Triangle area

[bigbill]

A triangle has side lengths x, x+8, x + 9 units, it's perimeter is 1/3 of it's area. Find x ? I used Heron's formula to find the area but it didn't work.I got 3x ^ 4 + 68 x ^3 + 286 x ^2 - 212 x -1,105. Thanks for any help. Is there a shortcut ? I KNOW X = 17. Don't seem able to reach it.

 

[Deleted member 4993]

A triangle has side lengths x, x+8, x + 9 units, it's perimeter is 1/3 of it's area. Find x ? I used Heron's formula to find the area but it didn't work.I got 3x ^ 4 + 68 x ^3 + 286 x ^2 - 212 x -1,105. Thanks for any help. Is there a shortcut ? I KNOW X = 17. Don't seem able to reach it.

How did you get a fourth degree polynomial? I am getting a cubic - which is solvable through rational-root-thorem.

 

[JeffM]

Hmmm...I also get a 4th degree, but different:
3x^4 + 68x^3 - 1010x^2 -14756x - 41905 = 0
which has these 4 solutions: 17, -29, -5, -17/3

Comes from (area per Heron's = 3*perimeter) :
SQRT[(3x+17)(x+17)(x+1)(x-1) / 16] = 3(3x + 17)

We await your verdict, Sir Khan

I got a cubic.

p = A/3. So 3p = A. But s (semiperimeter) = p / 2 so 3p = 6s.

Using Heron's formula $\displaystyle 6s = \sqrt{s(s - x)(s - x + 8)(s - x + 9)} \implies 36s^2 = s(s - x)(s - x + 8)(s - x + 9) \implies$

$\displaystyle 36s = (s - x)(s - x + 8)(s - x + 9).$ But this is cubic if x is a linear function of s.

$\displaystyle s = \dfrac{x + x + 8 + x + 9}{2} = \dfrac{3x + 17}{2} \implies x = \dfrac{2s - 17}{3}.$

Bad luck, denis, SK is right.

By the way, the cubic is ugly.

 

[Deleted member 4993]

We await your verdict, Sir Khan

Corner .... to the corner .....

 

[JeffM]

Sneeeeky...I like it!

Why do you say "ugly"? I don't find it ugly:
x^3 + 17x^2 - 433x - 2465 = 0

3 solutions are (same as 3 of mine!) : 17, -5, -29

There is something APPARENTLY weird about this problem. There seem to be two methods of solving it, one involving a quartic and one involving a cubic, and the quartic gives an extra solution. Now if the problem is viewed as geometric, the extra solution is meaningless because negative. If, however, the problem is viewed as simply algebraic, the extra solution is valid. What is going on?

When I divided through by s, I was implicitly assuming that s is not 0. That of course is a valid assumption given the geometric interpretation. But it is not a valid assumption under a purely algebraic interpretation. The possibility that s = 0 must then be addressed.

$\displaystyle 0 = s = \dfrac{3x + 17}{2} \implies 3x + 17 = 0 \implies x = - \dfrac{17}{3}$, which is denis's EXTRA solution.

So denis solved the algebraic problem. I think we can let him out of the corner if he promises to behave the rest of the day.

EDIT In fact, the geometric interpretation requires that x > 0, which implies that s > 0. If that implicit constraint is imposed on the algebraic problem, both the cubic and quartic methods give a single, identical result.