Maths Olympiad Prep

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Combinatorics Difficulty 4.8 AIME Find the answer Canada

In Figure 1, three unshaded dots are arranged to form an equilateral triangle, as shown.

Figure 0

Figure 1

Figure 2 is formed by arranging three copies of Figure 1 to form the outline of a larger equilateral triangle and then filling the resulting empty space with 1 shaded dot.

Figure 1

Figure 2
For each integer n gt;2\text{n gt;2}, Figure nn is formed by first arranging three copies of Figure n1n-1 to form the outline of a larger equilateral triangle and then filling the resulting empty space in the centre with an inverted triangle of shaded dots.

Figure 2

Figure 3

The smallest value of nn for which Figure nn includes at least 100 000 shaded dots is

88
99
1010
1111
1212

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since each Figure is formed by placing two copies of the previous Figure side-by-side along the base and then adding other pieces above, the number of dots in the base of each Figure is two times as many as in the previous Figure.

Since each Figure is an equilateral triangle, then the number of dots in the Figure equals the sum of the positive integers from 1 to the number of dots in the base, inclusive. In other words, if the base of a Figure consists of bb dots, then the Figure includes 1+2+3++(b1)+b1 + 2 + 3 + \cdots + (b-1) + b dots. This sum is equal to 12b(b+1)\frac{1}{2}b(b+1). (If this formula for the sum is unfamiliar, can you argue why it is true?)

Since each Figure is formed by using three copies of the previous Figure and any new dots added are shaded dots, the number of unshaded dots in each Figure is exactly three times the number of unshaded dots in the previous Figure.

Since each dot is either shaded or unshaded, the number of shaded dots equals the total number of dots minus the number of unshaded dots.

Using these statements, we construct a table:

FigureDots in baseDots in FigureUnshaded dotsShaded dots
12330
241091
3836279
4161368155
532528243285
66420807291351
7128825621876069
825632896656126335
951213132819683111645

Therefore, the smallest value of nn for which Figure nn includes at least 100 000 dots is n=9n=9.

We note that since the number of dots in the base of Figure 1 is 2 and the number of dots in the base of each subsequent Figure is double the number of dots in the previous Figure, then the number of dots in the base of Figure nn is equal to 2n2^n.

Since the number of unshaded dots in Figure 1 is 3 and the number of unshaded dots in each subsequent Figure is three times the number of unshaded dots in the previous Figure, then the number of unshaded dots in Figure nn is 3n3^n.

Therefore, a formula for the number of unshaded dots in Figure nn is 122n(2n+1)3n\frac{1}{2}2^n(2^n+1) - 3^n which can be re-written as 22n1+2n13n2^{2n-1} + 2^{n-1} - 3^n, which agrees with the numbers in the table above.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.