This is the only possible arrangement where three arrows are all pointing to the same vertex. You can always spin the tetrahedron so the head of the fourth face's arrow is on the bottom and pointing wherever you like.
Rather than trying to come up with a three-dimensional drawing that would be clear, I am going to use the graph-theoretical illustration of the tetrahedron smashed flat on a table. I am also indicating the arrows by red circles where the arrow's head would be, with the understanding that the bottom face's circle will be drawn "outside" the diagram. I will also call the faces of the tetrahedron "front", "left", "right", and "bottom"; if you imagine that the tetrahedron is lying on a table and you are looking down on it, it should be fairly evident what I'm talking about.
For this solution, we'll break the different tetrahedral paintings into four different classes. If you label each vertex by the number of arrows that are pointing to it, you'll get either (3,1,0,0), (2,2,0,0), (2,1,1,0), (1,1,1,1). Each of these cases is covered below.
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This is the only possible arrangement where three arrows are all pointing to the same vertex. You can always spin the tetrahedron so the head of the fourth face's arrow is on the bottom and pointing wherever you like. |
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With two arrows pointing to the same vertex and the remaining two arrows also pointing to the same vertex, there are two options. If you spin the tetrahedron so that the left and right faces point up, then the front face will either point left or right. |
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If two arrows are pointing to the same vertex and the remaining two arrows are pointing to different vertices, then fix the position of the tetrahedron so that the two arrows are pointing up and not on the front face. Then the arrow on the front face can either be on the left or right and the arrow on the bottom face has two choices as well. |
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If all four arrows are pointing at different vertices, then there are two options. Either the tails of the arrows are adjacent to two edges (left) or four edges (right). The image on the right is equivalent to its mirror image, as you can see for yourself if you imagine rotating the tetrahedron so that the right face is in front and the bottom face is still on the bottom. |
This is a total of 9 different paintings.
Notes: One of my (several) laments during the postmortem of this test is that I did guess that this would be the question that would be asked after reading the instructions, but I didn't take the time to solve it before the test. Like the math questions in last year's tests, I avoid them during the test, because I don't think that the stress of timed competition leads to worthwhile calculations. Even spending about a half-hour after the test on it, I still was not positive I had the right answer until I checked the answer key, and risking losing five points on the prospect of gaining ten points is a tough bet to take in my opinion.