Difference between revisions of "2003 Indonesia MO Problems/Problem 6"

Latest revision as of 09:37, 1 September 2025

Problem

A hall of a palace is in a shape of regular hexagon, where the sidelength is $6 \text{ m}$ . The floor of the hall is covered with an equilateral triangle-shaped tile with sidelength $50 \text{ cm}$ . Every tile is divided into $3$ congruent triangles (refer to the figure). Every triangle-region is colored with a certain color so that each tile has $3$ different colors. The King wants to ensure that no two tiles have the same color pattern. At least, how many colors are needed?

$[asy] pair C=(25,14.434); draw((0,0)--(50,0)--(25,43.301)--(0,0)); draw(C--(0,0)); draw(C--(50,0)); draw(C--(25,43.301)); [/asy]$

Solutions

Solution 1

First, count the number of tiles in the hall. The hall can be made up of six equilateral triangles with side length 600 cm. Since $600/50 = 12$ equilateral triangles with side length 50 cm can make up a side of the larger equilateral triangle, there are $12 + 2(11 + 10 + 9 + \cdots 1) = 144$ smaller triangles in each large triangle. In total, there are $144 \cdot 6 = 864$ tiles needed.

Let $x$ be the number of colors used. Given three colors, there are only two tiles using the color scheme that do not have the same color pattern. With this information, we can write an equation. $\begin{align*} 2 \binom{x}{3} &\ge 864 \\ 2 \cdot \frac{x(x-1)(x-2)}{6} &\ge 864 \\ x(x-1)(x-2) &\ge 2592 \end{align*}$ With some trial and error, we find that the King needs $\boxed{15}$ different colors.

Solution 2

We compute the area of the hexagon. The two smaller triangles each have area $9 \sqrt{3}$ , and the central rectangle has area $6 \cdot 6 \sqrt{3} = 36 \sqrt{3}$ . The total area is thus $54 \sqrt{3} m^2 = 54 \sqrt{3} \cdot 10^4 (cm)^2$ . We get: $54 \sqrt{3} \cdot 10^4 (cm)^2 = n \cdot \frac{\sqrt{3}}{4} \cdot 2500 (cm)^2 \implies n = 864$ Now, we have to color $864$ tiles. I don't know what the problem means by a color pattern. I assume that any tile that can be reflected or rotated to obtain another is the same color pattern. Suppose we have $k$ colors. We require the minimum $k$ satisfying $\binom{k}{3} \geq 864 \implies \boxed{k \geq 19 }$ I do believe that the above user has got it wrong, as there is no way to use $3$ colors and obtain 2 different colorings of the tile, because each is same under rotation and reflection.

@@ Line 11: / Line 11: @@
 </asy>
-==Solution==
+==Solutions==
+=== Solution 1 ===
 First, count the number of tiles in the hall.  The hall can be made up of six equilateral triangles with side length 600 cm.  Since <math>600/50 = 12</math> equilateral triangles with side length 50 cm can make up a side of the larger equilateral triangle, there are <math>12 + 2(11 + 10 + 9 + \cdots 1) = 144</math> smaller triangles in each large triangle.  In total, there are <math>144 \cdot 6 = 864</math> tiles needed.
@@ Line 24: / Line 24: @@
 With some trial and error, we find that the King needs <math>\boxed{15}</math> different colors.
-{{alternate solutions}}
+=== Solution 2 ===
+We compute the area of the hexagon. The two smaller triangles each have area <math>9 \sqrt{3}</math>, and the central rectangle has area <math> 6 \cdot 6 \sqrt{3} = 36 \sqrt{3}</math>. The total area is thus <math>54 \sqrt{3} m^2 = 54 \sqrt{3} \cdot 10^4 (cm)^2</math>. We get:
+<cmath> 54 \sqrt{3} \cdot 10^4 (cm)^2 = n \cdot \frac{\sqrt{3}}{4} \cdot 2500 (cm)^2 \implies n = 864</cmath>
+Now, we have to color <math>864</math> tiles. I don't know what the problem means by a color pattern. I assume that any tile that can be reflected or rotated to obtain another is the same color pattern. Suppose we have <math>k</math> colors. We require the minimum <math>k</math> satisfying <cmath> \binom{k}{3} \geq 864 \implies \boxed{k \geq 19 }</cmath>
+I do believe that the above user has got it wrong, as there is no way to use <math>3</math> colors and obtain 2 different colorings of the tile, because each is same under rotation and reflection.
 ==See Also==

2003 Indonesia MO (Problems)
Preceded by Problem 5	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 7
All Indonesia MO Problems and Solutions

Page

Toolbox

Search