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> | ||
| − | ==Solution== | + | ==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 01:02, 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.
Contents
Problem
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
![\[x^2 + (y - 2)^2 = y^2.\]](http://latex.artofproblemsolving.com/d/d/4/dd48499079a7d0fe961c38c747cdb79950c68aab.png)
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
If 
, then 
. Thus, 
and thus 
. Since 
must be an integer, we note that 
is a perfect square and by simple computations it is seen that 
works but 
doesn't. Therefore 
to 
work giving an answer of 
.
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. 
