2020 AMC 10 B

Solution 1:

There are 3 options here: 1. is the right angle. It's clear that there are points that fit this, one that's directly to the right of and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is. 2. is the right angle. Using the exact same reasoning, there are also solutions for this one. 3 . The new point is the right angle. (Diagram temporarily removed due to asymptote error) The diagram looks something like this. We know that the altitude to base must be since the area is . From here, we must see if there are valid triangles that satisfy the necessary requirements. First of all, area. Next, from the Pythagorean Theorem. From here, we must look to see if there are valid solutions. There are multiple ways to do this: Recognizing min and max: We know that the minimum value of is when . In this case, the equation becomes , which is LESS than . The equation becomes , which is obviously greater than 64 . We can conclude that there are values for and in between that satisfy the Pythagorean Theorem. And since , the triangle is not isoceles, meaning we could reflect it over and/or the line perpendicular to for a total of 4 triangles this case.

Solution 2:

Note that line segment can either be the shorter leg, longer leg or the hypotenuse. If it is the shorter leg, there are two possible points for that can satisfy the requirements - that being above or below . As such, there are 2 ways for this case. Similarly, one can find that there are also 2 ways for point to lie if is the longer leg. If it is a hypotenuse, then there are 4 possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is