Difference between revisions of "2007 AMC 12B Problems/Problem 23"
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We can factor <math>72</math> to arrive with <math>6</math> pairs of solutions: <math>(7, 78), (8,42), (9, 30), (10, 24), (12, 18),</math> and <math>(14, 15) \Rightarrow \mathrm{(A)}</math>. | We can factor <math>72</math> to arrive with <math>6</math> pairs of solutions: <math>(7, 78), (8,42), (9, 30), (10, 24), (12, 18),</math> and <math>(14, 15) \Rightarrow \mathrm{(A)}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | Let <math>a</math> and <math>b</math> be the two legs of the triangle, and <math>c</math> be the hypotenuse. | ||
+ | |||
+ | By using <math>Area = \frac{r}{2} (a+b+c)</math>, where <math>r</math> is the in-radius, we get: | ||
+ | |||
+ | <cmath>3(a+b+c) = \frac{r}{2} (a+b+c)</cmath> | ||
+ | <cmath>r=6</cmath> | ||
+ | |||
+ | In right triangle, <math>r = \frac{a+b-c}{2}</math> | ||
+ | <cmath>a+b-c = 12</cmath> | ||
+ | <cmath>c = a + b - 12</cmath> | ||
+ | |||
+ | |||
+ | By the triangle's area we get: | ||
+ | |||
+ | <cmath>\frac{ab}{2} = 6 \cdot \frac{a+b+c}{2}</cmath> | ||
+ | <cmath>ab = 6(a+b+c)</cmath> | ||
+ | |||
+ | By substituting <math>c</math> in: | ||
+ | |||
+ | <cmath>ab = 6(a+b+a + b - 12)</cmath> | ||
+ | <cmath>ab - 12a - 12b + 72 = 0</cmath> | ||
+ | <cmath>(a - 12)(b - 12) = 72</cmath> | ||
+ | |||
+ | As <math>72 = 2^3 \cdot 3^2</math>, there are <math>\frac{(3+1)(2+1)}{2} = 6</math> solutions, <math>\boxed{\textbf{(A) } 6}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == Solution 4 == | ||
+ | All pythagorean triples can be parametrized in the form <math>(a, b, c) = k(r^2 - s^2), k(2rs), k(r^2 + s^2)</math> for positive integers <math>k, r, s</math>. The area being triple the perimeter implies that <cmath>k^2(r^2 - s^2)rs = 3(k(r^2 - s^2) + k(2rs) + k(r^2 + s^2)).</cmath> This can be simplified to get <cmath>ks(r - s) = 6.</cmath> Now, we get the triples <cmath>(k, r, s) = (1, 7, 1), (1, 5, 2), (1, 5, 3), (1, 7, 6), (2, 4, 1), (2, 4, 3), (3, 3, 1), (3, 3, 2), (6, 2, 1).</cmath> However, the ones where <math>r</math> and <math>s</math> are not different signs and relatively prime are redundant, so we get <math>6</math> triples total. | ||
+ | |||
+ | |||
+ | ==Solution 5 (very cheesy)== | ||
+ | Well, obviously MAA would try to make the answer choices trap some people. One way they could do that is by thinking "non-congruent" would be ignored, so the answer would be multiplied by 2. The only answer choice that can be divided by 2 to create an existing answer is 12, so the answer is <math>\boxed{\textbf{(A) } 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}} |
Latest revision as of 18:59, 4 January 2024
Contents
[hide]Problem
How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to times their perimeters?
Solution 1
Let and be the two legs of the triangle.
We have .
Then .
We can complete the square under the root, and we get, .
Let and , we have .
After rearranging, squaring both sides, and simplifying, we have .
Putting back and , and after factoring using Simon's Favorite Factoring Trick, we've got .
Factoring 72, we get 6 pairs of and
And this gives us solutions .
Alternatively, note that . Then 72 has factors. However, half of these are repeats, so we have solutions.
Solution 2
We will proceed by using the fact that , where is the radius of the incircle and is the semiperimeter .
We are given .
The incircle of breaks the triangle's sides into segments such that , and . Since ABC is a right triangle, one of , and is equal to its radius, 6. Let's assume .
The side lengths then become , and . Plugging into Pythagorean's theorem:
We can factor to arrive with pairs of solutions: and .
Solution 3
Let and be the two legs of the triangle, and be the hypotenuse.
By using , where is the in-radius, we get:
In right triangle,
By the triangle's area we get:
By substituting in:
As , there are solutions, .
Solution 4
All pythagorean triples can be parametrized in the form for positive integers . The area being triple the perimeter implies that This can be simplified to get Now, we get the triples However, the ones where and are not different signs and relatively prime are redundant, so we get triples total.
Solution 5 (very cheesy)
Well, obviously MAA would try to make the answer choices trap some people. One way they could do that is by thinking "non-congruent" would be ignored, so the answer would be multiplied by 2. The only answer choice that can be divided by 2 to create an existing answer is 12, so the answer is .
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.