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

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Call <math>ED</math> y.
 
Call <math>ED</math> y.
 
By the Pythagorean Theorem,  <math>x^2 + y^2 = y + 2</math>.
 
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>
 
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>
+
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 11:54, 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.

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 $CD$. Call the intersection point of the perpindicular and $CD$ $E$.

$AE$'s length is $x$ as well. Call $ED$ y. By the Pythagorean Theorem, $x^2 + y^2 = y + 2$.

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 a $p$ 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 $\boxed{\textbf{(B) } 31}$

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


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