In the triangular pyramid $MABC$ all side edges equals $1$, $\angle AMB = \angle BMC = 60 ^\circ$, $\angle AMC = 45 ^\circ$.
Find:
1) square of the $\triangle ABC$;
2) dihedral angle on the $AB$ edge;
3) $\rho(M, ABC)$.
Help me, please.
P.S. sorry for my English, I am not a native speaker.
Wow the question recalls my old memory about high school. Actually you know $$AM = MC = 1$$ and because of the law of cosines $$cos(\angle AMC) = \frac{\sqrt{2}}{2} = \frac{1^2+1^2-AC^2}{2\cdot 1 \cdot 1}\Rightarrow AC = \sqrt{2-\sqrt{2}}.$$ The area can be computed by Heron's formula.
Notice $\triangle AMC = \triangle ABC$, so $\angle ABC = \pi/4.$ Draw a line(in plane $ABC$) through the middle point $D$ of $AB$ and it intersects $BC$ at $E$. We know $$AD = BD = DE = AB/2 = \frac{1}{2}\text{ and }BE = \frac{\sqrt{2}}{2}$$
And you can show $AB\perp \triangle MDE$ by $AB\perp MD, DE$. So $\angle MDE$ is what you want to solve. Again invoking the law of cosine for $\angle MBE$, $ME^2 = 1+\frac{1}{2}-\frac{\sqrt{2}}{2} = \frac{3-\sqrt{2}}{2}$. Now use the same law to $\angle MDE:$ $$\cos\angle MDE = \frac{\frac{3}{4}+\frac{1}{4}-ME^2}{2\cdot\frac{\sqrt{3}}{2}\cdot\frac{1}{2}} = \frac{3-\sqrt{3}}{3}.$$
Immediately from this result, draw a line $MF$ in plane $MDE$ intersecting $CD$ at $F$. Since $AB\perp MDE\Rightarrow AB\perp MF$ and $MF\perp CD$, we have $MF\perp ABC$, and $$\rho(M,ABC) = MD\cdot\sin\angle MDE.$$
