The question
Let $ABXY$ be a tetrahedron such that the triangles $ABX$ and $ABY$ are isosceles right angle in $A$ with legs of $2$ cm. Calculate the volume of the tetrahedron, knowing that there is a point $P$ such that $PA = PB=PX = PY = \sqrt{5} cm$.
The drawing
my idea
$AB=AX=AY=2$ and using Phytagoras theorem we get $BX=BY=2\sqrt{2}$
We can use Chasles Formula to write the volume $$\frac{a*b* dis(a,b)*sin(a,b)}{6}$$ .
The volume of the tetrahedron is also equal to $\frac{2*A(AXY)}{3}$
I don't know where the point $P$ should be put.
I hope one of you can help me solve this problem. Thank you!
Best Answer
From symmetry, point $P$ lies on the intersection of the perpendicular bisecting plane of $AB$, and the perpendicular bisecting plane of $XY$ as well as the perpendicular bisecting plane of $AX$.
Now let the base of the tetrahedron $\triangle AXY$ be in the $xy$ plane, with
$ A = (0,0,0), X = (2, 0, 0) , Y = (2 \cos(2\theta) , 2 \sin(2 \theta), 0) $ and finally $B = (0, 0, 2) $
The perpendicular bisecting plane of $AB$ has the equation $z = 1$, and the perpendicular bisecting plane of $XY$ has the equation $ - \sin(\theta) x + \cos \theta y = 0 $. Finally, the bisecting plane of $AX$ has the equation $x = 1$. Therefore, point $P$ is given by
$P = (x, y, z) = (1, y , 1 ) $
with $ 0 = - \sin \theta + \cos \theta \ y $, i.e.
$y = \tan \theta$
Since $P$ is $\sqrt{5}$ away from $A$ which is at the origin, then
$ 2 + y^2 = 5 $
Therefore, $ y = \sqrt{3} $
And $\theta = \dfrac{\pi}{3} $
Therefore, the volume of the tetrahedron is
$ V = \dfrac{1}{3} \cdot \dfrac{1}{2} (2)^2 \sin(2 \theta) (2) = \dfrac{4}{3} \sin( \dfrac{2 \pi}{3} ) = \dfrac{2 \sqrt{3}}{3} $