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 "2007 AMC 12B Problems/Problem 23"
(→‎Solution)
Line 29: Line 29:
  
 
And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>.
 
And this gives us <math>6</math> solutions <math>\Rightarrow \mathrm{(A)}</math>.
 +
 +
==Solution #2==
 +
We will proceed by using the fact that <math>[ABC] = r\cdot s</math>, where <math>r</math> is the radius of the incircle and <math>s</math> is the semiperimeter (<math>s = \frac{p}{2}</math>).
 +
 +
We are given <math>[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2007|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:43, 29 November 2014

Problem 23

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

$\mathrm {(A)} 6\qquad \mathrm {(B)} 7\qquad \mathrm {(C)} 8\qquad \mathrm {(D)} 10\qquad \mathrm {(E)} 12$

Solution

Let $a$ and $b$ be the two legs of the triangle.

We have $\frac{1}{2}ab = 3(a+b+c)$.

Then $ab=6\cdot (a+b+\sqrt {a^2 + b^2})$.

We can complete the square under the root, and we get, $ab=6\cdot (a+b+\sqrt {(a+b)^2 - 2ab})$.

Let $ab=p$ and $a+b=s$, we have $p=6\cdot (s+ \sqrt {s^2 - 2p})$.

After rearranging, squaring both sides, and simplifying, we have $p=12s-72$.


Putting back $a$ and $b$, and after factoring using $SFFT$, we've got $(a-12)\cdot (b-12)=72$.


Factoring 72, we get 6 pairs of $a$ and $b$


$(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).$


And this gives us $6$ solutions $\Rightarrow \mathrm{(A)}$.

Solution #2

We will proceed by using the fact that $[ABC] = r\cdot s$, where $r$ is the radius of the incircle and $s$ is the semiperimeter ($s = \frac{p}{2}$).

We are given $[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6$.

See Also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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=2007_AMC_12B_Problems/Problem_23&oldid=66391"