the question
In the triangle $ABC$ we consider $(AM$ the bisector of the angle $\angle A$ so that $MB=3MC, M\in (BC)$ and $N\in (AB)$ so that $BN=2NA$. On the plane of the triangle $ABC$, the perpendicular $DA$ rises.
a) Prove that the distance $d$ from point $A$ to plane $(DNC)$ verifies the relation $d< \frac{AB+3AD}{6\sqrt{2}}$
b) If $\angle A= 60, AC=a, AD=\frac{3a\sqrt{7}}{14}$. Determine the angle formed by the planes $\angle((ABC),(DMN))$
The drawing
The idea
As you can see I noted $BC=4x => BM=3x, MC=x$ and $BA=3y=> BN=2y, NA=y$
Using the thorem of the bisector: $\frac{AB}{AC}=\frac{BM}{MC}=3=> AB=3AC=> AC=y$
$AN=AC=y=>$ triangle $ANC$ is isoscells and $AM$ is the bisector of angle $A => AM \perp NC$
Let the intersection of $AM,NC$ be point $O$
$AO \perp NC$ and $AD\perp (ABC)$ by the theorem of the $3$ perpendiculars => $DO\perp NC$
Writing the volume of the tetrahedron $DANC$ in $2$ ways we get that the $d=\frac{AO*AD}{DO}$
We can also show that $DN=DC$
I don't know what to do forward…maybe the inequality of means would be helpful, but don't know. I hope one of you can help me! Thank you!
Best Answer
Let us call $z$ the length of $AD$ and $h$ the length of $AO$. For part (a), we know that the distance from $A$ to the plane is the altitude over the hypothenuse in $\triangle DAO$, right angled in $A$. That is,
\begin{align} d=\frac{zh}{\sqrt{z^2+h^2}}&=\left(z^{-2}+h^{-2}\right)^{-1/2}\\ &= \frac{1}{\sqrt{2}}\left(\frac{z^{-2}+h^{-2}}{2}\right)^{-1/2}\\ &= \frac{1}{\sqrt{2}} M_{-2}(z,h)~~, \end{align} where $M_{-2}$ is the power mean of $z$ and $h$ of order -2. A fairly general result is that the power means are monotonous in the power parameter, thus: $$M_{-2}(z,h)\le M_{1}(z,h) =\frac{z+h}{2}$$ Because $h$ is the altitude of isosceles $\triangle NAC$, we have that $h<y$ Therefore, we conclude that $$ d< \frac{1}{\sqrt{2}}\frac{y+z}{2}~~.$$ The result follows from noting that $AB=3y$.
For part (b), I'll keep using $y$ instead of $a$ for $AC$. I guess there are several ways, I found one not particularly inspired. You can start with the identity for the length of the bisector $$AM^2=AB.AC-BM.MC=3(y^2-x^2)~~.$$ Then, from cosine theorem in $\triangle ABC$ we get $BC^2=16x^2=AB^2+AC^2-2.AB.AC.cos(60^\circ)=9y^2+y^2-6y^2\times \frac{1}{2}=7y^2$. We conclude combining both relations that $$AM=\frac{3\sqrt{3}}{4}y~~.$$ Again using cosine theorem on $\triangle AMN$, we get that $$MN=\frac{5}{4}y~~.$$ The area of $\triangle AMN$ is $1/2.NO.AM$, with $NO=y/2$. Therefore, since $MN$ is known, we can obtain the altitude in $\triangle AMN$ from $A$ over $MN$, say $AL$, with $L$ being the foot of the perpendicular from $A$ over $MN$. This turns out to be $AL=\frac{3\sqrt{3}y}{2\sqrt{7}}$. We are almost done since the angle between the planes is equal to $\angle DLA$, whose tangent is $\frac{3\sqrt{7}y}{14}\times\frac{2\sqrt{7}}{3\sqrt{3}y}=\frac{1}{\sqrt{3}}$ which means that the angle is 30º.
A more direct derivation that $M_{-2}(h,z)\le M_{1}(h,z)$ is as follows: \begin{align} (h + z)^2 (h^2 + z^2) &= h^4 + z^4 + 2 h^2 z^2 + 2 hz (h^2+ z^2)\\ &\ge 2 h^2z^2 + 2 h^2 z^2 + 4 h^2z^2=8h^2z^2 \quad; \text{using}~a^2 + b^2\ge 2ab ~~.\\ \end{align} Thus, \begin{align} \frac{(h + z)^2 (h^2 + z^2)}{8h^2z^2}&\ge1\\ \left(\frac{z^{-2} + h^{-2}}{2}\right)\frac{(z + h)^2}{4}&\ge1\\ \rightarrow \frac{(z + h)^2}{4} &\ge \left(\frac{z^{-2} + h^{-2}}{2}\right)^{-1}\quad;\sqrt{\cdot}\\ \frac{(z + h)}{2} &\ge \left(\frac{z^{-2} + h^{-2}}{2}\right)^{-1/2}\\ \end{align}