Jigsaw Ingredients

Algebra Level 4

A typical jigsaw puzzle has 4 corner pieces, a number of edge pieces, and the rest interior pieces. (See the drawing above.)

A certain jigsaw puzzle has 50% more pieces horizontally than vertically, and 13.0% of the pieces are edges.

Calculate the horizontal dimension of the puzzle.


The answer is 36.

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1 solution

Let x , y x, y be the horizontal and vertical dimensions of the puzzle. Then we have

  • total number of pieces: N = x y N = xy ;

  • interior pieces: I = ( x 2 ) ( y 2 ) I = (x-2)(y-2) ;

  • edge pieces: E = N I 4 = x y ( x 2 ) ( y 2 ) 4 = 2 ( x + y ) 8 E = N - I - 4 = xy - (x-2)(y-2) - 4 = 2(x+y) - 8 .

We are told that x = 1 1 2 y E = 0.13 N , x = 1\tfrac12 y\ \ \ E = 0.13\cdot N, which gives the following equation for y y : 0.13 1 1 2 y y = 2 ( 1 1 2 y + y ) 8 0.195 y 2 = 5 y 8 0.195 y 2 5 y + 8 = 0. 0.13\cdot 1\tfrac12y \cdot y = 2(1\tfrac12y + y) - 8 \\ 0.195y^2 = 5y - 8 \\ 0.195y^2 - 5y + 8 = 0. The solutions of this quadratic equation are y = 5 ± 5 2 4 0.195 8 2 0.195 = 5 ± 18.76 0.39 23.93 or 1.71. y = \frac{5 \pm \sqrt{5^2 - 4\cdot 0.195\cdot 8}}{2\cdot 0.195} = \frac{5\pm\sqrt{18.76}}{0.39} \approx 23.93\ \ \text{or}\ \ 1.71. Clearly the latter solution makes no sense. We conclude that y = 24 y = 24 , so that x = 36 x = \boxed{36} .

Check: A 24 × 36 24\times36 puzzle has

  • N = 24 36 = 864 N = 24\cdot 36 = 864 pieces in total,

  • of which I = 22 34 = 748 I = 22\cdot 34 = 748 interior pieces,

  • and E = 864 748 4 = 112 E = 864 - 748 - 4 = 112 edge pieces,

so that the percentage of edge piece is E / N = 112 / 864 12.96 % . E/N = 112/864 \approx 12.96\%. .

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