# Menseki Meiro

Strategy 1:

If we have three collinear points A, B, and C, such that B lies between A and C, and we know any two of the lengths, then we can deduce the third. That's because AC = AB + BC, which we can rearrange as desired.

Strategy 2:

Use the area of a rectangle and one of its dimensions to deduce its other dimension. For example, if we have a rectangle that has an area of 42, and a width of 7, then its height must be 42 / 7 = 6.

Strategy 3:

If two rectangles have the same width, or height, then the proportion of their areas is equal to the proportion of their other dimension. For example, if two rectangles have the same width, one rectangle has area 13, the other, 26, then the height of the smaller rectangle must be half the height of the larger rectangle.

As a special case, if two rectangles have the same width, or height, and the same area, then the both their dimensions are the same.

Strategy 4:

If there is a rectangular region whose dimensions are known, draw the rectangle, and find its area.

Strategy 5:

If the width and part of the height, or the height and part of the width, is known for some rectangle, then cut the rectangle in two, such that the width and height are known for one of the rectangles. Then use the width and height to compute its area.

This puzzle can be solved easily using the strategies already laid out:

The 29 rectangle is half the 58 rectangle, but we don't know the height of either. Let's call them x and 2x respectively. Notice the bottom middle rectangle has the same height as the 29 rectangle, which has height x. The 27 rectangle is half the 57 rectangle. Call their heights y and 2y respectively. But then 3y = 3x. And dividing both sides by 3 tells us x = y. Thus, y and 2y can be rewritten to x and 2x, respectively. Notice the top middle rectangle has the same height as the 27 rectangle, which has height x. The rectangles above and below the 28 rectangle have height x, and the entire figure has height 3x, so the 28 rectangle must have height x.

Strategy 6:

Use variables to express how lengths are related.

The width of the 78 rectangle is 78 / 6 = 13. The width of the 65 rectangle is 65 / 5 = 13. Slide the 78 and ? rectangles up 2 in. Label the width of the 50 rectangle as x, and the ? rectangle as y. Then it's easy to see that 13 + x = 13 + y. But then, subtracting 13 from both sides, x = y. Thus, the width of the 50 rectangle is equal to the width of the ? rectangle. But it's also obvious from the current state of the diagram, that these rectangles also have the same height, and thus, the same area. Hence, the ? rectangle has area 50 in².

Strategy 7:

Move the rectangles around, to make edges of equal length.