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Find the lengths of the sides of these squares:
I put these equations in standard form:
And then I made a coefficient matrix:
And the constant matrix is
The determinant of the coefficient matrix is -7. Then I used Cramer's rule to find the values of the variables from the determinants of the altered coefficient matrices, each with one column replaced by the constant matrix. The are:
Oh, but this can't be the solution, because we were told all sides are integers. But the assignment of the length A=1 was arbitrary, remember? So we can multiply all the sides by 7 to get integers. In fact, we could have multiplied them by 14 or 21 to get other solutions. So the final solution to this problem is:
Alternative solution: I gave this puzzle to Caitlin, my daughter, and she assigned "x" to the side of square A and "y" to the side of square B. Then D=x+y, so E=x+(x+y), etc. Before too long she had written these facts on the puzzle (I show them here in the order she got them): A=x Now we want an equation that relates x and y, so let's see if we can express the top and bottom edges in two different ways. The top edge is J+I=15x and the bottom edge is H+F+G=15x. No help there. The left edge is H+J=20x-3y. The right edge is I+E+G=8x+4y. Aha! So 20x-3y=8x+4y, so The smallest integers, x and y, that satisfy this equation are x=7 and y=12, because 7 and 12 are coprime. Using these values of x and y, all the squares work out to the same answers that were listed, above. Related pages in this website |
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The webmaster and author of the Math
Help site is Graeme McRae.
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I
labeled the squares in ascending size order from A to J. Then I made
up some linear equations based on the diagram. It is easy to see,
for example, that the side of D is equal to the side of A plus the side of
B. In a similar way, I made up eight more equations. Then I
arbitrarily assigned a length of 1 to square A. Now I have 10
equations in 10 unknowns. Here they are:
