These are the following possibilities for the triangles: A. $\{ 3, 3 \}$ , B. $\{ 5, 5\}$ , C. $\{ 4, 5\}$ , D. $\{ 3,4\}$ . Hence, the only possibilities for $WXY$ and $WXZ$ are: A, B, or A, C, or B, D. Thus, we consider the the following cases:

Case 1. A, B. Without loss of generality, we have $WX = 2$ , $WY=3$ , $XY=3$ , $WZ=5$ , $XZ=5$ and so $YZ=4$ . Thus, $WYZ$ and $XYZ$ are both 3-4-5 triangles, hence right angled. This tetrahedron has base $WXY$ (with area $\frac {1}{2} \times 2 \times \sqrt{3^2-1^2}$ ) and height $YZ = 4$ , so has volume $\frac {1}{3} \times 4 \times \sqrt{8} = \frac {8\sqrt{2}}{3}$ .

Case 2. A, C. We have 2 subcases, depending on what $WZ, XZ$ are equal to. However, these tetrahedrons are reflections of each other, so have the same volume. Without loss of generality, we have $WX = 2$ , $WY=3$ , $XY=3$ , $WZ=4$ , $XZ=5$ and so $YZ = 5$ . Notice that this has the same base as Case 1, so let's consider the height of the tetrahedron. From $Z$ , drop a perpendicular to $WX$ , intersecting at $E$ . Since $2^2 + 4^2 > 5^2$ , $WXZ$ is an acute triangle. Thus, $WE < 4$ , which also means that the height of the tetrahedron is less than $4$ . Thus, the volume will be less than Case 1.

Case 3. B, D. Similarly, we have 2 sub cases, depending on what $WZ, XZ$ are equal to. However, these tetrahedrons are reflections of each other, so have the same volume. Without loss of generality, we have $WX = 2$ , $WY=5$ , $XY=5$ , $WZ=3$ , $XZ=4$ and so $YZ=3$ . Consider base $WZX$ with height $YZ$ , then the volume of the tetrahedron is $\frac {1}{3} [WZX] h_C \leq \frac {1}{3} \cdot ( \frac {1}{2} WX \times WZ ) \cdot YZ = 3$ . This is clearly less than Case 1.

Therefore the largest volume possible is $\frac {8\sqrt{2}}{3}$ . Hence $a + b + c = 8 + 2 + 3 = 13$ .