A pyramid $MABCD$ with $ABCD$ being a rhombus, and $MM_1$ being its altitude is such that $M_1$ lies on $AC$, and $AM_1 : M_1C = 1 : 3$. Knowing that $AB=a$, $\angle BAD=\alpha < 90^\circ$, and $MM_1=a$, find the angle between planes $(ABM)$ and $(CDM)$.
My attempt of the solution is the following:
As $M$ is a common point of two planes, we can drop perpendiculars from it: $MK\perp AB$ and $ML\perp CD$, and then find the angle between them. This is how we do it:
- We find the diagonals of the rhombus:
$AC=2a\cos\dfrac{\alpha}{2}$
$BD=2a\sin\dfrac{\alpha}{2}$
- As $AM_1 : M_1C = 1:3$, then consider $AM_1=x, M_1C=3x$. Then $AC=4x$. Thus: $$\begin{align*} AC &= AM_1+M_1C\\ 2a\cos\dfrac{\alpha}{2} &= x+3x \\ 2a\cos\dfrac{\alpha}{2} &= 4x \\ x &= \dfrac{a}{2}\cos\dfrac{\alpha}{2} \end{align*}$$
Therefore, $AM_1=\dfrac{a}{2}\cos\dfrac{\alpha}{2}$ and $M_1C=\dfrac{3a}{2}\cos\dfrac{\alpha}{2}$.
- We can now find hypotenuse $AM$ by applying Pythagorean Theorem:
$$\begin{align*} AM &= \sqrt{MM_1^2 + AM_1^2} \\ &= \sqrt{a^2+\left(\dfrac{a}{2}\cos\dfrac{\alpha}{2}\right)^2} \\ &= \sqrt{a^2+\dfrac{a^2}{4}\cos^2\dfrac{\alpha}{2}} \\ &= \sqrt{a^2\left(1+\dfrac{1}{4}\cos^2\dfrac{\alpha}{2}\right)} \\ &= a\sqrt{1+\dfrac{1}{4}\cdot\dfrac{1+\cos\alpha}{2}} \\ &= a\sqrt{\dfrac{8}{8}+\dfrac{1+\cos\alpha}{8}} \\ &= \dfrac{a\sqrt{9+\cos\alpha}}{2\sqrt{2}} \end{align*}$$
Notice that because, out of properties of rhombus, the intersection point $O$ of diagonals $AC$ and $BD$ bisects them, $AO=OC$, and $BO=OD$. This implies that, becuase $AM_1=\dfrac{1}{4}AC$, $M_1O = AO-AM_1 = AM_1 = \dfrac{a}{2}\cos\dfrac{\alpha}{2}$.
Consider the triangle $OBM_1$. It is a right triangle, because, from properties of rhombus, the diagonals of a rhombus are perpendicular. We know $BO=\dfrac{BD}{2}=a\sin\dfrac{\alpha}{2}$. We also know $OM_1=\dfrac{a}{2}\cos\dfrac{\alpha}{2}$. Then we can find hypotenuse $BM_1$ by applying Pythagorean Theorem: $$\begin{align*} BM_1 &= \sqrt{BO^2+M_1O^2} \\ &= \sqrt{\left(a\sin\dfrac{\alpha}{2}\right)^2+\left(\dfrac{a}{2}\cos\dfrac{\alpha}{2}\right)^2} \\ &= \sqrt{a^2\sin^2\dfrac{\alpha}{2}+\dfrac{a^2}{4}\cos^2\dfrac{\alpha}{2}} \\ &= \sqrt{a^2\left(\sin^2\dfrac{\alpha}{2}+\dfrac{1}{4}\cos^2\dfrac{\alpha}{2}\right)} \\ &= a\sqrt{\sin^2\dfrac{\alpha}{2}+\dfrac{1}{4}\cos^2\dfrac{\alpha}{2}} \end{align*}$$ We now apply the half angle trigonometric identities in order to simplify the expression inside the root: $$\begin{align*} \sin^2\dfrac{\alpha}{2}+\dfrac{1}{4}\cos^2\dfrac{\alpha}{2} &= \dfrac{1-\cos\alpha}{2}+\dfrac{1}{4}\cdot\dfrac{1+\cos\alpha}{2} \\ &= \dfrac{1-\cos\alpha}{2}+\dfrac{1+\cos\alpha}{8} \\ &= \dfrac{4(1-\cos\alpha)}{4\cdot 2}+\dfrac{1+\cos\alpha}{8} \\ &= \dfrac{4-4\cos\alpha}{8}+\dfrac{1+\cos\alpha}{8} \\ &= \dfrac{4-4\cos\alpha+1+\cos\alpha}{8} \\ &= \dfrac{5-3\cos\alpha}{8} \end{align*}$$ And this gives us the final form of an expression for the hypotenuse: $$BM_1 = a\sqrt{\dfrac{5-3\cos\alpha}{8}} = \dfrac{a\sqrt{5-3\cos\alpha}}{2\sqrt{2}}$$
Now consider another right triangle $BMM_1$ (it is a right triangle because $MM_1\perp (ABC)$, which implies it's perpendicular to each line in $(ABC)$). We know both of its legs, so we now apply Pythagorean Theorem to find the hypotenuse $BM$: $$\begin{align*} BM &= \sqrt{MM_1^2 + BM_1^2} \\ &= \sqrt{a^2+\left(\dfrac{a\sqrt{5-3\cos\alpha}}{2\sqrt{2}}\right)^2} \\ &= \sqrt{a^2+\dfrac{a^2(5-3\cos\alpha)}{8}} \\ &= \sqrt{a^2\left(1+\dfrac{5-3\cos\alpha}{8}\right)} \\ &= a\sqrt{\dfrac{13-3\cos\alpha}{8}} \\ &= \dfrac{a\sqrt{13-3\cos\alpha}}{2\sqrt{2}} \end{align*}$$
Consider the triangle $AMB$. Let's find its semiperimeter. $$\begin{align*} p &= \dfrac{1}{2}\cdot\left(a+\dfrac{a\sqrt{9+\cos\alpha}}{2\sqrt{2}}+\dfrac{a\sqrt{13-3\cos\alpha}}{2\sqrt{2}}\right) \\ &= \dfrac{a}{2}\cdot\left(1+\dfrac{\sqrt{9+\cos\alpha}}{2\sqrt{2}}+\dfrac{\sqrt{13-3\cos\alpha}}{2\sqrt{2}}\right) \\ &= \dfrac{a(2\sqrt{2}+\sqrt{9+\cos\alpha}+\sqrt{13-3\cos\alpha})}{4\sqrt{2}} \end{align*}$$
Now we apply Heron's formula to find its area: $$S = \sqrt{p(p-AB)(p-AM)(p-BM)}$$
To not overflow the whole place with the huge stack of TeX, we'll calculate and simplify each difference separately:
$$\begin{align*} p-AB &= \dfrac{a(2\sqrt{2}+\sqrt{9+\cos\alpha}+\sqrt{13-3\cos\alpha})}{4\sqrt{2}}-a \\ &= \dfrac{a}{4\sqrt{2}}\left(2\sqrt{2}+\sqrt{9+\cos\alpha}+\sqrt{13-3\cos\alpha}-4\sqrt{2}\right) \\ &= \dfrac{a}{4\sqrt{2}}\left(\sqrt{9+\cos\alpha}+\sqrt{13-3\cos\alpha}-2\sqrt{2}\right) \end{align*}$$
$$\begin{align*} p-AM &= \dfrac{a(2\sqrt{2}+\sqrt{9+\cos\alpha}+\sqrt{13-3\cos\alpha})}{4\sqrt{2}}-\dfrac{a\sqrt{9+\cos\alpha}}{2\sqrt{2}} \\ &= \dfrac{a}{4\sqrt{2}}\left(2\sqrt{2}+\sqrt{9+\cos\alpha}+\sqrt{13-3\cos\alpha}-2\sqrt{9+\cos\alpha}\right) \\ &= \dfrac{a}{4\sqrt{2}}\left(2\sqrt{2}+\sqrt{13-3\cos\alpha}-\sqrt{9+\cos\alpha}\right) \end{align*}$$
$$\begin{align*} p-BM &= \dfrac{a(2\sqrt{2}+\sqrt{9+\cos\alpha}+\sqrt{13-3\cos\alpha})}{4\sqrt{2}}-\dfrac{a\sqrt{13-3\cos\alpha}}{2\sqrt{2}} \\ &= \dfrac{a}{4\sqrt{2}}\left(2\sqrt{2}+\sqrt{9+\cos\alpha}+\sqrt{13-3\cos\alpha}-2\sqrt{13-3\cos\alpha}\right) \\ &= \dfrac{a}{4\sqrt{2}}\left(2\sqrt{2}+\sqrt{9+\cos\alpha}-\sqrt{13-3\cos\alpha}\right) \end{align*}$$
Now we find the simplified form of the expression under the square root:
$$\begin{align*} p(p-AB)(p-AM)(p-BM) &= \left(\dfrac{a}{4\sqrt{2}}\right)^4\left(272-16\cos^2\alpha\right) \\ &= \dfrac{a^4(272-16\cos^2\alpha)}{1024} \end{align*}$$
And now we find the area:
$$S = \sqrt{\dfrac{a^4(272-16\cos^2\alpha)}{1024}} = \dfrac{4a^2\sqrt{17-\cos^2\alpha}}{32} = \dfrac{a^2\sqrt{17-\cos^2\alpha}}{8}$$
A pretty nice number to see. I expected worse.
- Now we find the altitude $MK$ by using the altitude formula: $$MK = h_{AB} = \dfrac{2S}{AB} = \dfrac{2a^2\sqrt{17-\cos^2\alpha}}{8a} = \dfrac{a\sqrt{17-\cos^2\alpha}}{4}$$
One down!
As $MM_1\perp AC$ and $AC\perp BD$, then by theorem of three perpendiculars $MO\perp BD$. As $BO=BD$, the triangle $MBD$ has segment $MO$ as a median. But $MO\perp BD$, thus $MO$ is also the altitude. Therefore, triangle $MBD$ is isosceles with $DM=BM=\dfrac{a\sqrt{13-3\cos\alpha}}{2\sqrt{2}}$.
Now consider another right triangle $MM_1C$. We know both of its legs, so we now apply Pythagorean Theorem to find the hypotenuse $CM$:
$$\begin{align*} CM &= \sqrt{MM_1^2+M_1C^2} \\ &= \sqrt{a^2+\left(\dfrac{3a}{2}\cos\dfrac{\alpha}{2}\right)^2} \\ &= \sqrt{a^2+\dfrac{9a^2}{4}\cos^2\dfrac{\alpha}{2}} \\ &= \sqrt{a^2\left(1+\dfrac{9}{4}\cdot\dfrac{1+\cos\alpha}{2}\right)} \\ &= a\sqrt{\dfrac{8}{8}+\dfrac{9+9\cos\alpha}{8}} \\ &= \dfrac{a\sqrt{17+9\cos\alpha}}{2\sqrt{2}} \end{align*}$$
Consider the triangle $DMC$. We now find its semiperimeter: $$\begin{align*} p &= a+\dfrac{a\sqrt{17+9\cos\alpha}}{2\sqrt{2}}+\dfrac{a\sqrt{13-3\cos\alpha}}{2\sqrt{2}} \\ &= \dfrac{a}{2\sqrt{2}}\left(2\sqrt{2}+\sqrt{17+9\cos\alpha}+\sqrt{13-3\cos\alpha}\right) \\ &= \dfrac{a\left(2\sqrt{2}+\sqrt{17+9\cos\alpha}+\sqrt{13-3\cos\alpha}\right)}{2\sqrt{2}} \end{align*}$$
Now we apply Heron's formula to find its area: $$S = \sqrt{p(p-CD)(p-CM)(p-DM)}$$
Again, to not overflow the whole place with the huge stack of TeX, we'll calculate and simplify each difference separately:
$$\begin{align*} p-CD &= \dfrac{a\left(2\sqrt{2}+\sqrt{17+9\cos\alpha}+\sqrt{13-3\cos\alpha}\right)}{2\sqrt{2}}-a \\ &= \dfrac{a}{2\sqrt{2}}\left(2\sqrt{2}+\sqrt{17+9\cos\alpha}+\sqrt{13-3\cos\alpha}-2\sqrt{2}\right) \\ &= \dfrac{a}{2\sqrt{2}}\left(\sqrt{17+9\cos\alpha}+\sqrt{13-3\cos\alpha}\right) \end{align*}$$
$$\begin{align*} p-CM &= \dfrac{a\left(2\sqrt{2}+\sqrt{17+9\cos\alpha}+\sqrt{13-3\cos\alpha}\right)}{2\sqrt{2}}-\dfrac{a\sqrt{17+9\cos\alpha}}{2\sqrt{2}} \\ &= \dfrac{a}{2\sqrt{2}}\left(2\sqrt{2}+\sqrt{17+9\cos\alpha}+\sqrt{13-3\cos\alpha}-\sqrt{17+9\cos\alpha}\right) \\ &= \dfrac{a}{2\sqrt{2}}\left(2\sqrt{2}+\sqrt{13-3\cos\alpha}\right) \end{align*}$$
$$\begin{align*} p-DM &= \dfrac{a\left(2\sqrt{2}+\sqrt{17+9\cos\alpha}+\sqrt{13-3\cos\alpha}\right)}{2\sqrt{2}}-\dfrac{a\sqrt{13-3\cos\alpha}}{2\sqrt{2}} \\ &= \dfrac{a}{2\sqrt{2}}\left(2\sqrt{2}+\sqrt{17+9\cos\alpha}+\sqrt{13-3\cos\alpha}-\sqrt{13-3\cos\alpha}\right) \\ &= \dfrac{a}{2\sqrt{2}}\left(2\sqrt{2}+\sqrt{17+9\cos\alpha}\right) \end{align*}$$
...and here I am, stuck with this giant product under the square root without any real idea on how to simplify it nicely and/or continue. My idea was to find all sides of the triangle $KML$, and then use Cosine Theorem to find the angle $KML$, which would solve the problem.
...so, any ideas on how to continue this?