# Problem of the Week Problem C and Solution Shape On

## Problem

A necklace is to be created that contains only square shapes, circular shapes, and triangular shapes. A total of $$180$$ of these shapes will be strung on the necklace in the following sequence: $$1$$ square, $$1$$ circle, $$1$$ triangle, $$2$$ squares, $$2$$ circles, $$2$$ triangles, $$3$$ squares, $$3$$ circles, $$3$$ triangles, with the number of each shape type increasing by one every time a new group of shapes is placed. The diagram illustrates how the first $$18$$ shapes would be strung.

Once the necklace is completed, how many of each shape would the necklace contain?

## Solution

An equal number of square shapes, circular shapes and triangular shapes occur after \begin{aligned} 3(1)&=3 \text{ shapes are placed,}\\ 3(1)+3(2)=3+6&=9 \text{ shapes are placed,}\\ 3(1)+3(2)+3(3)=3+6+9&=18 \text{ shapes are placed, and so on.}\end{aligned}

The greatest total that can be placed with equal numbers of squares, circles and triangles is \begin{aligned} & 3(1)+3(2)+3(3)+3(4) + 3(5) + 3(6) + 3(7) + 3(8) +3(9)+3(10)\\ &=3+6+9+12 + 15 + 18 + 21 + 24+27+30\\ &=165\end{aligned}

At this point there are $$165\div 3=55$$ of each of the three shapes. We are at a point the last $$30$$ shapes placed are $$10$$ squares, $$10$$ circles and $$10$$ triangles, in that order.

The next group would have $$11$$ of each shape if all of the shapes could be placed. However, there are only $$180-165=15$$ shapes left to place. We are able to place $$11$$ square shapes leaving $$4$$ shapes left to place. At this point there are $$55+11=66$$ square shapes on the necklace. The final $$4$$ shapes would be circular. And at this point there would be $$59$$ circular shapes on the necklace. No more triangular shapes can be added so the total number of triangular shapes remains at $$55$$.

Therefore, the completed necklace will contain $$66$$ square shapes, $$59$$ circular shapes, and $$55$$ triangular shapes.