Difference between revisions of "2008 AMC 10B Problems/Problem 15"

Revision as of 00:34, 14 January 2011

How many right triangles have integer leg lengths $a$ and $b$ and a hypotenuse of length $b+1$ , where $b<100$ ?

$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$

By the Pythagorean theorem, $a^2+b^2=b^2+2b+1$

This means that $a^2=2b+1$ .

We know that $a,b>0$ , and that $b<100$ .

We also know that a must be odd, since the right

side is odd.

So $a=3,5,7,9,11,13$ , and the answer is $\boxed{A}$ .

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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

@@ Line 5: / Line 5: @@
 ==Solution==
-By the pytahagorean theorem, <math>a^2+b^2=b^2+2b+1</math>
+By the Pythagorean theorem, <math>a^2+b^2=b^2+2b+1</math>
 This means that <math>a^2=2b+1</math>.
@@ Line 15: / Line 15: @@
 side is odd.
 So <math>a=3,5,7,9,11,13</math>, and the answer is <math>\boxed{A}</math>.
 ==See also==
 {{AMC10 box|year=2008|ab=B|num-b=14|num-a=16}}