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 1== | ==Solution 1== | ||
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==Solution 2== | ==Solution 2== | ||
− | Let <math>BC = x</math> and <math>CD = AD = | + | 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. |
− | Call <math>ED</math> y. | + | Call <math>ED</math> <math>y</math>. |
− | By the Pythagorean Theorem, <math>x^2 + y^2 = y + 2</math>. | + | By the Pythagorean Theorem, <math>x^2 + y^2 = (y + 2)^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 | + | 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. |
+ | 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>. | ||
-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>, b is an integer | ||
+ | <math>b+1 \le 63</math> | ||
+ | <math>b \le 62</math> | ||
+ | |||
+ | From the triangle inequality: | ||
+ | <math>a-2+b>a</math> | ||
+ | <math>b>2</math> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2015amc10a/398 | ||
+ | |||
+ | ~ dolphin7 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/9DSv4zn7MyE | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Latest revision as of 08:24, 28 December 2021
- The following problem is from both the 2015 AMC 12A #19 and 2015 AMC 10A #24, so both problems redirect to this page.
Contents
Problem 24
For some positive integers , there is a quadrilateral with positive integer side lengths, perimeter , right angles at and , , and . How many different values of are possible?
Solution 1
Let and be positive integers. Drop a perpendicular from to to show that, using the Pythagorean Theorem, that Simplifying yields , so . Thus, is one more than a perfect square.
The perimeter must be less than 2015. Simple calculations demonstrate that is valid, but is not. On the lower side, does not work (because ), but does work. Hence, there are 31 valid (all such that for ), and so our answer is
Solution 2
Let and be positive integers. Drop a perpendicular from to . Denote the intersection point of the perpendicular and as .
's length is , as well. Call . By the Pythagorean Theorem, . And so: , or .
Writing this down and testing, it appears that this holds for all . 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, must be a multiple of 2 to let the side lengths be integers. We test, and soon reach 62. It gives us , which is less than 2015. However, 64 gives us , so we know 62 is the largest we can go up to. Count all the even numbers from 2 to 62, and we get .
-jackshi2006
Solution 3
Let . Construct point on line so that is perpendicular to , . , .
Because :
Note that from here we know that must be even.
We also know that :
Substituting in we get:
, b is an integer
From the triangle inequality:
But, does not have to exist. Quadrilateral could be a square, in that case .
So, and must be even. Count all the even numbers from to , .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc10a/398
~ dolphin7
Video Solution
~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.