How to prove in any triangle that the area $X$ is given by:
$$X=\frac{1}{4}(a+b+c)^2\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}$$
Asked
Active
Viewed 98 times
0
-
There is a hint for the product of the $\tan '$s:http://math.stackexchange.com/questions/477364/prove-that-tan-a-tan-b-tan-c-tan-a-tan-b-tan-c-abc-180-circ – K. Rmth Dec 13 '13 at 21:35
-
It isn't a hint ... It's better -> $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ – Ewin Dec 13 '13 at 21:42
-
Is this better?: http://en.wikipedia.org/wiki/Heron%27s_formula. Look at the "Proof using the Law of cotangents and the triple cotangent identity" paragraph. – K. Rmth Dec 13 '13 at 21:48
-
Yes, it's a hint now -.- – Ewin Dec 13 '13 at 21:53
3 Answers
1
assume that c is the longest side and that d is perpendicular to c forming 2 right triangles. The angle between a and d should be called D. The line along c from a to d is c'.
$$Area=\frac 14(a+b+c)^2\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}$$ $$Area=\frac 14(a+c'+d)^2\tan\frac{D}{2}\tan\frac{B}{2}\tan45+\frac 14(b+d+c-c')^2\tan\frac{A}{2}\tan\frac{C-D}{2}\tan45$$ $$\tan 45=1$$ $$d^2=a^2-c'^2=b^2-(c-c')^2$$ IFF we can show the following that works, then the full one should too! $$\frac 14(a+c'+d)^2\tan\frac{D}{2}\tan\frac{B}{2}=\frac 12c'd$$ $$B=90-D$$ $$\tan \frac B2 = \frac{\sin B}{1+\cos B}$$ $$\tan \frac D2 = \frac{\sin D}{1+\cos D}$$ $$\sin B = \cos D = \frac da$$ $$\sin D = \cos B = \frac {c'}a$$ $$\tan\frac{D}{2}\tan\frac{B}{2} = \frac{c'd}{a^2+ac'+ad+c'd}$$ $$\frac 14(a+c'+d)^2\frac{c'd}{a^2+ac'+ad+c'd}=\frac 12c'd$$ $$(a+c'+d)^2=2(a^2+ac'+ad+c'd)$$ $$a^2+c'^2+d^2+2ac'+2ad+2c'd=2a^2+2ac'+2ad+2c'd$$ $$a^2+c'^2+d^2=2a^2$$ $$c'^2+d^2=a^2$$ which is true.... if we repeat for the other right triangle this works.... Doing it from the Heron is easier but only, I guess, if you know the Heron and can prove it.This uses http://en.wikipedia.org/wiki/Tangent_half-angle_formula so .... yeah....
kaine
- 1,672
-
-
I guess http://math.ucsd.edu/~wgarner/math4c/textbook/chapter6/doublehalfangles.htm but I used it because it is the only half angle formula I know off the top of my head. – kaine Dec 13 '13 at 22:32
0
Use the popular identity: $$\tan\left(\frac A2\right)\tan\left(\frac B2\right)\tan\left(\frac C2\right) = \frac rp$$ $$RHS = \frac{1}{4}\times 4p^2*\frac rp = pr = A = LHS$$
-
1
-
-
-
1
-
-
When you say "popular identity", what do you mean? If you mean "the one the cool kids know", that may be true (for some definition of "cool"), but the OP is then not a cool kid, so explain to him/her and the rest of us. – Igor Rivin Dec 13 '13 at 21:55
-
@IgorRivin It's enough $tan(\frac A2)tan(\frac B2)tan(\frac C2) = \frac rp$ to prove. – Ewin Dec 13 '13 at 21:59
-
So I can do a demo on A = pr. So let AB = c, BC = a, and CA = b, then A = 1/2ar + 1/2br + 1/2*cr = 1/2(2p)r = pr. – DeepSea Dec 13 '13 at 22:02
-
1@Ewin what is it with you guys? If the OP knew this, do you think he would ask? – Igor Rivin Dec 13 '13 at 22:02
-
(tan(A/2)tan(B/2)tan(C/2))^2 = ((p-b)(p-c)/p(p-a))((p-a)(p-c)/p(p-b))((p-a)(p-b)/p(p-c))= (p-a)(p-b)(p-c)/p^3 = (A^2/p)/p^3 = A^2/p^4 = (pr)^2/p^4 = r^2/p^2. So taking square root we get the identity above. – DeepSea Dec 13 '13 at 22:10
-
1
-
-1
From this, $$\tan\frac A2=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\text{ where } s=\frac{a+b+c}2$$
This can be easily derived using Tangent half-angle formula and the Law of cosines
Now we know Heron's formula
lab bhattacharjee
- 274,582