19) In the tomb of Pharaoh Tetrankamon, an emerald was found, shaped into a tetrahedron (a pyramid with a triangular base) whose edges measure in millimeters $54, 32, 32, 29, 27, 20$. Indicating with $A, B, C, D$ the vertices of the tetrahedron and knowing that $A B$ is 54 mm long, how many millimeters is $C D$ long?
(A) 32
(B) 29
(C) 27
(D) 20
(E) cannot be determined.
19) The answer is (D)

The edges $A C$ and $B C$, as well as the edges $A D$ and $B D$, together with $A B$ form a triangle. Therefore, $A C + B C > 54$ and $A D + B D > 54$. Since the sum of 20 with any of the numbers 32, 32, 29, 27 is less than 54, the edge of length 20 cannot be any of $A C$, $B C$, $A D$, $B D$, and therefore must be $C D$ (the edge opposite to $A B$).

## SECOND SOLUTION

To prove that the side $C D$ is 20 units long, we show that no other side of the tetrahedron can be 20 units long. If this were the case, the side of length 20 would share an endpoint with the side $A B$ (note that $C D$ is the only side of the tetrahedron that does not share endpoints with $A B$). However, this would mean there is a face of the tetrahedron with one side of length 54, one side of length 20, and a third side whose length could be 32, 29, or 27. But this is not possible because in any triangle, the longest side must be greater than the sum of the other two.

## OBSERVATION

To convince oneself of the existence of a tetrahedron with the given measurements, one can perform the following construction. Consider a triangle $A B C$ with sides $A B = 54$, $A C = 32$, and $B C = 27$, and the triangle $A B D$ with sides $A B = 54$, $A D = 32$, and $B D = 29$. Suppose also that $C$ and $D$ are in different half-planes relative to the line $A B$. With some calculation (not trivial), it can be seen that in this configuration, the distance between $C$ and $D$ is $>20$. Now suppose we rotate the triangle $A B D$ 180 degrees around the line $A B$. At the end, the point $D$ will have moved to a point $D'$, located on the same side of $C$ relative to the line $A B$. In this configuration, it is (with some calculation) that $C D' < 20$. Therefore, during the rotation, there must be a position where the distance between $C$ and $D$ is exactly 20. In this position, the points $A B C D$ form the desired tetrahedron.