2011 JBMO Problems/Problem 3
Problem
Let be a positive integer. Equilateral triangle ABC is divided into
smaller congruent equilateral triangles (with sides parallel to its sides). Let
be the number of rhombuses that contain two small equilateral triangles and
the number of rhombuses that contain eight small equilateral triangles. Find the difference
in terms of
.
Solution
First we will show that the side lengths of the small triangles are of the original length. Then we can count the two rhombuses.
Lemma: Small Triangle is Length of Original Triangle
Let the side length of the triangle be , so the total area is
.
Since the big triangle is divided into congruent equilateral triangles, the area of each smaller equilateral triangle is
Thus, the length of each smaller triangle is
Since the lengths of each smaller triangle is and are parallel to the big triangle’s sides, the triangles are arranged like the below diagram. In diagram,
. Now we can count the rhombuses.
After dividing the triangles, for each line that is not in the border of the big equilateral triangle, there are two smaller equilateral triangles. The longest row in the triangle has lines and the shortest row has
line. Thus, there are
lines parallel to one side, resulting in a total of
lines or rhombuses with 2 equilateral triangles.
For a rhombus to have eight triangles, the center line must have two lines and the rows above and below must have one line. We can count the number of segments made of two lines that are not in the last row in the big triangle. The second longest row in the triangle has lines, while the second shortest row has
lines. Thus, there are
valid segments from two lines parallel to one side, resulting in a total of
rhombuses with 8 equilateral triangles.
Because and
,
See Also
2011 JBMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |