HongKong 2022-23 IMO Selection Tests

Answer: $\frac{5\sqrt{39}}{2}$

Let the vertices of the pyramid be $A$, $B$, $C$, $D$, where $BD=6$ and all other edges have length $5$. Let also $M$ be the midpoint of $BD$ and $N$ be the foot of perpendicular from $A$ to the base $BCD$. Of course we have $BM=MD=3$ and $CM=4$. Note that $$A{N}^2=A{B}^2-B{N}^2=A{C}^2-C{N}^2=A{D}^2-D{N}^2.$$ As $AB=AC=AD=5$, we have $BN=CN=DN$ and so $N$ is the circumcentre of $△BCD$. The circumradius of $△BCD$ is given by the extended sine formula $$BN=\frac{CD}{2\sin B}=\frac{5}{2\cdot \frac{4}{5}}=\frac{25}{8}.$$ The area of $△BCD$ is $\frac{1}{2}\cdot BD\cdot MC=12$. The volume of the pyramid is thus $$\frac{1}{3}\cdot 12\cdot AN=4\cdot \sqrt{5^2-{\left(\frac{25}{8}\right)}^2}=\frac{5\sqrt{39}}{2}.$$