The vertices of a plane quadrilateral are labelled $A, B, A'$ and $B'$, in clockwise order. A point $O$ lies in the same plane and within the quadrilateral. The angles $AOB$ and $A'OB'$ are right angles, and $OA = OB$ and $OA' = OB'$.
Use position vectors relative to $O$ to show that the midpoints of $AB$, $BA'$, $A'B'$ and $B'A$ are the vertices of a square.
Given that the lengths of $OA$ and $OA'$ are fixed (and the conditions of the first paragraph still hold), find the value of angle $BOA'$ for which the area of the square is greatest.
For the first one, I have tried letting the vertices $A,B,A´,B´$ be denoted by the vectors $a, b, a´, b´$ and then I computed the vectors between the midpoints. I got that opposite sides are equal in length, however, when trying to use the dot product to prove that adjacent sides are perpendicular, I was unable to do so.
For the second part I think I need to get the first part right, so therefore I am at a loss. Is there a way of expressing the midpoints in a simpler, more elegant way?
[Math] How to prove that a quadrilateral is a square with vectors
geometry
Best Answer
the mid points are:
$\frac a2 + \frac b2\\\frac b2 + \frac {a'}2\\\frac {a'}2 + \frac {b'}2\\\frac {b'}2 + \frac {a}2$
The vector from each vertex to the one clockwise from it:
$\frac b2 + \frac {a'}2\ - (\frac a2 + \frac b2) = \frac {a'}2\ - \frac a2\\ \frac {a'}2 + \frac {b'}2 -(\frac b2 + \frac {a'}2) = \frac {b'}2\ - \frac b2\\ \frac {b'}2 + \frac {a}2 - (\frac {a'}2 + \frac {b'}2)=\frac a2\ - \frac {a'}2\\ \frac a2 + \frac b2\ - (\frac {b'}2 + \frac {a}2) = \frac b2\ - \frac {b'}2$
$\frac {a'}2\ - \frac a2$ is parallel to and of equal magnitude as $\frac a2\ - \frac {a'}2$
Same for the other two sides.
Our quadrilateral is a parallelogram.
If $(\frac a2\ - \frac {a'}2)\cdot(\frac b2\ - \frac {b'}2) = 0,$ our vertices are right angles. $(\frac a2\ - \frac {a'}2)\cdot(\frac b2\ - \frac {b'}2) = \frac14 (a\cdot b-a\cdot b'+a'\cdot b+a'\cdot b')$
AOB and BOA are at right angles to one another $\implies a\cdot b = 0$ similarly for $a'\cdot b'$
$(a\cdot b-a\cdot b'+a'\cdot b+a'\cdot b') = (a\cdot b'+a'\cdot b) = ||a||b'||\cos\theta + ||a'||b||\cos\phi$
Where $\theta = m\angle AOB'$ and $\phi = m\angle A'OB$
$OA = OB \implies ||a|| = ||b||$ $OA' = OB' \implies ||a'|| = ||b'||$
$||a||a'||cos\theta + ||a'||a||cos\phi$
Since $\angle AOB$ and $\angle A'OB'$ are right angles, $\angle A'OB$ and$\angle AOB'$ are supplementary. $\cos\theta = - \cos\phi$
$||a||a'||cos\theta + ||a'||a||cos\phi = 0$
Lastly we need to show that:
$||\frac {a'}2\ - \frac a2|| = ||\frac {b'}2\ - \frac b2||$
$||\frac {a'}2\ - \frac a2||^2 = (\frac {a'}2\ - \frac a2)\cdot((\frac {a'}2\ - \frac a2) = \frac14(||a||^2 + ||a'||^2 - 2||a||||a'||\cos (90+\theta))\\ ||\frac {b'}2\ - \frac b2|| = \frac14(||b||^2 + ||b'||^2 - 2||b||||b'||\cos (90+\theta))$
And $||a|| = ||b||$ and $||a'|| = ||b'||$ as previously discussed.