Difference between revisions of "2007 AMC 12B Problems/Problem 23"

Revision as of 19:54, 25 December 2013

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

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

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 $\Rightarrow \mathrm{(A)}$ .

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

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 ==Solution==
-<math>\fracqwb = 3(a+bet)</math>
-qwetwqet
-qwet
-Using Euclid's formula for generating primitive triples:wqe
-<math>a = m^2-n^2</math>, <math>b=2we</math>, <math>c=m^2+n^2</math> whwqtere <math>m</math> and <math>n</math> are relatively prime positive integers, exactly one of which being eveet=2kmn<math>, </math>c=k(mqw^2+n^2)<math>
-wettqwwqetqwet
-</math>n(m-n)k = 6qwe<math>
-Now we dqweto some casework.
-For </math>k=1<math>et
-</math>n(m-n) = 6<math> wethich has solutions </math>(7,1)<math>, </math>(5,2)<math>, </math>(5,3)<math>, </math>(7,6)<math>
-ethe conditions of Euclid's formula, the only solutions are </math>(5,2)<math> and </math>(7,6et<math>
-For </math>k=2<math>
-</math>n(m-n)=3<math> has solutions </math>(4,qwet1)<math>, </math>(4,3)<math>, both of which are valid.
-For </math>k=3<math>
-</math>n(m-n)=2<math> has solutions </math>(3,1)<math>, </math>(3,2)<math> of which only </math>(3,2)<math> is valid.
-For </math>k=6<math>
-</math>n(m-n)=1<math> has solution </math>(1,2)<math>, which is valid.
-This means that the solutions for </math>(m,n,k)<math> are
-</math>(5,2,1), (7,6,1), (4,1,2), (4,3,2), (3,2,3), (1,2,6)<math>
-</math>6<math> solutions </math>\Rightarrow \mathrm{(A)}$
-==Solution 2==
 Let <math>a</math> and <math>b</math> be the two legs of the triangle.