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The problem is as follows:

From a rectangular piece of paper, $4$ straight cuts have been made. These cuts are parallel to the diagonals of the rectangle. After making the cuts, the 4 pieces are removed. The sum of the lengths of the four cuts made is $\textrm{80 cm}$. Find the perimeter of the piece of paper that is obtained.

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{292 cm}\\ 2.&\textrm{248 cm}\\ 3.&\textrm{276 cm}\\ 4.&\textrm{284 cm}\\ \end{array}$

For this particular situation. I'm stuck. Does it exist a way to get this perimeter?. The reason for why I'm stuck is that it isn't given the lengths of the corners of the paper neither full sides of the chunks which has been removed. Does it exist to get those, can this problem be solved?. Help please?

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  • $\begingroup$ Hint: use the Pythagorean triple $3-4-5$. $\endgroup$ Jan 6, 2021 at 3:01
  • $\begingroup$ Note that if you move the first and second cuts together left and right, the length cut gained by one exactly matches the length lost by the other. Similarly the third and fourth cuts moved together, the first and fourth cuts moved together, the second and third cuts moved together, and all four cuts moved together. The point is that you are focusing on particular measurements, but various totals are unchanged by manipulating the cuts, so the detailed values are less important than the totals. $\endgroup$ Jan 6, 2021 at 3:08
  • $\begingroup$ May I also invite or encourage you to perhaps accept answers that you feel best answer your question? I see that you actually have quite a long list of questions with only a small percentage of them marked as answered! $\endgroup$ Jan 6, 2021 at 11:22
  • $\begingroup$ @AndrewChin Yes, sorry. I'll do that. My health has not allowed me to review earlier problem in focus. $\endgroup$ Jan 6, 2021 at 17:11

2 Answers 2

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Recognize that if the cuts are parallel to the diagonals, then the side lengths of the triangles that have been cut out are in proportion to a $3-4-5$ right triangle.

When a cut is made, a length equal to the sum of the two legs of the right triangle will have been taken out of the perimeter, but the length equal to the hypotenuse will be added (i.e. for every $5cm$ cut, $7cm$ gets taken out of the perimeter, but $5cm$ will be added to the perimeter, giving a net loss of $2cm$ per $5cm$ cut).

Since we started with a perimeter of $280cm$, and $80cm$ of cuts have been made, this gives us a total loss of $32cm$, resulting in a perimeter of $248cm$.

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  • $\begingroup$ It's interesting method of solution. I must say that at first I had worried on why you did not took into an account the lengths of the edges of the rectangle but it turns out that what it matters here is not to focus in counting the left and right or up and down remaining edges of the figure in the resulting octagon. It is just a matter of focusing in the net loss as you indicated. For me the key was to spot the $3-4-5$ Pythagorean triple. Either if the $3k-4k$ sides increase in size it doesn't matter. In the end the net loss remains to be $2$. It took me time to realize it. $\endgroup$ Jan 6, 2021 at 17:16
  • $\begingroup$ Note that you are implicitly using AA rule in your first paragraph, but not mentioning it. $\endgroup$
    – Martund
    Jan 9, 2021 at 4:43
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Note that $$\text{New perimeter} = \text{Old perimeter}-\text{Lengths of corners}+\text{Lengths of new edges}$$ Since the cuts are parallel to the diagonal of the rectangle, and one of the angle is $90^\circ$, by AA rule, every cut triangle is similar to the triangle formed by cutting the rectangle along one of its diagonals. So the sides of the cut triangles are in the ratio $3:4:5$ (as shown in the following image). So the new perimeter is $$2(60+80)-3(x+y+z+w)-4(x+y+z+w)+5(x+y+z+w)$$ Using the constraint that $$5(x+y+z+w)=80$$ and putting in above expression, we get, the new perimeter to be $$280-2\times\dfrac{80}5=\boxed{248\ \mathrm{cm}}$$

enter image description here

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  • $\begingroup$ What do you mean with AA rule what's this? Are you perhaps referring to this? I'm assuming that's for similarity but I'm not sure because I'm not very savvy with this. Can you please explain? $\endgroup$ Jan 6, 2021 at 17:01
  • $\begingroup$ Yes @ChrisSteinbeckBell, this is precisely what I am referring to. If you can think of any other method of proving that the cut corners are triangles with sides in the ratio of $3:4:5$, that is also fine. $\endgroup$
    – Martund
    Jan 7, 2021 at 6:48

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