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If $\tan^3A+\tan^3B+\tan^3C=3\tan(A)\tan(B)\tan(C)$, prove triangle ABC is equilateral triangle

Now i remember a identity which was like if $a+b+c=0$,then $a^3+b^3+c^3=3abc$. So i have $\sum_{}^{} \tan(A)=0$. How do i proceed?

Thanks.

S.C.B.
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Taylor Ted
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  • No, you don't necessarily have $\tan(A)+\tan(B)+\tan(C)=0$. If $a,b,c\in\mathbb{R}$ are such that $a^3+b^3+c^3=3abc$, then either $a=b=c$ or $a+b+c=0$. Thus, you can have $\tan(A)=\tan(B)=\tan(C)$, which means $ABC$ is an equilateral triangle. – Batominovski Feb 07 '17 at 11:42
  • On the other hand, if $\tan(A)+\tan(B)+\tan(C)=0$, then we can assume WLOG that $A$ and $B$ are acute, so we get $\tan(A)+\tan(B)=-\tan(C)=\tan(A+B)=\frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$. You should know what to do next. – Batominovski Feb 07 '17 at 11:43
  • @N.S.John you could use the factorization of $a^3+b^3+c^3-3abc$, as in my answer. – S.C.B. Feb 07 '17 at 11:49
  • @Batominovski Proceeding as you suggested i got tanAtanB=1, so tanC=0 implies C=0. but C can be zero, where is contradiction – Taylor Ted Feb 08 '17 at 04:40

2 Answers2

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Note that if $A,B,C$ are angles of a triangle, we have that $$\tan A+\tan B +\tan C=\tan A \tan B \tan C$$ As seen here. If $$0=\tan A+\tan B +\tan C=\tan A \tan B \tan C$$ Then we have that for at least one angle among $A,B,C$ is $0$ or $\pi$. Thus, it is a contradiction. So $$\tan A+\tan B +\tan C \neq 0$$ Now exploit the fact $$a^3+b^3+c^3-3abc=0$$ is true only if $a=b=c$ given that $a+b+c \neq 0$. Thus $\tan A=\tan B= \tan C$. Thus $\triangle {ABC}$ is equilateral.

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S.C.B.
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  • Can you please tell me how is atleast one angle amone A,B,C is 0 or $\pi$ – Taylor Ted Feb 08 '17 at 04:28
  • also in a triangle we can have one angle to be 0 degrees which is why cos0 , sin0 makes sense? – Taylor Ted Feb 08 '17 at 04:35
  • @TaylorTed In Wikipedia note that the three points in a triangle are non-colinear. So if one of the angles is $0$ degrees, that makes all three point colinear. So it is impossible. – S.C.B. Feb 08 '17 at 05:32
  • @Taylor Also, $\sin 0$ is not defined using the right trinagle. Only acute angles (angles between $0$ and $\frac{\pi}{2}$) are defined that way. – S.C.B. Feb 08 '17 at 05:33
  • @TaylorTed A quick search reveals that $\sin x$ is defined as $$ \sin x=x-\frac{x^3}{6}+\frac{x^5}{120} -\dots$$ In modern times. – S.C.B. Feb 08 '17 at 05:34
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$$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc=\{(a+b)+c\}\{(a+b)^2-(a+b)c+c^2\}-3ab\{(a+b)+c\}$$

$$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

But $2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2$

Now if $a^3+b^3+c^3-3abc=0$

either $a+b+c=0$

But with $A+B+C=m\pi,\tan A+\tan B+\tan C=\tan A\tan B\tan C$

As $0<A,B,C<\pi; \tan A\tan B\tan C\ne0$

So, we need $(\tan A-\tan B)^2+(\tan B-\tan C)^2+(\tan C-\tan A)^2=0\ \ \ \ (1)$

As $\tan A,\tan B,\tan C$ are real, each of $(\tan A-\tan B)^2,(\tan B-\tan C)^2,(\tan C-\tan A)^2$ must be $\ge0$

$(1)\implies$ $$(\tan A-\tan B)^2=(\tan B-\tan C)^2=(\tan C-\tan A)^2=0$$

Observation : The proposition will hold if in the condition, $\tan$ is replaced with $\sin,\cos$

Can you prove them?

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lab bhattacharjee
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