The volume of a tetrahedron $ABCD$ is $\frac{1}{6}$. Determine $CD$ knowing that $\angle ACB=45$ and that $AD+BC+\frac{AC \sqrt{2}}{2}=3$.

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the question

The volume of a tetrahedron $ABCD$ is $\frac{1}{6}$. Determine $CD$ knowing that $\angle ACB=45°$ and that $AD+BC+\frac{AC \sqrt{2}}{2}=3$.

my drawing

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my idea

From the equality we can say that
$AC \sqrt{2}=6-2AD-2BC$

From the thorem of the cosine applied in the triangle ABC $AB^2=AC^2+BC^2-AC×BC×\sqrt{2}$

Also we can say that the area $A(ABC)=\frac{AC×BC×\sqrt{2}}{4}$ and the volume of the tetrahedron is $V(ABCD)=\frac{1}{6}=\frac{AD×BC×dis(AD,BC)×\sin(AD,BC)}{6}=\frac{AC ×BC ×\sqrt{2}×dis(D, (ABC) )}{12}$

Im pretty sure we need some auxiliar constructions but dont know which one. Hope one fo you can help me! Thank you!

Best Answer

Let's define: $x=BC$, $y=AC/\sqrt2$, $z=AD$. We obtain then, from the given relations: $$ x+y+z=3 $$ and $$ xyh=1, $$ where $h$ is the distance of $D$ from plane $ABC$. The tetrahedron can exist only if $z\ge h$, that is if: $$ xyz\ge1. $$ But from AM-GM inequality we have: $$ 1={x+y+z\over3}\ge\root3\of{xyz}\ge1, $$ hence arithmetic and geometric mean are the same, implying $$ x=y=z=1. $$ We also get $z=h=1$, that is $AD$ is perpendicular to plane $ABC$. From that it is easy to find $CD=\sqrt3$.

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