CEMC Banner

Problem of the Week
Problem C and Solution
Shape On


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?


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.