2020 AMC 10B Problems/Problem 8
Points and lie in a plane with . How many locations for point in this plane are there such that the triangle with vertices , , and is a right triangle with area square units?
There are 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.
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, because of the 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:
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 . 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 triangles this case.
Therefore, the answer is .
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 ways for this case. Similarly, one can find that there are also ways for point to lie if is the longer leg. If it is a hypotenuse, then there are 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 .
Solution 3 (Algebra)
Let a and b be the distances of R from P and Q, respectively. By the Pythagorean Theorem, we have . Since the area of this triangle is 12, we get . Thus . Now substitute this into the other equation to get . Multiplying by on both sides, we get . Now let . Substituting and rearranging, we get . We reduced this problem to a quadratic equation! By the quadratic formula, our solutions are . Now substitute back to get . All 4 of these solutions are rational and will work. But our answer is actually as we have only calculated the number of places where R could go above line segment PQ. We need to multiply our 4 by 2 to account for all the places R could go below segment PQ. ~mewto
Solution 4 (Illustration)
Let the brackets denote areas. We know that Since it follows that
All such locations for are shown below:
For note that
- If then by construction.
- If then by construction.
- If then by the Inscribed Angle Theorem.
Together, there are such locations for
- The reflections of about are respectively.
- The reflections of about the perpendicular bisector of are respectively.
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