Finding the angles of a non-equilateral $\triangle ABC$ with centroid $G$ such that $\angle GAB=\angle GCA=30^\circ$

Question

In the attached figures, $G$ is the centroid of $\triangle ABC$. When this triangle is equilateral ( fig 1) , it is obvious that each of the two angles shown in this figure measures $30^\circ$.

Using GeoGebra, I tried to find another triangle in which these same two angles are also equal to $30^\circ$ each. So after several attempts, it appeared to me that such a triangle does exist ( fig 2) , and one of its angles seems to be $120^\circ$.

So, I'm wondering how to calculate the exact angles of this amazing triangle ?!

Answer 1

The angles are $\angle C = \arctan(\sqrt{3}/2) \approx 40.9^\circ$, $\angle A = 120^\circ$, and $\angle B = 60^\circ- \angle C = \arcsin\left(\frac{\sqrt{3}}{2\sqrt{7}}\right)\approx 19.1 ^\circ$.

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Longer Answer:

We can use coordinate geometry. WLOG by shifting, rotating, and scaling, let $C = (0,0)$ and $A = (1,0)$ with $B = (x,y)$ unknown.

Then $G = \left(\frac{x+1}{3},\frac{y}{3}\right)$.

We have our two conditions as follows:

First, by the definition of the tangent function,

$$\angle GCA = 30^\circ \implies \frac{y/3}{(x+1)/3} = \tan(30^\circ) = \frac{1}{\sqrt{3}}$$

so we have

$$y = \frac{x+1}{\sqrt{3}}$$

Second, note that continuing the line $AG$ intersects the midpoint $D$ of $BC$, which is at coordinates $(x/2,y/2)$.

We use the dot-product formula

$$\sqrt{3}/2 = \cos(30^\circ) = \frac{AD \cdot AB}{||AD|| \hspace{5pt}||AB||}$$

where the vectors are

$$AD = (x/2 - 1, y/2)$$ $$AB = (x-1,y)$$

Squaring,

$$ \frac{3}{4} = \frac{\left((x/2 - 1)(x-1)+(y/2)y\right)^2}{((x/2-1)^2+(y/2)^2)((x-1)^2+y^2)}$$

Substituting in our above formula for $y$ and simplify through a lot of algebra,

$$\frac{3}{4} = \frac{16x^4-56x^3+105x^2-98x+49}{16x^4-56x^3+108x^2-92x+52}$$

Multiplying out the denominator, we get a degree-4 polynomial whose solutions are $x = 1/2$, $x=2$ and two complex solutions.

$x=1/2$ is the equilateral triangle, and $x=2$ corresponds to the other solution you found with $y = \sqrt{3}$.

Then the angles are $\angle C = \arctan(\sqrt{3}/2) \approx 40.9^\circ$, $\angle A = 120^\circ$, and $\angle B = 60^\circ- \angle C \approx 19.1 ^\circ$.

Answer 2

The "nontrivial" triangle is in fact related to the equilateral one through cevian division.

In equilateral $\triangle ABC$, draw cevians from each the vertices to rotationally equivalent trisection points on the respective opposing sides, thus defining an inner equilateral $\triangle A'B'C'$. $AC',A'C'$ and their rotational equivalents are all congruent and taken to measure one unit, then (with the 120° exterior angles at the inner equilateral triangle) the sides of the outer $\triangle ABC$ are easily seen to measure $\sqrt{1^2+(1×2)+2^2}=\sqrt7$.

$\overline{BC'}$ is the base of an isosceles triangle with unit legs and apex angle measuring 120°, so it measures $\sqrt3$ and that makes $\triangle A'BC$ right with right angle at $A$. So the perpendicular to $CC'$ dropped from $B'$ to $A''$ is a median of $\triangle BB'C$, and $\overline{CA'}$ is another median. Now $\angle A'CB'$ measures 30° because it is a base angle of an isosceles triangle with a 120° apex angle, and $\angle A'B'A''$ measures 30° by difference between the right angle and the 120° angle at $B'$. Therefore

$\triangle BB'C$ meets the problem requirements with a 120°angle at $B'$.

This triangle agrees in shape and side lengths with the one identified through coordinate geometry by mathperson314.

Answer 3

In case someone disagrees about what exactly constitutes an "amazing" triangle, I'll note that, generally, taking $$p := \cot\angle GAB \qquad q := \cot\angle GCA $$ a bit of work can show potentially four solution triangles $$\begin{align} a^2:b^2:c^2 &= 4 + \left( p +3 \sigma_1 q +\sigma_2 t\right)^2 \\[4pt] &\,:\, 4 \left(1+q^2\right) \\[4pt] &\,:\, 16 + \left( p +\sigma_1 q + \sigma_2 t\right)^2 \end{align} \tag1$$ where $$t := \sqrt{p^2 + q^2 + 10\sigma_1 pq-24} \qquad \sigma_1:=\pm 1 \qquad \sigma_2:=\pm 1$$ with the signs of $\sigma_1$ and $\sigma_2$ chosen independently.

In particular, when $p=q$, we can ignore $\sigma_1=-1$ (for making $t$ imaginary), and we have $$\begin{align} a^2:b^2:c^2 &= 1 + \left(2 p \pm \sqrt{3 (p^2 - 2)}\right)^2 \\[4pt] &\,:\,1 + p^2 \\[4pt] &\,:\,4 + \left(1 p \pm \sqrt{3 (p^2 - 2)}\right)^2 \end{align} \tag2$$ where the "$\pm$"s match. Interestingly, $a^2+b^2=2c^2$ for any $p$.

In particular-particular, if $p=q=\cot 30^\circ=\sqrt{3}$, this reduces to $$a^2:b^2:c^2 \;=\; 8 \pm 6 : 2 : 5 \pm 3 \;=\; 7:1:4 \;\text{or}\; 1:1:1 \tag3$$ The latter option corresponds to the equilateral triangle, while the former has an angle at $A$ satisfying $$\cos A = \frac{-a^2+b^2+c^2}{2bc} = -\frac12 \quad\to\quad A = 120^\circ$$

Answer 4

Let vertices of the triangle be $A\left(0;0\right)$, $B\left(x;y\right)$, $C\left(-1;0\right)$. Then $G\left(\frac{x-1}{3};\frac{y}{3}\right)$, and the possible locations of the point $B$ such that $\angle BAG=\angle ACG=\frac{\pi}{6}$ can be found as the two intersections of the line $$ y\cdot\sqrt{3}=x+2 $$ (which is "responsible" for $\angle ACG=\frac{\pi}{6}$) and the circle $$ \left(2\cdot x-1\right)^2+\left(2\cdot y-\sqrt{3}\right)^2=4 $$ (which is "responsible" for $\angle BAG=\frac{\pi}{6}$):

When $B$ is on the circle $\left(2\cdot x-1\right)^2+\left(2\cdot y-\sqrt{3}\right)^2=4$, one can rewrite $y^2=\sqrt{3}\cdot y-x^2+x$ and

$$ \begin{align} \left(\cos\angle BAG\right)^2 &= \left(\frac{\vec{AB}\cdot\vec{AG}}{\left|AB\right|\cdot\left|AG\right|}\right)^2 = \frac{\left(x\cdot\frac{x-1}{3}+y\cdot\frac{y}{3}\right)^2}{\left(x^2+y^2\right)\cdot\left(\left(\frac{x-1}{3}\right)^2+\left(\frac{y}{3}\right)^2\right)} =\\ &= \frac{\left(x\cdot\left(x-1\right)+y^2\right)^2}{\left(x^2+y^2\right)\cdot\left(\left(x-1\right)^2+y^2\right)} =\\ &= \frac{\left(x\cdot\left(x-1\right)+\sqrt{3}\cdot y-x^2+x\right)^2}{\left(x^2+\sqrt{3}\cdot y-x^2+x\right)\cdot\left(\left(x-1\right)^2+\sqrt{3}\cdot y-x^2+x\right)} =\\ &= \frac{\left(\sqrt{3}\cdot y\right)^2}{\left(\sqrt{3}\cdot y+x\right)\cdot\left(-x+1+\sqrt{3}\cdot y\right)} =\\ &= \frac{3\cdot y^2}{\sqrt{3}\cdot y+3\cdot y^2+x-x^2} =\\ &= \frac{3\cdot\left(\sqrt{3}\cdot y-x^2+x\right)}{\sqrt{3}\cdot y+3\cdot\left(\sqrt{3}\cdot y-x^2+x\right)+x-x^2} =\\ &= \frac{3\cdot\left(\sqrt{3}\cdot y-x^2+x\right)}{4\cdot\left(\sqrt{3}\cdot y-x^2+x\right)} = \frac{3}{4} \end{align} $$