Has anyone seen a problem like this before?

MathTutor25

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I started tutoring a student in advanced geometry. This was the only question that really had me stumped. How do you draw a parallelogram based solely on the perimeter? Am I missing something? Any ideas would be helpful. (See image attached)
 

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I think you missed the significance of skeeter's assumption, which actually summarizes a number of assumptions.

Unless you assume that: (1) the dots in each row are colinear, (2) each dot in a row is separated from an immediately adjacent dot in the same row by a distance of one unit, (3) the dots in each column are colinear, (4) each dot in a column is separated from an immediately adjacent dot in the same column, and (5) the lines forming rows and columns are orthogonal, the problem cannot be solved. So it is a badly expressed problem.

If you make those assumptions, which result in the formation of unit squares as skeeter pointed out, ask your self how a rectangle (which is a type of paraellelogram) connecting the dots has anything but a perimeter that is a sum of integers.

Still trying to stay simple, let's say two sides lie on rows and have length r. Note that r must be a positive integer. Why?

Then the length of the either of the other two sides is

[MATH](6\sqrt{2} + 8 - 2r) \div 2 = 3\sqrt{2} + 4 - r.[/MATH]
The length of that side is the hypotenuse of a right triangle. Let the length of the other two sides of that right triangle be x and y. Note that x and y must be positive integers. Why?

[MATH]\therefore x^2 + y^2 = (3\sqrt{2} + 4 - r)^2 = \\ (3\sqrt{2} + 4)^2 - 2r(3\sqrt{2} + 4) + r^2 =\\ 34 - 8r + r^2 + (24 - 6r)\sqrt{2}.[/MATH]Therefore 24 - 6r must equal 0. Why?

[MATH]\therefore r = 4 \implies x^2 + y^2 = 34 - 32 + 16 = 18.[/MATH]
The above is a constrained Diophantine equation because both x and y are positive integers less than 5. Some simple numeric experimentation will show that x = 3 = y.

EDIT: Not only is there a solution, it is a unique solution. I just wish the assumptions had been stated clearly.

I don't think that purely geometric thinking will solve this, but the problem is one that the ancient Greeks would have loved because it eliminates those irrationals.

SECOND EDIT:

I didn't check my answer.

Each sloping side has a length of [MATH]\sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}[/MATH]
Thus, the perimeter is [MATH]3\sqrt{2} + 4 + 3\sqrt{2} + 4 = 6\sqrt{2} + 8. \ \checkmark[/MATH]
 
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Considering this was the only question that stood out to me during the tutoring session, I'm sure I'm doing fine after not touching geometry in over a decade. Please offer help in relation to the question next time you post.
We are volunteers on this forum and we get to say whatever we want. I too think that being a tutor and not being able to do a problem is unacceptable. Get another job.

Also, instead of writing negative posts maybe you should be studying geometry instead.

For the record I have not done any geometry since 10th grade (1975) and I recently tutored a student for a whole semester in Geometry and I could do every problem.

By the way, I have never seen a problem like this. Do you think that I could do it?
 
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