International Mathematical Olympiad Shortlisted Problems

Let $K,L,M$, and $N$ be the vertices of the rhombus lying on the sides $AE,ED,DB$, and $BA$, respectively. Denote by $d(X,YZ)$ the distance from a point $X$ to a line $YZ$. Since $D$ and $E$ are the feet of the bisectors, we have $d(D,AB)=d(D,AC),d(E,AB)=d(E,BC)$, and $d(D,BC)=d(E,AC)=0$, which implies $$d(D,AC)+d(D,BC)=d(D,AB)\text{ and }d(E,AC)+d(E,BC)=d(E,AB).$$ Since $L$ lies on the segment $DE$ and the relation $d(X,AC)+d(X,BC)=d(X,AB)$ is linear in $X$ inside the triangle, these two relations imply $$\begin{array}{c}d(L,AC)+d(L,BC)=d(L,AB)\end{array}\text{\hspace{1em}(1)}$$ Denote the angles as in the figure below, and denote $a=KL$. Then we have $d(L,AC)=a\sin \mu$ and $d(L,BC)=a\sin \nu$. Since $KLMN$ is a parallelogram lying on one side of $AB$, we get $$d(L,AB)=d(L,AB)+d(N,AB)=d(K,AB)+d(M,AB)=a(\sin \delta +\sin \epsilon )$$ Thus the condition (1) reads $$\begin{array}{c}\sin \mu +\sin \nu =\sin \delta +\sin \epsilon .\end{array}\text{\hspace{1em}(2)}$$ If one of the angles $\alpha$ and $\beta$ is non-acute, then the desired inequality is trivial. So we assume that $\alpha ,\beta <\pi /2$. It suffices to show then that $\psi =\angle NKL⩽\max \{\alpha ,\beta \}$. Assume, to the contrary, that $\psi >\max \{\alpha ,\beta \}$. Since $\mu +\psi =\angle CKN=\alpha +\delta$, by our assumption we obtain $\mu =(\alpha -\psi )+\delta <\delta$. Similarly, $\nu <\epsilon$. Next, since $KN\Vert ML$, we have $\beta =\delta +\nu$, so $\delta <\beta <\pi /2$. Similarly, $\epsilon <\pi /2$. Finally, by $\mu <\delta <\pi /2$ and $\nu <\epsilon <\pi /2$, we obtain $$\sin \mu <\sin \delta \text{ and }\sin \nu <\sin \epsilon$$ This contradicts (2).