Let $D, E, F$ be points on the sides $BC, CA, AB$ of a triangle $ABC$, respectively such that the lines $AD, BE, CF$ are concurrent at the point $P$. Let a line $\ell$ through $A$ intersect the rays $[DE$ and $[DF$ at the points $Q$ and $R$, respectively. Let $M$ and $N$ be points on the rays $[DB$ and $[DC$, respectively such that the equation \[ \frac{QN^2}{DN}+\frac{RM^2}{DM}=\frac{(DQ+DR)^2-2\cdot RQ^2+2\cdot DM\cdot DN}{MN} \] holds. Show that the lines $AD$ and $BC$ are perpendicular to each other.
Problem
Source: Turkey National Olympiad 2014 P3
Tags: inequalities, geometric inequality, geometry proposed, geometry
Mathematicalx
20.11.2014 13:26
I think first step of solution is applying CS to LHS.
benpigchu
21.11.2014 10:14
Maybe some information is lost while translating.
Mathematicalx
21.11.2014 13:37
No, problem is true.
crazyfehmy
23.11.2014 01:54
The difficulty about this problem is to realize that this is actually a geometric inequality problem. Try to show that $LHS \geq RHS$ and find the equality case.
randomusername
31.03.2016 13:01
Let's attack the equation first. It involves only points $D,Q,R,M,N$, and must mean something special. Applying the Law of Cosines on $\triangle DRM$ and $\triangle DQN$, we find that \begin{align*} QN^2 &= DQ^2+DN^2-2\cdot DN\cdot DQ\cdot \cos \angle QDN, \\ RM^2 &= DR^2+DM^2-2\cdot DM\cdot DR\cdot \cos \angle RDM. \end{align*}Also, note that if $U,V$ are the projections of $Q,R$ on $MN$, respectively, then regardless of whether $\angle QDN$ or $\angle RDM$ is acute or not, \[ DQ\cos \angle QDN+DR\cos \angle RDM=|\overrightarrow{DU}-\overrightarrow{DV}|=UV. \]It follows that \begin{align*} \frac{QN^2}{DN}+\frac{RM^2}{DM}&=\frac{DQ^2}{DN}+\frac{DR^2}{DM}+DN+DM-2\cdot UV\ge \\ &\ge \frac{(DQ+DR)^2}{MN}+MN-2\cdot RQ, \end{align*}where we used Titu's lemma, $MN=DM+DN$ and the trivial estimate $UV\le RQ$. However, we also have \begin{align*} MN-2\cdot RQ &\ge \frac{-2\cdot RQ^2+2\cdot DM\cdot DN}{MN},\\ 2\cdot MN^2-4\cdot MN\cdot RQ &\ge -4\cdot RQ^2+4\cdot DM\cdot DN,\\ (DM-DN)^2+(2\cdot RQ-MN)^2 &\ge 0. \end{align*}Combining our two estimates, we find that $\text{RHS}\ge \text{LHS}$, and the equality may only hold under the following conditions: (1) $\frac{DQ}{DN}=\frac{DR}{DM}$, (2) $RQ=UV$ (i.e., $RQ$ and $UV$ are parallel), (3) $DM=DN$ and $2\cdot RQ=MN$. Now (3) implies $DM=DN=RQ$, from which using (2) we conclude that $MDQR$ and $NDRQ$ are parallelograms, and thus $\triangle MDR\cong \triangle QRD\cong \triangle DNQ$. However, (1) and (3) tell us $DQ=DR$, hence these triangles are isosceles. Consequently, $\angle QDN=\angle RDM$. Returning to the original diagram, we see a harmonic pencil $(DQ,DR,DA,BC)$. It follows that $DA$ is the internal bisector of $\angle QDR$, which is thus perpendicular to the external bisector $BC$.