User:Jeff10

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PROBLEM I have constructed a rectangular figure. After drawing both diagonal, I will have constructed four congruent triangles. The length of the hypotenuses are both 17 units. Given that the length and width of this rectangle are both integer lengths, what is the area of one of the four triangles created? After you have found that, find the lengths and angles of the triangle. You may use a calculator if necessary.

SOLUTION

Given the hypotenuse is 17 units, you may want to list all of the possibilities such that a square and a number added together is 289. A better way is to use the difference of squares and actually hunt for the values, which are length of 15 and width of 8. When you multiply them together, you will get the area of the rectangle, which is 120, but you want the area of one of the four congruent triangles, which is 30. With the knowledge of trigonmetry, you can use sine or cosine to find the angles.

Graph Theory Solutions

1. Let's turn this into a graph. Let's draw a graph with 10 vertices and we want to start adding as few edges as possible such that we can draw any one of the 252 sets of 5 so that this subgraph would already contain 2 edges (2 pairs of vertices and 1 alone vertex). Now obviously I could just draw every single edge, but that's too easy and boring. Of course, just removing all of the edges of one vertex is also way too boring, because of course there is going to be that isolated vertex and two guaranteed pairs of vertices. Now if we choose any 5 vertices among the 10, at least 4 of them have to be a member of the group (not the isolated vertex), so a minimum of 6 edges is needed. Now let's disconnect all of the edges of another vertex, leaving us with our united group of 8 and 2 isolated vertices. So choosing any 5 of the vertices would mean at least 3 of them have to be in the united group. (This graph has 28 edges because we removed 9 edges from the first vertex and 8 from the second.) Now, do you think this is it? I don't think so. Let's go even further! Bleh, I am getting tired of having to go through all of this, but let's not give up. Suppose we remove the 7 edges of another vertex to get only 21 edges. Now this means that now we have 3 isolated vertices, and this would not be good, so let's connect two of the three isolated vertices together to form another edge. Now we have our two required edges given that we go into the extreme case of the 2 in the group and 3 excluded vertices. Now we have 22 edges as a minimum, which is a record. I really don't want to have to do all of this work for 9 million hours! This time, I am going to start grouping, and I am going to hope for a better result. Let's say I grouped so I would try to derive the fewest number edges. Right now, the two main candidates are 4+4+2 and 4+3+3. Now I have 6+6+1=13 for the first case and 6+3+3=12 for the second case, so it seems like 12 edges is the best record that is even possible to go for, so I might as well just say 12 is my answer. Now, I do have to prove it, but that's probably going to take very long.

2. We need to find the degree of each of the 16 squares. Now there are 3 different types of squares in this 4 by 4 checkerboard. The corner squares, the edge squares, and the center squares. Each of the corner squares have a degree of 2. This does look like we are pretty clear, but what screws this whole thing up is that the edge and center squares all have degree 3, because there are 3 squares from each of the 12 squares that you (the knight) can jump to. Therefore, by what we saw earlier about the integer number of paths, we cannot jump to every single square on the 4 by 4 checkerboard as a knight.

Sidenote: If you are able to draw a specific graph without taking your pencil off the paper, it is called a Eulerian path. That will be something you can find.