3D geometry problem concerning a quadrilateral inside a cube

Question

Please see question 2(a) in the following link:-

https://bmos.ukmt.org.uk/home/bmo1-1995.pdf (sorry I couldn't copy the question in here for some reason)

Here's my approach:-

Length of $AM=MH=HN=NA$ (ignoring the trivial proof)

Now, length of $MN=\sqrt{(2^2+2^2)}=\sqrt{8}$

Length of $AM=AN=\sqrt{(2^2+1^2)}=\sqrt5$

If angle MAN were to be a right angle, $MN=\sqrt{(AM^2+AN^2)}=\sqrt{10}$ However, as MN is NOT $\sqrt10$, by contradiction, we can conclude that $\angle MAN \neq 90^\circ$

Combining the $2$ facts that $AM=MH=HN=NA$ and $\angle MAN \neq 90^\circ$ we conclude that quadrilateral AMHN is a rhombus.

Hence, the area is $NH\cdot NX$(where $NX$ is perpendicular to $AM$) =$\sqrt{(5)}*2 = 2\sqrt5$ units.

Firstly, is my approach correct?

Secondly, I have calculated the area in this problem by making small deductions helping me to understand the shape of the quadrilateral. However, I am unable to imagine it inside the cube. How can this be done? In other words, how can I see the quadrilateral visually in the context of the cube?

Answer 1

Your approach seems correct to me, until your result. What is the $X$ point you introduce? I suspect it is the middle of the $AM$ segment, then you could just stick with $AM$ for simplicity.

The area you're looking for is $\frac{MN*AH}{2}$.

You already know that $MN = \sqrt{8}$, and if you draw the square $AFHC$ you'll see that $AH = \sqrt{12} = 2\sqrt{3}$.

Therefore the surface of the rhombus $AMHN$ is $\frac{2\sqrt{3}\sqrt{8}}{2}=\sqrt{3*8}=2\sqrt{6}$.

As for the second question, I don't understand where your problem lies. Have you tried drawing the cube and placing the points $M$ and $N$? If you do, the quadrilateral appears, and is not difficult to visualize.

Answer 2

Let $\vec{FG}=\vec{u}$, $\vec{FE}=\vec{v}$ and $\vec{FA}=\vec{w}.$

Thus, $$\vec{AM}=\vec{AB}+\vec{BM}=\vec{u}+\frac{1}{2}\vec{v}$$ and $$\vec{NH}=\vec{NE}+\vec{EH}=\frac{1}{2}\vec{v}+\vec{u},$$ which says that $AMHN$ is a parallelogram.

Now, $$AM=AN=\sqrt{2^2+1^2}=\sqrt5$$ and $$\cos\measuredangle NAM=\frac{\vec{AM}\vec{AN}}{AM\cdot AN}=\frac{\left(\vec{u}+\frac{1}{2}\vec{v}\right)\left(-\vec{w}+\frac{1}{2}\vec{v}\right)}{\sqrt5\cdot\sqrt5}=\frac{\frac{1}{4}\cdot2^2}{5}=\frac{1}{5}.$$ Id est, $$S_{AMHN}=AN\cdot AM\cdot\sin\measuredangle NAM=\sqrt5\cdot\sqrt5\cdot\sqrt{1-\frac{1}{25}}=2\sqrt6.$$ Also, since $AMHN$ is a rhombus, we obtain $$S_{AMHN}=\frac{AH\cdot MN}{2}=\frac{\sqrt{2^2+2^2+2^2}\cdot\sqrt{2^2+2^2}}{2}=2\sqrt6.$$ Let $R$ be a midpoint of $DC$ and $Z\in AR$ such that $AZ:ZR=2:1.$

Thus, $ZY||AF$, $ZY=AF$ and $ZX=\frac{1}{3}BD$.

Now, by Pythagoras for $\Delta YZX$ we obtain: $$XY=\sqrt{AF^2+\left(\frac{1}{3}BD\right)^2}=\sqrt{2^2+\left(\frac{1}{3}\cdot2\sqrt2\right)^2}=\frac{2\sqrt{11}}{3}.$$