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

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{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #19]] and [[2015 AMC 10A Problems|2015 AMC 10A #24]]}}
 
{{duplicate|[[2015 AMC 12A Problems|2015 AMC 12A #19]] and [[2015 AMC 10A Problems|2015 AMC 10A #24]]}}
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==Problem==
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==Problem 24==
 
For some positive integers <math>p</math>, there is a quadrilateral <math>ABCD</math> with positive integer side lengths, perimeter <math>p</math>, right angles at <math>B</math> and <math>C</math>, <math>AB=2</math>, and <math>CD=AD</math>.  How many different values of <math>p<2015</math> are possible?
 
For some positive integers <math>p</math>, there is a quadrilateral <math>ABCD</math> with positive integer side lengths, perimeter <math>p</math>, right angles at <math>B</math> and <math>C</math>, <math>AB=2</math>, and <math>CD=AD</math>.  How many different values of <math>p<2015</math> are possible?
  
 
<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 1==
 
==Solution 1==
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==Solution 2==
 
==Solution 2==
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Let <math>BC = x</math> and <math>CD = AD = z</math> be positive integers. Drop a perpendicular from <math>A</math> to <math>CD</math>. Denote the intersection point of the perpendicular and <math>CD</math> as <math>E</math>.
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<math>AE</math>'s length is <math>x</math>, as well.
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Call <math>ED</math> <math>y</math>.
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By the Pythagorean Theorem,  <math>x^2 + y^2 = (y + 2)^2</math>.
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And so: <math>x^2 = 4y + 4</math>, or <math>y = (x^2-4)/4</math>.
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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.
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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>.
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-jackshi2006
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== Video Solution by Richard Rusczyk ==
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https://artofproblemsolving.com/videos/amc/2015amc10a/398
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~ dolphin7
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==Video Solution==
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https://youtu.be/9DSv4zn7MyE
  
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If <math>AD = CD = x</math>, then <math>BC = \sqrt{x^2-(2-x)^2} = 2\sqrt{x-1}</math>. Thus, <math>p = 2x+2\sqrt{x-1} + 2 < 2015</math> and thus <math>2x + 2\sqrt{x-1} < 2013</math>. Since <math>\sqrt{x-1}</math> must be an integer, we note that <math>x-1</math> is a perfect square and by simple computations it is seen that <math>x = 31^2+1</math> works but <math>x = 32^2+1</math> doesn't. Therefore <math>x = 1</math> to <math>31</math> work giving an answer of <math>\boxed{\textbf{(B) } 31}</math>.
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~savannahsolver
  
 
== See Also ==
 
== See Also ==
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{{MAA Notice}}
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{{MAA Notice}}      
  
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[[Category: Introductory Geometry Problems]]
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[[Category: Intermediate Geometry Problems]]

Latest revision as of 16:35, 26 May 2024

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 $CD = AD = z$ 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

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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