Difference between revisions of "2015 AMC 10A Problems/Problem 24"

Revision as of 21:31, 7 June 2021

The following problem is from both the 2015 AMC 12A #19 and 2015 AMC 10A #24, so both problems redirect to this page.

Problem 24

For some positive integers $p$ , there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$ , right angles at $B$ and $C$ , $AB=2$ , and $CD=AD$ . How many different values of $p<2015$ are possible?

$\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$

Solution 1

Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ to show that, using the Pythagorean Theorem, that $x^2 + (y - 2)^2 = y^2.$ Simplifying yields $x^2 - 4y + 4 = 0$ , so $x^2 = 4(y - 1)$ . Thus, $y$ is one more than a perfect square.

The perimeter $p = 2 + x + 2y = 2y + 2\sqrt{y - 1} + 2$ must be less than 2015. Simple calculations demonstrate that $y = 31^2 + 1 = 962$ is valid, but $y = 32^2 + 1 = 1025$ is not. On the lower side, $y = 1$ does not work (because $x > 0$ ), but $y = 1^2 + 1$ does work. Hence, there are 31 valid $y$ (all $y$ such that $y = n^2 + 1$ for $1 \le n \le 31$ ), and so our answer is $\boxed{\textbf{(B) } 31}$

Solution 2

Let $BC = x$ and $CD = AD = y$ be positive integers. Drop a perpendicular from $A$ to $CD$ . Denote the intersection point of the perpendicular and $CD$ as $E$ .

$AE$ 's length is $x$ , as well. Call $ED$ $y$ . By the Pythagorean Theorem, $x^2 + y^2 = (y + 2)^2$ . And so: $x^2 = 4y + 4$ , or $y = (x^2-4)/4$ .

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, $x$ must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us $p = 1988$ , which is less than 2015. However, 64 gives us $2116 > 2015$ , so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get $\boxed{\textbf{(B) } 31}$ .

-jackshi2006

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2015amc10a/398

~ dolphin7

Video Solution

https://youtu.be/9DSv4zn7MyE

~savannahsolver

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.

@@ Line 21: / Line 21: @@
 Writing this down and testing, it appears that this holds for all <math>x</math>. 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 <math>p = 1988</math>, which is less than 2015. However, 64 gives us <math>2116 > 2015</math>, so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get <math>\boxed{\textbf{(B) } 31}</math>
+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 <math>p = 1988</math>, which is less than 2015. However, 64 gives us <math>2116 > 2015</math>, so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get <math>\boxed{\textbf{(B) } 31}</math>.
 -jackshi2006

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