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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>
  
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==Solution==
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==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>
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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>
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==Solution 2==
 
==Solution 2==
  

Revision as of 02:02, 1 February 2016

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 $p$, 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

If $AD = CD = x$, then $BC = \sqrt{x^2-(2-x)^2} = 2\sqrt{x-1}$. Thus, $p = 2x+2\sqrt{x-1} + 2 < 2015$ and thus $2x + 2\sqrt{x-1} < 2013$. Since $\sqrt{x-1}$ must be an integer, we note that $x-1$ is a perfect square and by simple computations it is seen that $x = 31^2+1$ works but $x = 32^2+1$ doesn't. Therefore $x = 1$ to $31$ work giving an answer of $\boxed{31}$.

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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