math question - Printable Version +- HP Forums (https://archived.hpcalc.org/museumforum) +-- Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum-1.html) +--- Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum-2.html) +--- Thread: math question (/thread-218908.html) math question - Don Shepherd - 04-24-2012 I am having my students write a Basic program to input the length and width of a rectangle, then calculate and display the area and perimeter of the rectangle. I noticed that, for a certain number pair, area = perimeter. I tried a few other pairs of length/width and couldn't find a repeat of that relationship. If x is length and y is width, we have: ```xy = 2x + 2y ``` How can I determine if there are other number pairs that satisfy this condition? Re: math question - Eric Smith - 04-24-2012 Use algebra to transform your equation into y=2x/(x-2). Now for any x you can find the corresponding y that makes the perimeter equal to the area. There are an infinite number of solutions. If you're only interested in integer solutions, there are only two unique solutions, considering x and y to be interchangeable. Edited: 24 Apr 2012, 8:29 p.m. Re: math question - Gerson W. Barbosa - 04-24-2012 Or use Dario Alpern's Solver and fill in the blanks with 0, -1, 0, 2, 2 ,0, then solve it :-) Positive integer solutions: a) x = 0, y = 0; b) x = 3, y = 6; c) x = 4, y = 4. P.S.: I assume you haven't counted the trivial one. Edited: 24 Apr 2012, 8:44 p.m. Re: math question - Peter Murphy (Livermore) - 04-24-2012 Be sure to tell your students that equating area and perimeter is, formally speaking, nonsense: they're quantities with different units. As a relationship between pure numbers, there's no such objection, of course. Re: math question - Eduardo Duenez - 04-24-2012 Rewrite as ```(x-2)(y-2)=4. ``` From here you can easily find all solutions. Assuming you are after positive integer solutions, there are only two: (4,4) and (3,6) (and (6,3), I guess). These come from factoring 4=2x2=1x4. Eduardo Re: math question - Eric Smith - 04-24-2012 Right. I assumed that a figure with sides of length zero was not a rectangle. Re: math question - Gerson W. Barbosa - 04-24-2012 I tend to see the Euclidean point as any of the infinite n-zero-lenght-side regular polygons, a square with null sides being one of them. This doesn't mean I am right, of course! Edited: 24 Apr 2012, 9:17 p.m. Re: math question - Gilles Carpentier - 04-25-2012 Well done Eduardo ! Very nice Re: math question - Don Shepherd - 04-25-2012 Yes, I see that. Very clever! Re: math question - Alexander Oestert - 04-25-2012 You're pulling our respective legs, aren't you? ;-) Re: math question - Don Shepherd - 04-25-2012 No, this actually happened. I'm giving a test to my students to write a Basic program to calculate area and perimeter of a rectangle from inputs of length and width, so I wrote my little program first and then tested it and I happened to enter length/width 3 and 6 and I noticed that area and perimeter were the same. I wondered if any other number pair would result in that situation, but I couldn't create a system of equations with xy = 2x + 2y, so I posted here for help, and I knew someone more clever than I am would figure it out. And several did! Re: math question - Dominic Richens - 04-25-2012 Is a rectangle if you lim x->0 & y > 0 :D Re: math question - Marcus von Cube, Germany - 04-25-2012 I have a related question which bogs me for some time now. I try to explain it with a small ASCII picture: ``` ___________ | | |___ | | | | | |___ | | | | |_______|___| ___________ |_ | | |_ | | |_ | | |_ | | |_ | |_________|_| ``` The zig-zag line from the top left corner to the bottom right corner is an approximation to the diagonal but its length is just twice the side length. You can refine it ad infinitum and it will become indistinguishable from the diagonal but the total length is constant in each iteration. Strange, isn't it? Re: math question - bill platt - 04-25-2012 Fractals at their simple best. Re: math question - Frank Boehm (Germany) - 04-25-2012 As a student I had to do some wicked (as I thought, but what do I konw ) presentation of some topics on Linear Algebra, this one reminds me of it. I think a "Koch-Kurve" might deliver some hints. Re: math question - Roman P Flesch (Berlin) - 04-25-2012 ... it's interesting that for a VERY long rectangle the short side will approach the limit y=2 if xy=2(x+y). It is trivial, but I wasn't aware of that. Regards, Roman Re: math question - Gilles Carpentier - 04-25-2012 Edited: 25 Apr 2012, 6:48 p.m. Re: math question - BruceH - 04-25-2012 Quote: Strange, isn't it? What makes it strange is that school maths drums into you that a piece of paper only has two dimensions, X & Y. And that any points or lines drawn on that paper can always be represented by X,Y coordinates. Yet diagonal lines, as you point out, 'break the rules' of two dimensions because they have a length that is less than the sum of the steps along the X,Y dimensions, no matter how small those steps are. I'm not sure that fractals are the whole answer. Sure, the stepped line from corner to corner has a fractal dimension. But the diagonal is straight and therefore not fractal. It's as if a 2-dimensional plane has 3 degrees of freedom - left-right only, up-down only and combined. Perhaps we should confuse the world by calling that 3 dimensional ;-) Re: math question - Carey - 04-25-2012 Hi Marcus, Great question! Vectors may be helpful in resolving the paradox. ```1. Convert all line segments into vectors by inserting arrows. 2. For each pair of horizontal and vertical vectors (i.e., for each zig and zag along the path) draw diagonal resultant vectors, forming little triangles. 3. For each little triangle, the difference between the distance along the zig-zag (the sum of 2 sides) and the displacement (the length of the diagonal) is the triangle error. 4. The total error is the sum of all triangle errors along the path. The total error is the same whether there are few zig-zags (a few large errors) or many little zig-zags (many little errors). This can be seen by adding all the horizontal arrows and, separately, adding all the vertical arrows. The vector sums are constant no matter the size of the zig-zags (if they weren't constant, some paths would undershoot or overshoot the destination). ``` Paths of many small zig-zags look smoother and are better approximations to the diagonal in a least squares sense (distance of zig-zag path to diagonal path). However the total error in approximating the length of the diagonal is the same for paths of many large zig-zags or many small zig-zags (few big errors vs many little errors). A hallmark of a fractal (self-similar, irregular) curve (e.g., a coastline) is that its curve length is not a definite value but depends on the scale used in measuring it. In this problem: (1) the zig-zag curves are not self-similar (i.e., when you zoom in on the coarse zig-zag curve you do not find little zig-zag curves) and, (2) the length of the zig-zag curves is definite and not dependent on the scale used in measuring them. Hence, this does not appear to be a fractal problem. Edited: 26 Apr 2012, 9:08 a.m. Re: math question - Allen - 04-25-2012 Don- what an interesting observation and excercise!! Would your students be open to a discussion about units? e.g. although their magnitude may be equal sometimes, they can check their formulas by thinking of each side of the equation's geometrically. Although a pure maths person may say the answer is y=2x/(x-2) leaving y alone as a unit-less number). An engineer might look at the problem statement and, by following the units, suggest the perimeter (distance) is never equal area (distance^2) because the units don't match. :) I realize it's a small point of order, but it's a great skill for the students to have later in life! Re: math question - Don Shepherd - 04-25-2012 Thanks Allen. I just happened to notice that the length/width pair 3/6 yields the same quantity for both area and perimeter, and I wondered if there was a different pair like that, and it turns out there is (4/4). I wanted to build and solve a system of equations but it was not possible in this case, and I really like what Eduardo did with factoring to solve the problem. That's the beauty of the forum. When I teach math (I'm teaching "technology" this year) and we discuss area and linear measures like perimeter, of course I do distinguish between inches and square inches (or feet). I generally bring in my tape measure and we measure the distance along the four walls (perimeter), and they always understand that well. Then for area I'll bring in some 1 foot by 1 foot floor tiles and start to lay them on the floor so they can see what area and square feet really mean. Kids always appreciate real physical stuff, and they tend to understand concepts like square units better when you show them the real thing. But until they get out in the real world and have to do things like buy carpet for their living room, they tend to forget what they've learned about square units, so we teach it again the next year. Repitition helps the learning process. Re: math question - David Hayden - 04-26-2012 Quote: You can refine it ad infinitum and it will become indistinguishable from the diagonal. Not really. If it seems indistinguishable, you just need a bigger magnifying glass. I zoomed in on your indistinguishable line and it still straddles the diagonal. Problems similar to this can be used to show how one's intuitive sense of math breaks down when you do "ad infinitim" problems. It's a nice introduction to some of the theories about infinity and calculus. Re: math question - Dominic Richens - 04-26-2012 Similarly ignoring that you can "prove" Pi = 4 www.youtube.com/watch?v=D2xYjiL8yyE