Equilateral triangle?

Relevant wiki: Jensen's Inequality

(Notice that all the angle in this solution is degree not radians.) Let $\angle{BAK}=c,\angle{CBK}=a,\angle{ACK}=b$

By ceva's theorem in trigonometric form, (Ceva thoerem states that $sin\angle{ABK}sin\angle{BCK}sin\angle{CAK}=sin\angle{BAK}sin\angle{CBK}sin\angle{ACK}$ ), $\frac{(sin30)^3}{\sin{a}\sin{b}\sin{c}}=1$ $\sin{a}\sin{b}\sin{c}=\frac{1}{8}$

By AM-GM inequality, $\begin{aligned} \sin{a}+\sin{b}+\sin{c} &\geq 3\sqrt[3]{\sin{a}\sin{b}\sin{c}} \\ &\geq\frac{3}{2} \end{aligned}$

By Jensen inequality,(a+b+c+3 x 30=180) $\begin{aligned} \sin{a}+\sin{b}+\sin{c}&\leq3\sin{\frac{180-3\times30}{3}}\\ &\leq\frac{3}{2} \end{aligned}$

Hence, $\sin{a}+\sin{b}+\sin{c}=\frac{3}{2}$ As AM-GM hold equality when all variables are same, which mean $\sin{a}=\sin{b}=\sin{c}$ $a=b=c$ Hence, all the angle of triangle ABC are same, which means triangle ABC is equilateral triange.

Drop perpendiculars from $K$ to the side lengths of the triangle, as shown above. Since $\triangle AYK, BZK, CXK$ are all right triangles, we have

$KX = KC \sin 30^{\circ} = \dfrac{1}{2} KC \\ KY = KA \sin 30^{\circ} = \dfrac{1}{2} KA \\ KZ = KB \sin 30^{\circ} = \dfrac{1}{2} KB. \\$

Thus,

$KX + KY + KZ = \dfrac{1}{2}(KA + KB + KC).$

However, the Erdös-Mordell Inequality states that $KX + KY + KZ \leq \frac{1}{2}(KA + KB + KC),$ with equality holding if and only if $\triangle ABC$ is equilateral. In this scenario, equality holds, so we conclude that yes , $\triangle ABC$ must be equilateral.

Let $\angle BAK=α, \angle CBK=β, \angle ACK=γ$

We have $α+β+γ=90^{\circ}$ and from Ceva's Theorem in trigonometric form, $\frac{\sin 30^{\circ}}{\sin α} · \frac{\sin 30^{\circ}}{\sin β} · \frac{\sin 30^{\circ}}{\sin γ} = 1$

Rearranging, $\sin α \sin β \sin γ = \frac{1}{8}$

Adding the natural logarithm to both sides, $\ln \sin α + \ln \sin β + \ln \sin γ = \ln \frac{1}{8}$

Now denote the function $f:(0,\frac{π}{2}) \mapsto \mathbb{R}^{-}, f(x) = \ln \sin x$ , since $f'(x) = \frac{1}{\sin x} · \cos x = \cot x$ and $f''(x) = - \csc^2 x < 0$ , $f(x)$ is a concave function. Therefore by Jensen's inequality, $\frac{f(α)+f(β)+f(γ)}{3} \leqslant f(\frac{α+β+γ}{3})$

$\frac{\ln \sin α + \ln \sin β + \ln \sin γ}{3} \leqslant \ln \sin \frac{α+β+γ}{3} = \ln \frac{1}{2}$

$\therefore \ln \sin α + \ln \sin β + \ln \sin γ \leqslant \ln \frac{1}{8}$

Equality holds true if and only if $α=β=γ$ or $f(x)$ is linear, the latter of which is not the case as $f''(x)<0$

Thus $α=β=γ=30^{\circ}$ and $\triangle ABC$ is obviously an equilateral triangle.

Note: With some modifications, the same method can also be applied to solve P5 of IMO 1991.

Another approach is to use straight-forward analytic geometry.

Without loss of generality, put B at the origin, C on the x-axis one unit away (this distance simply sets the length scale), and the point A at $\{a_x, a_y\}$ . Further, restrict A to the first quadrant.

Then we can represent rays from each of the vertices towards the point K. First, consider the ray from B towards K. The direction for this ray is the vector from B to A rotated 30 degrees clockwise, and $\beta$ is a parameter along this ray.

$\left\{\frac{1}{2} \beta \left(\sqrt{3} a_x+a_y\right),-\frac{1}{2} \beta \left(a_x-\sqrt{3} a_y\right)\right\}$

From A towards K, $\alpha$ is a parameter along the ray.

$\left\{-\frac{1}{2} \sqrt{3} \alpha \left(a_x-1\right)+a_x-\frac{\alpha a_y}{2},\frac{\alpha a_x}{2}-\frac{1}{2} \alpha \left(\sqrt{3} a_y+1\right)+a_y\right\}$

From C towards K, $\gamma$ is a parameter along the ray.

$\left\{1-\frac{\sqrt{3} \gamma }{2},\frac{\gamma }{2}\right\}$

The point K is the intersection of these three rays, and we demand this be a single point. Setting these three rays equal to each other, and eliminating the parameters $\alpha$ , $\beta$ , and $\gamma$ , yields the following relationship:

$a_x^2+a_y^2+1=a_x+\sqrt{3} a_y$

which can be rewritten, by completing the square, into the suggestive form

$\left(a_x-\frac{1}{2}\right){}^2+\left(a_y-\frac{\sqrt{3}}{2}\right){}^2 = 0$

This is a circle centered at $\{\frac{1}{2}, \frac{\sqrt{3}}{2} \}$ , of radius 0. In other words, a single point. But this point makes the triangle equilateral.

Draw circles AKC, AKB, BKC, then we will see that angle KBC=KAB=KCA.

It appears to be that, since all mentioned angles (ABK, BCK, CAK) are equal to 30 degrees and an equilateral triangle has equal angles at all corners, we can infer that point K must be within the plane of an equilateral triangle.

Inscribed in a circle we take an arbitrary triangle ABC and from vertexes A, B and C draw lines forming an arbitrary angle w.
In our case w=30 these three new lines will form another triangle A1B1C1 congruent with the original one ABC . Then since we look for a triangle A1B1C1 which area must be null, then points A1, B1, C1 and K must been all coincident, the only solution is that our original points A,B,C inscribed in the circle form an equilateral triangle. Of course K is also the Fermat point

If angles' sum of a triangle is 180° so it's equilateral.