Part 2 of this series.

**Table of Contents:**

**Problem:**

In isosceles triangle , is located at the origin and is located at (20,0). Point is in the first quadrant with and angle . If triangle is rotated counterclockwise about point until the image of lies on the positive -axis, the area of the region common to the original and the rotated triangle is in the form , where are integers. Find .

**My Attempt****:**

When I sketched it out, it looked something like this:

That quadrilateral bit right there in the middle is the area we are trying to find.

I found out that the length of the isosceles triangle is . I also found some right triangles, but nothing I did helped me much. It seemed that the best thing to do was to subtract a big triangle from a little triangle.

I’m going to denote the triangle as the rotated triangle, the intersection of and as $E$, the intersection of and as , and the intersection of and as . We are trying to find the area of . We could find it by either getting or .

The problem with the subtraction is that I don’t know , , , or . It’s possible to find them with a bit of similar triangle bashing, but that’s not really what I want to do right now. There must be a clever way to find the area of , and I think that’s what the official solution will have. I also think it will have something to do with the fact that is a 30-60-90 triangle.

**Solution****:**

The “official” AoPS solution says that we are trying to find the area of .

To find the area of , we observe that is a 45-60-75 triangle. Before today, I didn’t know this had a significance, but in this case, it helps us find the altitude (and therefore the area) of the triangle. This makes me feel dumb, as I tried to find the area of it by using . Drawing the altitude divides the triangle into a 30-60-90 triangle and a 45-45-90 triangle. Since , the altitude is . Calculation shows that the base is , so the area of is .

Now, we need to find the area of . This is easier; I already calculated it before I read the solution. Since is a right triangle, , and . So . Therefore, . Multiplying these together and dividing by 2, we get that the area of triangle is . Subtracting, we get , meaning the answer is .

**Conclusion****:**

I didn’t solve it. I thought the official solution was going to use the 30-60-90 triangle , but instead they used another, more clever, 30-60-90 triangle. There are other solutions, but they are either a remix of the one above or a bash. I believe the one above is the most elegant solution.

I thought this problem required a *lot* of calculation, but I still enjoyed it overall.

And the time is , which is prime.

## One thought on “AIME I 2007 Problem 12 pt. 2”