Solve[{12 ==1/a Sqrt[2] \[Sqrt]((a + b + c) (1/2 (a + b + c) - a) (1/2 (a + b + c) - b) (1/2 (a + b + c) - c)), 14 == 1/b Sqrt[2] \[Sqrt]((a + b + c) (1/2 (a + b + c) - a) (1/2 (a + b + c) - b) (1/2 (a + b + c) - c)),
83 == 1/c Sqrt[2] \[Sqrt]((a + b + c) (1/2 (a + b + c) - a) (1/2 (a + b + c) - b) (1/2 (a + b + c) - c)), 12 == (b*c)/(2 R)}, {a, b, c, R}] // N

Only had to try 84 and 83.

bobbym

2013-06-01 22:20:44

Hi;

What is it, please post what you have.

ElainaVW

2013-06-01 22:19:55

I have a different solution.

anonimnystefy

2013-05-31 05:17:45

Okay, see you!

bobbym

2013-05-31 05:02:38

Hi;

Okay, see you later.

anonimnystefy

2013-05-31 04:58:27

Well, it is because a*ha=b*hb=c*hc. From this we can get a:b:c=(1/ha):(1/hb):(1/hc), which means that, if a, b and c can form a triangle, 1/ha, 1/hb and 1/hc must be able to form a triangle as well.

bobbym

2013-05-31 04:54:38

I do not get it but it works so the problem is done.

anonimnystefy

2013-05-31 04:47:46

Yes, but, as I already said, 1/ha=1/12, 1/hb=1/14 and 1/hc must be length of some triangle in order to be a valid set of altitudes.

bobbym

2013-05-31 04:46:32

What triangle? That inequality is for the sides. You have altitudes there.

anonimnystefy

2013-05-31 04:44:31

The triangle inequality states that 1/hc+1/14>1/12.

bobbym

2013-05-31 04:41:33

Why the minus and not a plus?

anonimnystefy

2013-05-31 04:40:26

What about it?

bobbym

2013-05-31 04:38:33

Okay but what about this?

1/hc>1/12-1/14

anonimnystefy

2013-05-31 04:32:01

Well, I had the fact that the lengths 1/12, 1/14 and 1/hc must form a triangle, where 1/hc is as small as possible. Because of the triangle inequality we have that 1/hc>1/12-1/14=1/84. So, hc<84. The maximum possible length is 83.