Difference between revisions of "2015 AMC 10A Problems/Problem 24"
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<math>\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63</math> | <math>\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63</math> | ||
− | ==Solution | + | ==Solution == |
Let <math>BC = x</math> and <math>CD = AD = y</math> be positive integers. Drop a perpendicular from <math>A</math> to <math>CD</math> to show that, using the Pythagorean Theorem, that | Let <math>BC = x</math> and <math>CD = AD = y</math> be positive integers. Drop a perpendicular from <math>A</math> to <math>CD</math> to show that, using the Pythagorean Theorem, that | ||
<cmath>x^2 + (y - 2)^2 = y^2.</cmath> | <cmath>x^2 + (y - 2)^2 = y^2.</cmath> |
Revision as of 17:50, 1 January 2018
- The following problem is from both the 2015 AMC 12A #19 and 2015 AMC 10A #24, so both problems redirect to this page.
Problem
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?
Solution
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
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.