Let $ABC$ be a triangle and $D$ the midpoint of the side $BC$. Define points $E$ and $F$ on $AC$ and $B$, respectively, such that $DE=DF$ and $\angle EDF =\angle BAC$. Prove that $$DE\geq \frac {AB+AC} 4.$$
Problem
Source: Serbia TST 2017, Day 1, Problem 1
Tags: inequalities, geometry
rmtf1111
22.05.2017 11:06
Wlog $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ , $AC$ respectively. Note that $\angle{MDF}=\angle{EDN}$. From sine law we have: $$\frac{MF}{\sin{MDF}}=\frac{FD}{\sin{FMD}}=\frac{FD}{\sin{BAC}}=\frac{FE}{\sin{BAC}}=\frac{FE}{\sin{END}}=\frac{EN}{\sin{EDN}} \implies MF \equiv EN$$Now let $X$ be a point such that $DFXE$ is a parallelogram. It is obvious that $FXAE$ is circumscribed in a circle $\omega$. Because $FX \equiv EF$ we have that $X$ is the midpoint of $\overarc{FXE}$ in $\omega \implies$ $$FD+EF = FX+EX \ge AF+AE = AM+MF+AE-EN = AM+AN= \frac {AB+AC} 4$$
square_root_of_3
10.06.2017 23:10
Trigonometry kills it. My idea was to express $DF$ as a function of angle $\angle BFD$ by using law of sines in triangles $BFD$ and $DEC$, then with a bit of algebra we obtain the value of $sin \angle BFD$ and then we plug it in our inequality and do some more algebra.