Calculate the volume of the tetrahedron...

Question

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!

Answer 1

Let the midpoint of A and B to M, the point where the line perpendicular to surface AXY and goes through P meets surface AXY to H. Since PA=PB=PX=PY , A,B,Y and X are vertices of the pyramid P is the circumscribed sphere.

Since PB=PA ∠PMB=90

PM = √(BP^2-BM^2)=2

Since ∠BAX = ∠BAY= 90 and X,Y and H share the safe surface ∠BAH = 90

Therefore AB∥PH and PM∥AH so PM = AH = 2

Since AB∥PH , P is centre of sphere and H is in the same surface as A,X,Y H is the centre of the circumcircle of △AXY .

Therefore AH = XH = YH = 2

Since AX=AY=2 △AXH and △AYH are regular triangles.

Therefore ∠XAH and ∠YAH = 60 and ∠XAY = 120

Therefor XY = 2√3 and △AXY =√3 Therefor volume of pyramid = △AXY*AB *1/3 =2√3/3 I think this is much simple then using sin like others.

Answer 2

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} $