How to prove that ABC is isosceles where $\angle{BAC}=20^\circ$ and $\sqrt[3]{a^3+b^3+c^3-3abc}= \min \{b,c\}$

Question

Let $\bigtriangleup{ABC}, \angle{BAC}=20^\circ$ and $\sqrt[3]{a^3+b^3+c^3-3abc}= \min \{b,c\}$. How to prove that ABC is isosceles? I try to use that $a^3\cos(B−C)+b^3\cos(C−A)+c^3\cos(A−B)=3abc$.

Answer 1

First, let's prove the lemma below.

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Lemma: The minimum value of $f(x)=\frac{\sin^2x}{\sin(20^{\circ}+x)},$ where $x \in [80^{\circ}, 160^{\circ})$, is $\sin20^{\circ}(1+2\cos20^{\circ}).$

Proof: $f(x)$ is continuous, and we have:

$$f'(x)=\frac{2\sin x\cos x\sin(20^{\circ}+x)-\sin^2x\cos(20^{\circ}+x)}{\sin^2(20^{\circ}+x)}\\=\frac{\sin x(2\cos x\sin(20^{\circ}+x)-\sin x\cos(20^{\circ}+x))}{\sin^2(20^{\circ}+x)}\\=\frac{\sin x(\cos x\sin(20^{\circ}+x)+\sin 20^{\circ})}{\sin^2(20^{\circ}+x)}\\=\frac{0.5\sin x (\sin (20^{\circ}+2x)+3\sin 20^{\circ})}{\sin^2(20^{\circ}+x)}>0,$$

where $x \in [80^{\circ}, 160^{\circ})$ because: $$\sin (20^{\circ}+2x)+3\sin 20^{\circ} \geq -1+3\sin 20^{\circ} >0.$$

Therefore, $$f(x) \geq f(80^{\circ})=\sin 80^{\circ}=\sin20^{\circ}(1+2\cos20^{\circ}).$$

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Now, assume $c\geq b$, so $160^{\circ}> \angle C \geq 80^{\circ}$.

And,

$$a^3+c^3=3abc \implies a(b^2+c^2-2bc \cos 20^{\circ})+c^3=3abc\\ \implies ab^2+ac^2+c^3=abc(3+2\cos 20^{\circ}) \\ \implies \frac{b}{c}+\frac{c}{b}+\frac{c^2}{ab}=3+2\cos 20^{\circ}.$$

But on the other hand by the lemma above and a very simple application of AM-GM:

$$\frac{b}{c}+\frac{c}{b} \geq 2, \\ \frac{c^2}{ab}=\frac{\sin ^2 \angle C}{\sin 20^{\circ} \sin (20^{\circ}+\angle C)} \geq 1+2\cos20^{\circ},$$

which gives $b=c$ and $\angle C=80^{\circ}.$

We are done.

Answer 2

Without loss of generality let $b\le c$. We have $\cos 20^\circ = \dfrac{b^2+c^2-a^2}{2bc}$. Using the triple-angle formula $\cos 3\varphi = 4\cos^3 \varphi - 3 \cos \varphi$ we find $$\frac 12 = \cos 60^\circ = 4\cos^3 20^\circ - 3\cos 20^\circ = 4\left(\frac{b^2+c^2-a^2}{2bc}\right)^3-3\cdot \frac{b^2+c^2-a^2}{2bc}.$$ Clearing denominators we obtain $$0=a^6-3a^4(b^2+c^2)+3a^2(b^2+bc+c^2)-b^6+b^3c^3-c^6.$$ We are given $a^3=3abc-c^3$. Substituting this to RHS we get \begin{align*} 0&=(3abc-c^3)^2-3a(3abc-c^3)(b^2+c^2)+3a^2(b^4+b^2c^2+c^4)-b^6+b^3c^3-c^6 \\ &=3a^2(b^4-3b^3c+4b^2c^2-3bc^3+c^4)+3ac^3(b^2-2bc+c^2)+b^3(c^3-b^3)\\ &=3a^2(b-c)^2(b^2-bc+c^2)+3ac^3(b-c)^2+b^3(c^3-b^3) \\ &\ge 0 \end{align*} with equality if and only if $b=c$.

Answer 3

The difficulty of this problem, (at least for me who has spent a lot of time solving it) consists of wanting to use the equality $b=\dfrac{a^3+c^3}{3ac}$ to derive the result, which leads to complicated calculations. The matter turned out to be the other way around: we can show that, in every isosceles triangle with a minor angle equal to twenty degrees, the property is verified. We are done if we have previously underlined the fact that in the capable arc corresponding to side BC there is only one isosceles triangle.

In the attached figure, the corresponding (red) arc capable of the segment $\overline{BC}$ which play the role of side $a$ of our triangle $\triangle {ABC}$. We have $$\overline{AC}=2r\cos(10^{\circ})\\\overline{BC}=2r\sin(20^{\circ})$$ and we finish verifying that $$\frac{(\overline{AC})^3+(\overline{BC})^3}{3\overline{AC}\cdot\overline{BC}}=\overline{AC}$$ i.e. $$\frac{(2r\cos(10^{\circ}))^3+(2r\sin(20^{\circ})^3}{3\cdot2r\cos(10^{\circ})\cdot2r\sin(20^{\circ})}=2r\cos(10^{\circ})$$ which reduces to $$1+8\sin^3(10^{\circ})=6\sin(10^{\circ})$$ This is true each member being equal to $1.041889066$.

Thus it is a property that characterizes every isosceles triangle whose minor angle measures twenty degrees.