Baltic Way 2019

We denote the angle between the planes $AND$ and $ACD$ by $\alpha$, and the angle between the planes $AMD$ and $AND$ by $\beta$. Furthermore, when $P,Q,R$ and $S$ are four points in space which do not all lie on a plane, we denote the volume of the tetrahedron $PQRS$ by $|PQRS|$. The triangles $△MBD$, $△MCD$, $△NBD$ and $△NCD$ have a common height, which is the distance of the point $D$ from the line $BC$. Therefore $$\frac{MB}{MC}+\frac{NB}{NC}=\frac{|△MBD|}{|△MCD|}+\frac{|△NBD|}{|△NCD|}.$$ In the same vein, since the points $M,B,C$ and $D$ lie on a common plane, which does not contain the point $A$, we must have $$\frac{|△MBD|}{|△MCD|}=\frac{|MBDA|}{|MCDA|}.$$ Similarly, $$\frac{|△NBD|}{|△NCD|}=\frac{|NBDA|}{|NCDA|}.$$ Let us consider the volume of the tetrahedron $MBDA$. Naturally $$|MBDA|=\frac{1}{3}\cdot |△ABD|\cdot h,$$ where $h$ is the distance of $M$ from the plane $ABD$. Let $P$ be the projection of $M$ to the line $AD$ and let $Q$ be the projection of $M$ to the plane $ABD$. Then $$h=MQ,\angle QPM=\alpha ,\text{and}|△AMD|=\frac{1}{2}\cdot AD\cdot PM.$$ Because in addition $h=PM\cdot \sin \alpha$, we reach the conclusion $$|MBDA|=\frac{2\cdot |△ABD|\cdot |△AMD|\sin \alpha }{3\cdot AD}.$$

$\frac{MB}{MC}+\frac{NB}{NC}=\frac{|MBDA|}{|MCDA|}+\frac{|NBDA|}{|NCDA|}\ge 2\sqrt{\frac{|MBDA|\cdot |NBDA|}{|MCDA|\cdot |NCDA|}}=2\cdot \sqrt{\frac{\frac{2\cdot |△ABD|\cdot |△AMD|\sin \alpha }{3\cdot AD}\cdot \frac{2\cdot |△ABD|\cdot |△AND|\sin (\alpha +\beta )}{3\cdot AD}}{\frac{2\cdot |△ACD|\cdot |△AMD|\sin (\alpha +\beta )}{3\cdot AD}\cdot \frac{2\cdot |△ACD|\cdot |△AND|\sin \alpha }{3\cdot AD}}}=2\cdot \frac{|△ABD|}{|△ACD|}$

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Alternative solution.

(prepared by the Organizing Committee) Let $B^{\prime }$ be a point so that $B{B}^{\prime }\parallel AD$ and $B^{\prime }C\perp AD$. Let $M^{\prime },N^{\prime }$ be points on $B^{\prime }C$ so that $M{M}^{\prime }\parallel N{N}^{\prime }\parallel B{B}^{\prime }$. Let $E$ be a point on $AD$ such that the plane $B^{\prime }CE$ is perpendicular to $AD$. Then $\frac{MB}{MC}=\frac{M^{\prime }{B}^{\prime }}{M^{\prime }C}$, $\frac{NB}{NC}=\frac{N^{\prime }{B}^{\prime }}{N^{\prime }C}$, and $\frac{|\Delta ABD|}{|\Delta ACD|}=\frac{|\Delta A{B}^{\prime }D|}{|\Delta ACD|}=\frac{B^{\prime }E}{CE}$. Since the angle between planes $AND$, $ACD$ is equal to the angle between planes $AMD$, $ABD$, it follows that lines $E{M}^{\prime }$, $E{N}^{\prime }$ are isogonal in the angle $B^{\prime }EC$. It is well-known that $\frac{M^{\prime }{B}^{\prime }}{M^{\prime }C}\cdot \frac{N^{\prime }{B}^{\prime }}{N^{\prime }C}=\frac{B^{\prime }{E}^2}{C{E}^2}$. Hence by AM-GM we obtain $$\frac{MB}{MC}+\frac{NB}{NC}=\frac{M^{\prime }{B}^{\prime }}{M^{\prime }C}+\frac{N^{\prime }{B}^{\prime }}{N^{\prime }C}\ge 2\sqrt{\frac{M^{\prime }{B}^{\prime }}{M^{\prime }C}\cdot \frac{N^{\prime }{B}^{\prime }}{N^{\prime }C}}=2\cdot \frac{B^{\prime }E}{CE}=2\cdot \frac{|\Delta ABD|}{|\Delta ACD|}$$