pohoatza wrote: Note. If the pentagon is labeled $ ABCDE$, the adjacent vertices of $ A$ are $ B$ and $ E$, the ones of $ B$ are $ A$ and $ C$ etc. Got it It seemed that a combinatorial solution for this question was out of reach, altough some students mentioned (successful) approaches by repeated projections of sides on diagonals & careful area estimations - maybe w/ convex combinations, and even with some term smashing (being required a strict inequality...). Here follows my (polished) solution from the contest: Denote by $ [\mathcal{P}]$ the area of the polygon (polygonal surface) $ \mathcal{P}$, $ S : = [ABCDE]$, and by lowercase letters the area of the convex figure spanned by the sides at the corresponding vertex (thus, $ a = [EAB]$, and so on). From now on i will denote cyclic sums by $ \sum$. Assume that $ a = \min\{a, b, c, d, e\}$. We will prove that the vertex $ A$ itself satisfies the hyphotesis, id est, $ d(A, CD) < d(B, CD) + d(E, CD)$. Since it is difficult to handle distances, multiply by $ CD / 2$ to get an inequality between areas: $ [ACD] < [BCD] + [ECD]$, or equivalently, $ S < b + c + d + e$. Assume, by contradiction, that the inequality fails. Clearly, $ S > a$, hence $ (S - a)\left(S - (b + c + d + e) \right) \ge 0$, or, by expanding everything, $ S^2 - S \sum a + a(b + c + d + e) \ge 0$. However, by the minimality of $ a$, $ a(b + c + d + e) \le ab + ea + bc + ed < \sum ab$. This yields $ 0 \le S^2 - S \sum a + a(b + c + d + e) < S^2 - S \sum a + \sum ab = 0$, a contradiction in the view of the lemma in the sequel (which, in fact, renders these computations straightforward). $ \blacksquare$ Lemma. With the notations above, the following (not so, it seems) well-known identity holds $ \boxed{S^2 - S \sum a + \sum ab = 0}$. What follows is a condensed exposition of the solution in Prasolov's book. However, there are other approaches available, such as... analytic geometry (the way I proceeded in the contest ). In what follows I will use $ \mathbf{boldface}$ letters to denote vectors. (Those familiarized with the "Vectors" chapter in the aforementioned book may skip to the next paragraph without missing anything). Introduction. Denote by $ \prec \cdot, \cdot \succ$ the pseudoinner product of two nonzero vectors, namely $ \prec \mathrm{u}, \mathrm{v} \succ : = |\mathrm{u}| \cdot |\mathrm{v}| \cdot \sin \left( \angle ( \mathrm{u}, \mathrm{v} ) \right)$, (the signed area of the parralelogram spanned by $ \mathrm{u}, \mathrm{v}$), and extend it such that it maps a pair to zero when one of the vectors is the zero vector. If we denote by $ < \mathrm{u}, \mathrm{v} >$ the inner product of $ \mathrm{u}, \mathrm{v}$, we get the relation $ < \mathrm{u}, \mathrm{v} > ^2 + \prec \mathrm{u}, \mathrm{v} \succ^2 = (|\mathrm{u}| \cdot |\mathrm{v}|)^2$, whence, if $ \mathrm{u} = (a_1, a_2)$ and $ \mathrm{v} = (b_1, b_2)$, $ \prec \mathrm{u}, \mathrm{v} \succ = a_1b_2 - a_2b_1$. Now, it is easy to check that this application is anticommutative and "well-behaved", namely, For any scalar $ \lambda \in \mathbb{R}$, $ \prec \lambda\mathrm{u}, \mathrm{v} \succ = \lambda \prec \mathrm{u}, \mathrm{v} \succ$, and $ \prec \mathrm{u}, \mathrm{v} + \mathrm{w} \succ = \prec \mathrm{u}, \mathrm{v} \succ + \prec \mathrm{u}, \mathrm{w} \succ$. Now, assign to any triple of points $ (A, B, C)$ its oriented area, namely $ S(A, B, C) : = \frac {1}{2} \prec \overrightarrow{AB}, \overrightarrow{AC} \succ$. Clearly, $ [ABC] = |S(A, B, C)|$ and $ S(A, B, C) = S(B, C, A) = - S(B, A, C)$. Solution (for the Lemma). Assume that $ \{\mathrm{e_1}, \mathrm{e_2}\}$ generates the vectors in the plane, and pick two reals $ x_1, x_2$. Let $ \mathrm{x} : = x_1\mathrm{e_1} + x_2\mathrm{e_2}$. Taking the pseudoinner product to the right with $ \mathrm{e_1}$, respectively $ \mathrm{e_2}$, we get $ \prec \mathrm{x}, \mathrm{e_1} \succ = x_2 \prec \mathrm{e_2}, \mathrm{e_1} \succ$, and $ \prec \mathrm{x}, \mathrm{e_2} \succ = x_1 \prec \mathrm{e_1}, \mathrm{e_2} \succ$, and now, multiplying by $ \mathrm{e_2}$, respectively $ \mathrm{e_1}$, and substracting $ \prec \mathrm{e_1}, \mathrm{e_2} \succ \mathrm{x} = \prec \mathrm{x}, \mathrm{e_2} \succ e_1 + \prec \mathrm{e_1}, \mathrm{x} \succ e_2$. Finally, take our product to the right with some vector $ \mathrm{y}$, which yields $ \prec \mathrm{e_1}, \mathrm{e_2} \succ \prec \mathrm{x}, \mathrm{y} \succ + \prec \mathrm{x}, \mathrm{e_2} \succ \prec \mathrm{y}, \mathrm{e_1} \succ + \prec \mathrm{e_1}, \mathrm{x} \succ \prec \mathrm{y}, \mathrm{e_2} \succ = 0$. Now, the problem is over. Assume that $ A, B, C, D, E$ is the order of the points, counter-clockwise (i.e., $ S(A, B, C) > 0$). Take $ \mathrm{e_1} = \overrightarrow{AB}$, $ \mathrm{e_2} = \overrightarrow{AE}$, $ \mathrm{x} = \overrightarrow{AC}$, $ \mathrm{y} = \overrightarrow{AD}$. The obtained relation reduces to $ a(S - b - e) - (S - b - d)(S - c - e) + be = 0$, and this ends the proof. $ \square$