An inscribed $n$-gon ($n > 3$), is divided into $n-2$ triangles by diagonals which meet only in vertices. What is the maximum possible number of congruent triangles obtained? (An inscribed $n$-gon is an $n$-gon where all its vertices lie on a circle) Proposed by Boris Frenkin - Russia
Problem
Source: IGO 2024 Elementary Level - Problem 4
Tags: geometry, combinatorial geometry
15.11.2024 15:55
The answer is \( \lfloor \frac{n}{2} \rfloor \). The construction is straightforward: simply assume the points are equally spaced and create \( \lfloor \frac{n}{2} \rfloor \) congruent isosceles triangles along the border of the polygon. The remaining part of the polygon can be tiled arbitrarily. Below, the construction is illustrated for \( n = 9 \). [asy][asy] // Asymptote code for a unit circle with an inscribed nonagon, segments, and shaded triangles size(150); // Set canvas size // Draw the unit circle draw(circle((0,0), 1), blue); // Circle centered at (0,0) with radius 1 dot((0,0)); // Mark the center of the circle label("$O$", (0,0), SW); // Label the center // Define the vertices of the nonagon pair[] P; // Array to store points int n = 9; // Number of sides of the nonagon for (int i = 0; i < n; ++i) { P[i] = dir(360*i/n); // Points evenly distributed around the circle } // Label the vertices string[] labels = {"A", "B", "C", "D", "E", "F", "G", "H", "I"}; for (int i = 0; i < n; ++i) { dot(P[i]); label("$"+labels[i]+"$", P[i], dir(360*i/n)); // Adjust label direction } // Draw the sides of the nonagon for (int i = 0; i < n; ++i) { draw(P[i]--P[(i+1)%n], gray); // Connect each vertex to the next } // Draw the specified segments draw(P[0]--P[2], red); // Segment AC draw(P[2]--P[4], red); // Segment CE draw(P[4]--P[6], red); // Segment EG draw(P[6]--P[8], red); // Segment GI draw(P[0]--P[4], dashed+blue); // Dashed segment AE draw(P[0]--P[6], dashed+blue); // Dashed segment AG // Shade the interiors of the triangles fill(P[0]--P[1]--P[2]--cycle, green+opacity(0.3)); // Triangle ABC fill(P[2]--P[3]--P[4]--cycle, yellow+opacity(0.3)); // Triangle CDE fill(P[4]--P[5]--P[6]--cycle, cyan+opacity(0.3)); // Triangle EFG fill(P[6]--P[7]--P[8]--cycle, magenta+opacity(0.3)); // Triangle GHI [/asy][/asy] Now we prove that this is the best possible bound. If a triangle is acute, it must contain the center of the circle in its interior. Therefore, there can be at most \( 1 \) acute triangle. Similarly, there can be at most \( 2 \) right triangles. [asy][asy] // Asymptote code for a unit circle with an inscribed triangle having an obtuse angle size(150); // Set canvas size // Draw the unit circle draw(circle((0,0), 1), blue); // Circle centered at (0,0) with radius 1 dot((0,0)); // Mark the center of the circle label("$O$", (0,0), SW); // Label the center // Define points A, B, and C on the circle pair A = dir(100); // Point A at 30 degrees pair B = dir(150); // Point B at 150 degrees pair C = dir(30); // Point C at 270 degrees // Draw the triangle ABC draw(A--B--C--cycle, red); // Draw triangle ABC with red edges // Label the points dot(A); label("$B$", A, N); dot(B); label("$A$", B, NW); dot(C); label("$C$", C, NE); [/asy][/asy] Thus, assume that the triangles are congruent to \( \triangle ABC \), where \( \angle ABC \) is obtuse. Notice that we cannot fit another congruent version of this triangle within the arcs \( AB \) or \( BC \). Consequently, at least two of the \( n \) arcs created by the \( n \) points must be uniquely assigned to each congruent copy of \( \triangle ABC \). This establishes the upper bound.