0
$\begingroup$

How can one show that if in a tetrahedron all plane angles in every vertex separately are equal, then this tetrahedron is regular?

What can I use here?

$\endgroup$
1
$\begingroup$

If $a$, $b$, $c$, $d$ are the plane angles about respective vertices $A$, $B$, $C$, $D$, then just we just solve the linear system $$\begin{align} a + b + c &= 180^\circ \qquad \text{(from } \triangle ABC \text{ )} \qquad (1)\\ a + c + d &= 180^\circ \qquad \text{(from } \triangle ACD \text{ )} \qquad (2)\\ a + d + b &= 180^\circ \qquad \text{(from } \triangle ADB \text{ )} \qquad (3)\\ b + c + d &= 180^\circ \qquad \text{(from } \triangle BCD \text{ )} \qquad (4) \end{align}$$

Note that subtracting any equation from any other reveals a pair of angles to be equal; in particular, $$\begin{align} (1)-(2) &\qquad\to\qquad b = d \\ (1)-(3) &\qquad\to\qquad c = d \\ (1)-(4) &\qquad\to\qquad a = d \\ \end{align}$$ Therefore, $a=b=c=d = 60^\circ$, so that each face is equilateral (and necessarily congruent to the others), making the tetrahedron regular. $\square$

$\endgroup$
2
$\begingroup$

Let the tetrahedron have vertices $A$, $B$, $C$, and $D$. Let the angles at $A$, $B$, $C$, and $D$ be $a$, $b$, $c$, and $d$ respectively. Then because $ABC$ and $ABD$ form triangles, we have $c=d=180-a-b$. Now consider triangle $ACD$, from this we get $180=a+c+d=a+2c$, but considering triangle $ABC$ again we have $180=a+b+c=a+2c$. Hence $b=c=d$. Considering triangle $BCD$ now gives us $180=b+c+d=3c$, so $b=c=d=60$. Considering triangle $ABC$ one last time, we have $180=a+b+c=a+120$. Hence $a=b=c=d=60$, so every face of the tetrahedron is an equilateral triangle, so the tetrahedron must be regular and we are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.