2015 AMC 10A Problems/Problem 24
- The following problem is from both the 2015 AMC 12A #19 and 2015 AMC 10A #24, so both problems redirect to this page.
For some positive integers , there is a quadrilateral with positive integer side lengths, perimeter , right angles at and , , and . How many different values of are possible?
Let and be positive integers. Drop a perpendicular from to to show that, using the Pythagorean Theorem, that Simplifying yields , so . Thus, is one more than a perfect square.
The perimeter must be less than 2015. Simple calculations demonstrate that is valid, but is not. On the lower side, does not work (because ), but does work. Hence, there are 31 valid (all such that for ), and so our answer is
Let and be positive integers. Drop a perpendicular from to . Denote the intersection point of the perpendicular and as .
's length is , as well. Call . By the Pythagorean Theorem, . And so: , or .
Writing this down and testing, it appears that this holds for all . However, since there is a dividend of 4, the numerator must be divisible by 4 to conform to the criteria that the side lengths are positive integers. In effect, must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us , which is less than 2015. However, 64 gives us , so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get .
Let . Construct point on line so that is perpendicular to , . , .
Note that from here we know that must be even.
We also know that :
Substituting in we get:
, b is an integer
From the triangle inequality:
But, does not have to exist. Quadrilateral could be a square, in that case .
So, and must be even. Count all the even numbers from to , .
Video Solution by Richard Rusczyk
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