Recently on a math test, I was given a the following problem.
Show that the points: $O=(0, 0, 0)$, $A=(6, 0, 0)$, $B=(6, -\sqrt{24}, \sqrt{12})$, $C=(0, -\sqrt{24}, \sqrt{12})$ form a square.
The "correct" method involved showing that the magnitudes of $OA$, $AB$, $BC$, and $CO$ are equal, and that one angle is perpendicular, but since the points aren't shown to be coplanar, this shouldn't be valid without also showing that they are.
My method, which was apparently incorrect, was this:
- Show that $OB$ and $AC$ have the same magnitudes.
- Show that $OA$ and $BC$ are parallel and have the same magnitude.
This should correctly show that the points form a square and are coplanar, as the second part shows that the shape is a parallelogram, and a parallelogram with equal diagonals is a square.
Since the problem came from the IB Math HL test in 2014, it seems unlikely that the given method would be invalid, but I can't see how it is right. Why is it not necessary to show that all the points are coplanar?
Your objection is valid.
For example, consider the $4$ points $O(0,0,0),\;A(2,0,0),\;B(1,1,\sqrt{2}),\;C(0, 2, 0)$.
Then $OA = AB = BC = CO = 2$, and $\angle COA = 90^\circ$.
But $OACB$ is definitely not a square, since $O,A,B,C$ are not coplanar.
But your method is also incorrect, since it wouldn't distinguish between a square and a rectangle.
A rhombus with equal diagonals is a square. A parallelogram with equal diagonals is a rectangle, but not a square, unless the parallelogram is actually a rhombus.