I worn down counting all the rectangles I can see, yet that didn"t work. How do I technique this?
Go action by step.
First Picture: 1 rectangle
Second Picture: 2 extr rectangles. The small rectangle, which has been included and the huge one, which includes the two small rectangles.
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Third picture: The large rectangle. Then two rectangles, which includes 2 little linked rectangles. And the little rectangle, which has actually been added
Fourth picture: only one tiny rectangle.
Fifth Picture: The rectangle, which includes the two tiny rectangle and the tiny additional rectangle.
You go on favor this. Then sum the quantity of rectangles.
answer Jul 20 "14 in ~ 15:34
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Each rectangle has actually two upright lines and two horizontal lines.
There are five vertical present in the picture, we deserve to label lock 1, 2, 3, 4, 5.
If the outward edge is 1: then the top and also bottom room uniquely determined, and also it is simple to see that 3 or 4 need to be the best edge. 2 options.
If the leftmost edge is 2: then the rightmost edge is 3 or 4 (2 choices), and also in either situation there are 3 horizontal segments that can serve as the top/bottom($inom32 =3$ choices). So this provides $2 cdot 3 = 6$. 6 options.
If the leftmost edge is 3: If the rightmost leaf is 5 there is just one rectangle. If the rightmost sheet is 4, there are 5 horizontal segments for top and bottom, for this reason $inom52 = 10$ choices. Therefore 11 options.
If the leftmost edge is 4: then the rightmost leaf is 5, and also there are four horizontal segments yielding $inom42 = 6$ feasible rectanges. 6 options
The full is 2 + 6 + 11 + 6 = 25.
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reply Apr 8 "16 in ~ 22:41
Alex ZornAlex Zorn
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