**Problem:**

A square pyramid with base and vertex has eight edges of length 4. A plane passes through the midpoints of , , and . The plane’s intersection with the pyramid has an area that can be expressed as . Find .

**My Attempt:**

(Remember pyramids?)

I didn’t think this method would turn out bashy, but it was. Oh no.

Let’s make a coordinate system to the pyramid where and . Then, is the height of an equilateral triangle with a length of 4, so . Let be the center of the square. Then, , so . This means,

Using midpoint formulas. Since and , the cross product is which can be simplified to . So we know that the equation for the plane containing , , and is in the form . Plugging in the value of the point of shows that the equation of the plane is

.

Now we have to intersect that with and to see what the other points are that bound the figure whose area we have to find. First, . The vector , and , so , , and . Plugging in those values for the equation of the plane, we get that and intersect when , or at . We’ll call that point . Similarly, and intersect at . We’ll call that point .

Now, time to find the area of . This divides into 1) an isosceles triangle with lengths , , and , and 2) an isosceles trapezoid with parallel lengths and , the other sides both being . Some “intense” calculations reveal that the area of this is , so the answer is . Let’s see if we are .

**Solution**

We’re right! And both solutions mentioned use coordinates, just as we did. Theirs had less calculations, though, finding the height of the triangle a much faster way (I guess). Ok, it *seems* shorter the way they did it, but I think they pretty much did it the same way. I would call that a win.

**Conclusion**

Coordinate bashing problems are the type that I could kind of do, but I don’t like them, as they don’t inspire any sort of creativity. I hope the next problem will be more clever.

And the time is , which is prime.

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