Show some proofs in $ABCD$, a regular tetrahedron

Question

My question:

Let $ABCD$ be a regular tetrahedron and let $O$ be the centre of the base $BCD$.

We consider the points $M \in (AO)$, $N \in (BC)$ and $P \in (BD)$ such that $(MNP) \parallel (ACD)$ and $\frac{OM}{OA} = k$

a) Determine, depending on $k$, the ratio between the area of the triangle $MNP$ and the area of the triangle $ACD$.

b) Determine $k \in \mathbb{R}$ so that the triangle $MNP$ is right-angled.

c) Is there $k \in \mathbb{R}$ such that the triangle $MNP$ is equilateral?

The drawing:

My idea

$(MNP) \parallel (ACD)=> MN,NP,MP \parallel (ACD)$

Using the theorem: "If a line $d$ is parallel to a plane $\alpha$ then any other plane $\beta$ that contains this line intersects the initial plane $\alpha$ after a line $g$ that is parallel to the initial line $d$ or with the plane $\alpha$"

We get that $NP \parallel CD$, $MP \parallel AD$ , $MN \parallel AC$.

Using Thales theorem, we get $\frac{CN}{NB}=\frac{PD}{BP}$,also we get that $NPB$ is equilateral

I don't know what to do forward. Hope one of you can help me! Thank you!

Answer 1

Let $E$ be the midpoint of $CD$ and $F$ the midpoint of $NP$. From the similarity of triangles $AOE$ and $MOF$ we get: $$ {OF\over OE}={OM\over OA}={MF\over AE}=k. $$ Moreover, from the similarity of $BNP$ and $BCD$ we have: $$ {PN\over CD}={BF\over BE}={2OE+OF\over 3OE}={2+k\over3}. $$ Hence: $$ {area_{MNP}\over area_{ACD}}={PN\cdot MF\over CD\cdot AE}= {k(2+k)\over3}. $$ Note then that $\angle NMP=90°$ only if $PN=2MF$, that is: $$ {2+k\over3}CD=2kAE=k{\sqrt3}\,CD. $$ This gives the equation $2+k=3k\sqrt3$, with solution: $$ k={3\sqrt3+1\over13}. $$ Finally, from $MF={\sqrt3\over2}PN$ we can find the obvious result that $\angle NMP=60°$ if $k=1$.

Answer 2

Taking the three posted questions in reverse order:

c) If plane $PMN$ remains parallel to plane $DAC$ as $M$ moves from $A$ down to $O$, then triangle $PMN$ changes its shape: from equilateral when $M$ coincides with $A$ to degenerate triangle $PMN$ (now a straight line $=\frac{2}{3}\cdot CD$) when $M$ coincides with $O$.

Thus the vertex angle of isosceles $\triangle PMN$ increases from $60^o$ to $180^o$, while the sum of the base angles decreases from $120^o$ to $0^o$. Triangle $PMN$ is equilateral, then, only when $M$ coincides with $A$, i.e. when$$\frac{OM}{OA}=k=\frac{1}{1}$$

b) Approximate solution: Make $PN\parallel DC$ and such that $PN=OA$, and let the side of the tetrahedron $s=1$. Hence the altitude of the tetrahedron $OA=\frac{\sqrt 2}{\sqrt 3}$, and $OE=\frac {1}{3}\cdot BE=\frac{\sqrt 3}{6}$.

Draw $QM\parallel EA$, join $MP$, $MN$, and draw $PR\parallel BE$.

Since$$\frac{BE}{DE}=\frac{PR}{DR}=\sqrt 3$$and$$DR=\frac{DC-OA}{2}=\frac{1-\frac{\sqrt 2}{\sqrt 3}}{2}=\frac{\sqrt 3-\sqrt 2}{2\sqrt3}$$and$$PR=DR\cdot \sqrt 3=\frac{\sqrt 3-\sqrt2}{2}$$then$$OQ=OE-PR=\frac{\sqrt 3}{6}-\frac{\sqrt 3-\sqrt2}{2}=\frac{3\sqrt 2-2\sqrt 3}{6}$$And since$$AE=BE=3\cdot OE$$then by similar triangles$$MQ=3\cdot OQ=\frac{3\sqrt 2-2\sqrt 3}{2}\approx .4$$But$$PQ=\frac{\sqrt 2}{2\sqrt 3}\approx .4$$Therefore, isosceles triangle $PMN$, in a plane parallel to $\triangle DAC$, has approximately a right angle at $M$ when base $PN=OA$.

And since $OM=\sqrt{MQ^2-OQ^2}=\sqrt{(3OQ)^2-OQ^2}$, then in that case$$k=\frac{OM}{OA}=\frac{\sqrt{\left(\frac{3\sqrt 2-2\sqrt 3}{2}\right)^2-\left(\frac{3\sqrt 2-2\sqrt 3}{6}\right)^2}}{\frac{\sqrt 2}{\sqrt 3}}=…=\sqrt 6-2$$

Note: But again, although $k=\sqrt 6-2$ is exact, $MQ=PQ=.4$ only approximately for $NP=OA=\frac{\sqrt 2}{\sqrt 3}$, making $\angle PMN\approx90^o$ when $NP=OA$.

a) For all triangles $PMN$ parallel to the plane of $\triangle DAC$, since$$k=\frac{OM}{OA}=\frac{QM}{EA}$$and the areas of triangles are proportional to the ratio of their heights multiplied by the ratio of their bases, then in area$$\frac{\triangle NMP}{\triangle CAD}=k\cdot \frac{NP}{CD}$$