Is it possible that in a pyramid ABCD ∠BAC=∠CAD=60° and ∠BAD=120°?

Question

Is it possible that in a pyramid ABCD ∠BAC=∠CAD=60° and ∠BAD=120°? If it's impossible it would be great if you would prove it.

Answer 1

This is impossible. Three vectors emanating from vertex $A$ are $\overrightarrow {AB}$ , $\overrightarrow {AC}$ and $\overrightarrow {AD}$ which we denote unit vectors in their directions by $v_1$ , $v_2$ and $v_3$ respectively. Showing $\vec v_1$ , $\vec v_2$ and $\vec v_3$ are coplanar is equivalent to show that: $$\vec v_1.(\vec v_2 \times \vec v_3)=0$$ where $|\vec v_1|=|\vec v_2|=|\vec v_3|=1$. The angle between two vectors $\vec a$ and $\vec b$ can be found using $\cos\theta=\frac{\vec a . \vec b}{|\vec a||\vec b|}$ therefore by appropriate substitution due to our problem we have: $$\vec v_1.\vec v_2=\frac{1}{2}\\ \vec v_1.\vec v_3=\frac{1}{2}\\\vec v_2.\vec v_3=-\frac{1}{2}$$ Since rotation is isometry we can arbitrarily take $\vec v_1=\hat i$ where $\hat i$ is unit vector along $x-axis$. Therefore by substituting $\vec v_1=(1,0,0)$ , $\vec v_2=(v_{2x},v_{2y},v_{2z})$ and $\vec v_3=(v_{3x},v_{3y},v_{3z})$ we obtain:$$v_{2x}=v_{3x}=\frac{1}{2} \\ v_{2y}v_{3y}+v_{2z}v_{3z}=-\frac{3}{4}$$ or $$2v_{2y}v_{3y}+2v_{2z}v_{3z}=-\frac{3}{2}$$ and knowing that $v_{2y}^2+v_{2z}^2=\frac{3}{4}$ and $v_{3y}^2+v_{3z}^2=\frac{3}{4}$ we get: $$v_{2y}^2+v_{2z}^2+v_{3y}^2+v_{3z}^2+2v_{2y}v_{3y}+2v_{2z}v_{3z}=0$$ or $$(v_{2y}+v_{3y})^2+(v_{2z}+v_{3z})^2=0$$ which implies that: $$v_{2y}=-v_{3y} \\ v_{2z}=-v_{3z}$$ Now for calculating $\vec v_1.(\vec v_2 \times \vec v_3)$ we can write: $$\vec v_1.(\vec v_2 \times \vec v_3)=v_{2y}v_{3z}-v_{3y}v_{2z}$$ that equals to zero by substitution and implies on impossibility of existence of such pyramid.