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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 24=
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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?
  
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==Solution 2==
 
==Solution 2==
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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>. Denote the intersection point of the perpendicular and <math>CD</math> as <math>E</math>.
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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>.
  
 
<math>AE</math>'s length is <math>x</math>, as well.
 
<math>AE</math>'s length is <math>x</math>, as well.
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-jackshi2006
 
-jackshi2006
  
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==Solution 3==
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==Solution 3 (Less Counting)==
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Let <math>BC = n</math> and <math>CD = AD = m</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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Let <math>AD=CD=a</math>. Construct point <math>E</math> on line <math>CD</math> so that <math>AE</math> is perpendicular to <math>CD</math>, <math>AE=BC=b</math>. <math>CE=AB=2</math>, <math>DE=a-2</math>.
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<math>AE</math>'s length is <math>n</math>, and <math>ED = m-2</math>
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By the Pythagorean Theorem,  <math>m^2 = (m-2)^2 + n^2</math>
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, therefore <math>n^2 = 4(m-1)</math>
  
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Because <math>\angle AED={90}^\circ</math>:
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Notice that 4 is already a perfect square and <math>n</math> is an integer, meaning <math>m-1</math> also has to be a perfect square.
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<math>(a-2)^2+b^2=a^2</math>
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We can denote <math>m-1</math> as <math>x^2</math>, therefore we have <math>m = x^2 + 1</math>, <math>n = 2x</math>, and each <math>x</math> will correspond to a pair of <math>(n, m)</math>.
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<math>a^2-4a+4+b^2=a^2</math>
 
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<math>b^2+4=4a</math>
 
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Note that from here we know that <math>b</math> must be even.
 
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<math>a=\frac{b^2+4}{4}</math>
 
  
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We also know that <math>p < 2015</math>:
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Since <math>p<2015</math>, <math>2 + n + m + m<2015</math> which gives us <math>n+2m<2013</math>. Substituting <math>x</math> will give us <math>2x+2(x^2 + 1) < 2013</math>, simplify this will again give us <math>x^2 + x + 1 < 1006.5</math>
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<math>p=AB+BC+CD+AD</math>
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Try <math>x = 32</math>, <math>x^2 + x + 1 = 1057 > 1006.5</math>, exceeded
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<math>p=2+b+a+a</math>
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so <math>x</math> must be less than <math>32</math>
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<math>p=2a+b+2</math>
 
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<math>2a+b+2<2015</math>
 
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<math>2a+b<2013</math>
 
  
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Substituting <math>a</math> in we get:
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When <math>x=31</math>, <math>x^2 + x + 1 = 993 < 1006.5</math>, <math>x=31</math> works for us.
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<math>\frac{b^2+4}{2}+b<2013</math>
 
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<math>b^2+4+2b<4026</math>
 
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<math>b^2+2b+1<4023</math>
 
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<math>(b+1)^2<4023</math>
 
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<math>b+1<\sqrt{4023}</math>, b is an integer
 
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<math>b+1 \le 63</math>
 
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<math>b \le 62</math>
 
  
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From the triangle inequality:
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For <math>x</math> in <math>(0, \infty)</math>, <math>f(x) = x^2 + x + 1</math> increases, therefore for any <math>x<31</math>, <math>f(x) < f(31) < 1006.5</math>
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<math>a-2+b>a</math>
 
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<math>b>2</math>
 
  
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But, <math>\triangle ADE</math> does not have to exist. Quadrilateral <math>ABCD</math> could be a square, in that case <math>b=2</math>.
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Thus, any <math>x</math> in <math>[1, 31]</math> satisfy the condition that <math>x^2 + x + 1 < 1006.5</math>. <math>x</math> needs to be integer, and there are <math>31</math> integers in the interval <math>[1, 31]</math>.  
  
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So, <math>2 \le b \le 62</math> and <math>b</math> must be even. Count all the even numbers from <math>2</math> to <math>62</math>, <math>\frac{62-2}{2}+1=\boxed{\textbf{(B) } 31}</math>.
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Each <math>x</math> corresponds to a <math>(n, m)</math>, meaning there are <math>31</math> possible value for different pairs of <math>(n, m)</math> and lead to <math>\boxed{\textbf{(B) } 31}</math> possible values for the perimeter <math>p</math>.  
  
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~isabelchen
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~ [https://artofproblemsolving.com/wiki/index.php/User:Andy_li0805 Andy_li0805]
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Latest revision as of 10:54, 14 September 2025

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

Solution 3 (Less Counting)

Let $BC = n$ and $CD = AD = m$ 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 $n$, and $ED = m-2$ By the Pythagorean Theorem, $m^2 = (m-2)^2 + n^2$ , therefore $n^2 = 4(m-1)$

Notice that 4 is already a perfect square and $n$ is an integer, meaning $m-1$ also has to be a perfect square. We can denote $m-1$ as $x^2$, therefore we have $m = x^2 + 1$, $n = 2x$, and each $x$ will correspond to a pair of $(n, m)$.

Since $p<2015$, $2 + n + m + m<2015$ which gives us $n+2m<2013$. Substituting $x$ will give us $2x+2(x^2 + 1) < 2013$, simplify this will again give us $x^2 + x + 1 < 1006.5$ Try $x = 32$, $x^2 + x + 1 = 1057 > 1006.5$, exceeded so $x$ must be less than $32$

When $x=31$, $x^2 + x + 1 = 993 < 1006.5$, $x=31$ works for us.

For $x$ in $(0, \infty)$, $f(x) = x^2 + x + 1$ increases, therefore for any $x<31$, $f(x) < f(31) < 1006.5$

Thus, any $x$ in $[1, 31]$ satisfy the condition that $x^2 + x + 1 < 1006.5$. $x$ needs to be integer, and there are $31$ integers in the interval $[1, 31]$.

Each $x$ corresponds to a $(n, m)$, meaning there are $31$ possible value for different pairs of $(n, m)$ and lead to $\boxed{\textbf{(B) } 31}$ possible values for the perimeter $p$.

~ Andy_li0805

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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