2016 AMC 12B Problems/Problem 21
Let be a unit square. Let be the midpoint of . For let be the intersection of and , and let be the foot of the perpendicular from to . What is
We are tasked with finding the sum of the areas of every where is a positive integer. We can start by finding the area of the first triangle, . This is equal to ⋅ ⋅ . Notice that since triangle is similar to triangle in a 1 : 2 ratio, must equal (since we are dealing with a unit square whose side lengths are 1). is of course equal to as it is the mid-point of CD. Thus, the area of the first triangle is ⋅ ⋅ .
The second triangle has a base equal to that of (see that ~ ) and using the same similar triangle logic as with the first triangle, we find the area to be ⋅ ⋅ . If we continue and test the next few triangles, we will find that the sum of all is equal to or
This is known as a telescoping series because we can see that every term after the first is going to cancel out. Thus, the summation is equal to and after multiplying by the half out in front, we find that the answer is .
Note that . So
We compute because as .
We plot the figure on a coordinate plane with and in the positive y-direction from the origin. If for some , then the line can be represented as . The intersection of this and , which is the line , is
As is the projection of onto the x-axis, it lies at . We have thus established that moving from to is equivalent to the transformation on the x-coordinate. The closed form of of the x-coordinate of can be deduced to be , which can be determined empirically and proven via induction on the initial case . Now
suggesting that is equivalent to . The sum of this from to is a classic telescoping sequence as in Solution 1 and is equal to .
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