# geometry algebra

• Dec 30th 2005, 04:56 AM
yiyayiyayo
Help
A triangle ABC,a=AC, b=AC, c=AB , and the
angle C≥60°
• Jan 2nd 2006, 12:19 AM
yiyayiyayo
Do I fail to express the question?
• Jan 2nd 2006, 11:15 AM
ThePerfectHacker
I want to help you but I do not understand your question, sorry. Can you draw it?
• Jan 2nd 2006, 10:59 PM
yiyayiyayo
a triangle:......A..:o
.....................:o....:o
................b..:o.........:o..c
..................:o...............:o
...............C:o:o:o:o:o:o:o:o:o:o ...B
............................a

I hope you can understand my question now. :)
Looking forward to your help,thank you.
• Jan 3rd 2006, 06:50 PM
ThePerfectHacker
Is it isoseles? Because in the picture it is not and in the first post it is. If it is isosoles I have found a proof.
• Jan 4th 2006, 03:03 AM
yiyayiyayo
Yes
• Jan 4th 2006, 03:03 PM
ThePerfectHacker
If I understand the problem correctly the angle can be any angle not just more than 60 and the inequality is really and equality. Let me explain:

Since $\displaystyle a=b$ the problem reduces to proving
$\displaystyle 2a(\frac{2}{a}+\frac{1}{c}) \geq 4+\frac{1}{\sin (C/2)}$. Thus,
$\displaystyle 4+\frac{2a}{c}\geq 4+\frac{1}{\sin (C/2)}$. Thus, (if and only if)
$\displaystyle \frac{2a}{c}\geq \frac{1}{\sin C/2}$.
But this is always equal to eachother, thus,
$\displaystyle \frac{2a}{c}= \frac{1}{\sin (C/2)}$.

Because if an isoseles triangle has sides $\displaystyle a$ with angle between them $\displaystyle C$ and $\displaystyle c$ is its third side then $\displaystyle 2a\sin (C/2)=c$.
The reason why is simple, by the law of cosines, $\displaystyle a^2+a^2-2a^2\cos C=c^2$ thus, $\displaystyle 2a^2(1-\cos C)=c^2$ by the half-angle identity $\displaystyle 4a^2\sin^2(C/2)=c^2$ thus $\displaystyle 2a\sin(c/2)=a$ thus, $\displaystyle \frac{2a}{c}=\frac{1}{\sin(C/2)}$. Thus your inequality is in reality always equal, thus the limitation for the angle were not needed.
Thus, your problem was tricky in that it was making me believe the angle needs to be greater than 60 which is not true.

Here are some examples which show what you are saying is an equality. A right triangle with sides $\displaystyle 1,1,\sqrt{2}$, here the angle is 90. An equilateral triangle with sides $\displaystyle 1,1,1$ here the angle is 60.
• Jan 5th 2006, 02:52 AM
yiyayiyayo
But after this question, there is another one asking me to guess when a and b are not equal, can it still be right. I've put many sets of numbers in it, which shows it is always right. But I can't prove it.
I think I have to trouble again.

just like this:
A triangle ABC,a=BC, b=AC, c=AB , and the
angle C≥60°
• Jan 5th 2006, 06:02 PM
ThePerfectHacker
I am trying to solve your problem for all triangles but I did not find a proof yet. Are you certain it works for cases that you tried it for? Where did you even get this problem?
Any ideas which I could use in my proof?
• Jan 6th 2006, 04:00 AM
yiyayiyayo
Now I'm sure that when angle C≥60°,it is always the case.
I've looked for this problem in many reference books, one of which says that it is true, but it doesn't offer a proof.

A triangle ABC,a=BC, b=AC, c=AB , and the
angle C≥60°

This is my first time to turn to a foreigner for help,and your patient assistance shows that you're very kind. I do hope my poor English didn't make you confused.