Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
Line 1: Line 1:
 
{{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]]}}
−
==Problem 24==
+
=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>
 +
 +
  
 
==Solution 1==
 
==Solution 1==
Line 24: Line 26:
  
 
-jackshi2006
 
-jackshi2006
 +
 +
==Solution 3==
 +
 +
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>.
 +
 +
Because <math>\angle AED={90}^\circ</math>:
 +
<math>(a-2)^2+b^2=a^2</math>
 +
<math>a^2-4a+4+b^2=a^2</math>
 +
<math>b^2+4=4a</math>
 +
Note that from here we know that <math>b</math> must be even.
 +
<math>a=\frac{b^2+4}{4}</math>
 +
 +
We also know that <math>p < 2015</math>:
 +
<math>p=AB+BC+CD+AD</math>
 +
<math>p=2+b+a+a</math>
 +
<math>p=2a+b+2</math>
 +
<math>2a+b+2<2015</math>
 +
<math>2a+b<2013</math>
 +
 +
Substituting <math>a</math> in we get:
 +
<math>\frac{b^2+4}{2}+b<2013</math>
 +
<math>b^2+4+2b<4026</math>
 +
<math>b^2+2b+1<4023</math>
 +
<math>(b+1)^2<4023</math>
 +
<math>b+1<\sqrt{4023}</math>
 +
<math>b+1<63.4</math>
 +
<math>b<62.4</math>
 +
 +
From previous equations and triangle inequality:
 +
<math>a=\frac{b^2+4}{4}</math>
 +
<math>a>b</math>
 +
<math>\frac{b^2+4}{4}>b</math>
 +
<math>b^2+4>4b</math>
 +
<math>b^2-4b+4>0</math>
 +
<math>(b-2)^2>0</math>
 +
<math>b-2>0</math>
 +
<math>b>2</math>
 +
 +
So, <math>2< b \le 62</math> and b must be even. Count all the even numbers from 2 to 62, we get <math>\boxed{\textbf{(B) } 31}</math>.
 +
 +
 +
~isabelchen
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Revision as of 04:23, 4 October 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

Solution 3

Let $AD=CD=a$. Construct point $E$ on line $CD$ so that $AE$ is perpendicular to $CD$, $AE=BC=b$. $CE=AB=2$, $DE=a-2$.

Because $\angle AED={90}^\circ$: 

$(a-2)^2+b^2=a^2$
$a^2-4a+4+b^2=a^2$
$b^2+4=4a$
Note that from here we know that $b$ must be even.
$a=\frac{b^2+4}{4}$

We also know that $p < 2015$: 

$p=AB+BC+CD+AD$
$p=2+b+a+a$
$p=2a+b+2$
$2a+b+2<2015$
$2a+b<2013$

Substituting $a$ in we get: 

$\frac{b^2+4}{2}+b<2013$
$b^2+4+2b<4026$
$b^2+2b+1<4023$
$(b+1)^2<4023$
$b+1<\sqrt{4023}$
$b+1<63.4$
$b<62.4$

From previous equations and triangle inequality: 

$a=\frac{b^2+4}{4}$
$a>b$
$\frac{b^2+4}{4}>b$
$b^2+4>4b$
$b^2-4b+4>0$
$(b-2)^2>0$
$b-2>0$
$b>2$

So, $2< b \le 62$ and b must be even. Count all the even numbers from 2 to 62, we get $\boxed{\textbf{(B) } 31}$.


~isabelchen

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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10A_Problems/Problem_24&oldid=163120"