How many non congruent right triangles with positive integer leg lengths have areas that are numerically equal to 3 times their perimeters?
Let $a$ and $b$ be the two legs of the triangle.
We have $\frac{1}{2}ab = 3(a+b+c)$.
Then $ab=6 \left(a+b+\sqrt {a^2 + b^2}\right)$.
We can complete the square under the root, and we get, $ab=6 \left(a+b+\sqrt {(a+b)^2 - 2ab}\right)$.
Let $ab=p$ and $a+b=s$, we have $p=6 \left(s+ \sqrt {s^2 - 2p}\right)$.
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 $(a-12)(b-12)=72$.
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 $\Rightarrow \mathrm{(A)}$.
Alternatively, note that $72 = 2^3 \cdot 3^2$. Then 72 has $(3+1)(2+1) = (4)(3) = 12$ factors. However, half of these are repeats, so there we have $\frac{12}{2} = 6$ solutions.
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 $\left(s = \frac{p}{2}\right)$.
We are given $[ABC] = 3p = 6s \Rightarrow rs = 6s \Rightarrow r = 6$.
The incircle of $ABC$ breaks the triangle's sides into segments such that $AB = x + y$, $BC = x + z$ and $AC = y + z$. Since ABC is a right triangle, one of $x$, $y$ and $z$ is equal to its radius, 6. Let's assume $z = 6$.
The side lengths then become $AB = x + y$, $BC = x + 6$ and $AC = y + 6$. Plugging into Pythagorean's theorem:
$(x + y)^2 = (x+6)^2 + (y + 6)^2$
$x^2 + 2xy + y^2 = x^2 + 12x + 36 + y^2 + 12y + 36$
$2xy - 12x - 12y = 72$
$xy - 6x - 6y = 36$
$(x - 6)(y - 6) - 36 = 36$
$(x - 6)(y - 6) = 72$
We can factor $72$ to arrive with $6$ pairs of solutions: $(7, 78), (8,42), (9, 30), (10, 24), (12, 18),$ and $(14, 15) \Rightarrow \mathrm{(A)}$.
To find the number of non-congruent right triangles with positive integer leg lengths that have areas equal to 3 times their perimeters, we can use a systematic approach.
Let's assume that the legs of the right triangle are "a" and "b", with "a" being the shorter leg. The hypotenuse can be represented as "c".
First, let's find the expressions for the area and perimeter of the right triangle:
Area: 1/2 * a * b
Perimeter: a + b + c
Given that the area is equal to 3 times the perimeter, we can write the equation:
1/2 * a * b = 3 * (a + b + c)
Now, let's simplify the equation:
a * b = 6 * (a + b + c)
Next, we need to express "c" in terms of "a" and "b" using the Pythagorean theorem:
c^2 = a^2 + b^2
Since we are looking for positive integer solutions, let's consider the following possibilities:
Case 1: a = 1
If we assume a = 1, the equation becomes:
b = 6 * (1 + b + c)
Simplifying further:
b = 6 + 6b + 6c
-5b - 6c = 6
From here, we can deduce that b and c must be negative, which is not possible since we are looking for positive leg lengths. Therefore, there are no solutions if a = 1.
Case 2: a = 2
Similarly, if we assume a = 2, the equation becomes:
2b = 6 * (2 + b + c)
Simplifying further:
2b = 12 + 6b + 6c
-4b - 6c = 12
Again, this equation does not have integer solutions for positive b and c.
Case 3: a = 3
Assuming a = 3:
3b = 6 * (3 + b + c)
3b = 18 + 6b + 6c
-3b - 6c = 18
This equation simplifies to:
-4b - 6c = 18
By inspection, we can find one solution where b = 6 and c = 1. This corresponds to a right triangle with leg lengths 3, 6, and 1.
Case 4: a = 4
Assuming a = 4:
4b = 6 * (4 + b + c)
4b = 24 + 6b + 6c
-2b - 6c = 24
There are no integer solutions for this equation.
From the above analysis, we can conclude that there is only one non-congruent right triangle with positive integer leg lengths that has an area numerically equal to 3 times its perimeter. This triangle has leg lengths 3, 6, and 1.
To find the number of non-congruent right triangles with positive integer leg lengths having areas equal to 3 times their perimeters, we can use a systematic approach.
Let's assume the legs of the right triangle have lengths a and b, and the hypotenuse has length c. Since these are positive integers, we can write the Pythagorean theorem as:
a^2 + b^2 = c^2
The perimeter of the triangle is given by:
Perimeter = a + b + c
The area of the triangle is given by:
Area = 0.5 * a * b
Now, we have the condition that the area is equal to 3 times the perimeter:
0.5 * a * b = 3 * (a + b + c)
Simplifying the equation, we get:
a * b = 6 * (a + b + c)
Now, since a, b, and c are positive integers, we can try different possible values for a, b, and c to satisfy the equation and find the solutions.
Since we have a quadratic equation a * b - 6 * (a + b) - 6c = 0, we can rearrange it to form a quadratic equation: a * b - 6a - 6b - 6c = 0.
We can use the quadratic formula to find the possible values of a and b:
a = (-b - sqrt(b^2 - 4 * (-6 * b - 6c)))/2
a = (-b + sqrt(b^2 - 4 * (-6 * b - 6c)))/2
For each value of a, we can determine possible integer values of b based on the equation. Then, we can check if the corresponding c value satisfies the Pythagorean theorem.
By systematically trying different values of a and b, we can find the non-congruent right triangles that satisfy the given conditions.