Serbian Mathematical Olympiad

Solution:

Let $KLMN$ be the rhombus with $K \in AB$, $L \in BD$ and $N \in EA$. Consider the trapezoid $PQRS$ with $P, Q \in AB$, $PQ \parallel RS$ and $\angle PQR=\angle QPS=\angle KNM=\varphi$ such that $K, L, M, N$ are on $PQ, QR, RS, SP$ respectively. Suppose that $\varphi>\alpha, \beta$. Then $R$ and $S$ lie outside $\triangle ABC$.

Since $\angle SNM=180^\circ-\varphi-\angle NMS=\angle RML$ and $\angle LRM=\angle MSN$, triangles $LRM$ and $MSN$ are congruent, so $LR=MS$. Similarly, if $Q'$ is the point on $AB$ such that $\angle LQ'B=\varphi$, triangles $LRM$ and $KQ'L$ are congruent, so $MR=LQ'=LQ$. It follows that $RQ=RL+LQ=SM+MR=RS$.

Now $d(R, AB)=RQ \sin \varphi>RS \sin \alpha>d(R, AC)$ and analogously $d(S, AB)>d(S, BC)$, which means that points $R$ and $S$ lie above the lines $AD$ and $BE$ respectively (i.e., in the half-planes containing $C$), from which it follows that both points lie above the line $DE$. This is impossible since segments $RS$ and $DE$ intersect at $M$.

Second solution. The distance from point $X$ to line $p$ is denoted by $d(X, p)$.

Lemma. For an arbitrary point $M$ on segment $DE$ we have $d(M, AB)=d(M, AC)+d(M, BC)$.

Proof. If $\frac{DM}{DE}=k$, then $d(M, AB)=k d(E, AB)+(1-k) d(D, AB)=k d(E, BC)+(1-k) d(D, AC)=d(M, BC)+d(M, AC)$.

We use the same notation as in the first solution. Let $a$ be the side of the rhombus and $O$ its center. We have

$$\begin{aligned} d(M, AC)+d(M, BC) & =a(\sin \angle MNC+\sin \angle MLC), \\ \text{и}\quad d(M, AB) & =2 d(O, AB)=d(L, AB)+d(N, AB) \\ & =a(\sin \angle NKA+\sin \angle LKB) . \end{aligned}$$

However, if $\varphi>\alpha, \beta$, then $\angle NKA=\angle MNC+\varphi-\alpha>\angle MNC$ and analogously $\angle LKB>\angle MLC$, so from the equations above it follows that $d(M, AB)>d(M, AC)+d(M, BC)$, a contradiction.