Difference between revisions of "2015 AMC 10A Problems/Problem 24"
m |
Jackshi2006 (talk | contribs) (→Solution) |
||
Line 5: | Line 5: | ||
<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 1== |
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> | ||
Line 11: | Line 11: | ||
The perimeter <math>p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2</math> must be less than 2015. Simple calculations demonstrate that <math>y = 31^2 + 1 = 962</math> is valid, but <math>y = 32^2 + 1 = 1025</math> is not. On the lower side, <math>y = 1</math> does not work (because <math>x > 0</math>), but <math>y = 1^2 + 1</math> does work. Hence, there are 31 valid <math>y</math> (all <math>y</math> such that <math>y = n^2 + 1</math> for <math>1 \le n \le 31</math>), and so our answer is <math>\boxed{\textbf{(B) } 31}</math> | The perimeter <math>p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2</math> must be less than 2015. Simple calculations demonstrate that <math>y = 31^2 + 1 = 962</math> is valid, but <math>y = 32^2 + 1 = 1025</math> is not. On the lower side, <math>y = 1</math> does not work (because <math>x > 0</math>), but <math>y = 1^2 + 1</math> does work. Hence, there are 31 valid <math>y</math> (all <math>y</math> such that <math>y = n^2 + 1</math> for <math>1 \le n \le 31</math>), and so our answer is <math>\boxed{\textbf{(B) } 31}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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>. Call the intersection point of the perpindicular and <math>CD</math> <math>E</math>. | ||
+ | |||
+ | <math>AE</math>'s length is <math>x</math> as well. | ||
+ | Call <math>ED</math> y. | ||
+ | By the Pythagorean Theorem, <math>x^2 + y^2 = y + 2</math>. | ||
+ | And so: <math>x^2 = 4y + 4</math>, or <math>y = (x^2-4)/4</math> | ||
+ | Writing this down and testing it appears that this holds for all x. 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, <math>x</math> must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us a <math>p</math> 1988, which is less than 2015. While 64 gives us 2116, so we know 62 is the largest we can go. Count all the even numbers from 2 to 62, and you get <math>\boxed{\textbf{(B) } 31}</math> | ||
== See Also == | == See Also == |
Revision as of 11:53, 28 January 2020
- The following problem is from both the 2015 AMC 12A #19 and 2015 AMC 10A #24, so both problems redirect to this page.
Contents
Problem 24
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 1
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
Solution 2
Let and be positive integers. Drop a perpendicular from to . Call the intersection point of the perpindicular and .
's length is as well. Call y. By the Pythagorean Theorem, . And so: , or Writing this down and testing it appears that this holds for all x. 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 a 1988, which is less than 2015. While 64 gives us 2116, so we know 62 is the largest we can go. Count all the even numbers from 2 to 62, and you get
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.