When I was in third grade, I was playing with rectangles and diagonal lines, and discovered something very interesting with fractions. I've shown several math teachers and professors over the years, and never got an answer. Just a few, "Wow, that's neat!"

Draw a rectangle. Draw a line from the top left corner to the bottom right corner. Then draw a line from the top right corner to the bottom left corner. The intersection obviously becomes 1/2 units of the rectangle's width.

Now draw a line from the last intersection to the bottom line of the rectangle, and then from that point to the top right corner of the rectangle. The new intersection becomes 1/3 units of the rectangle's width.

Keep doing this and the denominator of the fraction increases by one each time to infinite. Why does this happen? I don't know how to prove why this happens, but it would be interesting if someone could. Can you? I never became a mathematician to prove it, but if it's easy, please forgive my mathematical ignorance. I tried this several years ago with AutoCAD and it does in fact work out.

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    $\begingroup$ If you look at the lower left half of the entire rectangle, it is precisely the perspective projection of an infinite tape marked like |/|/|/|/|/|... with the vanishing point on the right. Can you see it? Surely there's a way to turn this into an actual proof. $\endgroup$ – Rahul Aug 6 '12 at 7:42

If we take the top-right corner as the origin and the $x$ and $y$ axes leftward and downward, respectively, and take the side lengths of the rectangle as the respective units, you're intersecting the lines $y=nx$ with the line $y=1-x$. You get the intersections by equating the two right-hand sides, $nx=1-x$, and that yields $(n+1)x=1$, and thus $x=1/(n+1)$.

  • $\begingroup$ +1 .. I'll mark an answer in a few days. I want to see what some other mathematics think of. Thank you! Very cool! This looks a bit like an algebra proof (not a geometry proof), but I'm not criticizing. Haha! Kidding. $\endgroup$ – MacGyver Aug 6 '12 at 5:27
  • $\begingroup$ @MacGyver: You're welcome. I liked the question, too :-) $\endgroup$ – joriki Aug 6 '12 at 5:29
  • $\begingroup$ I'm not even sure what tag to add for an algebra proof. $\endgroup$ – MacGyver Aug 6 '12 at 5:30
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    $\begingroup$ @MacGyver: How about analytic-geometry? $\endgroup$ – joriki Aug 6 '12 at 5:33

Joriki's proof is very beautiful. I just wanted to add other way to proof the relation via similar triangles. I drew 2 steps. If someone follows that way, the result can be gotten easily .

enter image description here enter image description here enter image description here enter image description here

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    $\begingroup$ Your proof is much more beautiful than mine :-) $\endgroup$ – joriki Aug 6 '12 at 16:15

This is essentially the same as Mathlover:

enter image description here

By similar triangles:

$$\frac{S}{T} = \frac{B}{H} \hspace{2cm} \frac{A+B}{T} = \frac{A}{H}$$

Then, replacing $T$ and setting $S=1$, we get

$$A = B (A+B)$$

or, setting $C = A+B$ :

$$B = \frac{C}{C+1}$$

So $$C=A+B=\frac{1}{n} \Rightarrow B=\frac{1}{n+1}$$

In general, if you $C=p/q$ then $B=p/(p+q)$. So, for example, if we start from $1$ we get the (original) sequence $(1, \,1/2, \,1/3, \,1/4 \cdots)$, if we start from $2/3$ we get $(2/3,\, 2/5,\, 2/7 \cdots)$, etc.

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    $\begingroup$ we may never need to use a ruler again $\endgroup$ – MacGyver Aug 8 '12 at 4:12

To address all those talking about projections, I can contribute a little here. This also constitutes a geometric proof which relies on projective geometry. Firstly, we can formalize the your method for drawing these lines using projections. Call the bottom of the rectangle M. Call the diagonal from the top left to bottom right L. Call the top right corner O. The projectivity you define is a product of two perspectivities. First is M to L with center of perspectivity O. Then, you go from L to M using the relevant point at infinity (where all the vertical lines meet). This tells us your method relates points on M projectively.

The heart of this proof relies on the fact that given four points, ABCD with distances $\frac{1}{n}, \frac{1}{n+1}, \frac{1}{n+2}, \frac{1}{n+3}$, from some fixed point (such as the bottom right corner), the cross ratio of (AC,BD) is constant: $\frac{-1}{3}$ in fact. You can check it if you want.

Note that I use two well known facts from projective geometry. The proofs for these are on Wikipedia.

1) Projectivites preserve cross ratios

2) Given three collinear points and a cross ratio, there exists a unique point satisfying the cross ratio equation.

Now, we induct. This is less painful than it sounds. You proved the case for $n = 1$. So base case is done.

Suppose that our our claim is true for some fixed n. Then take points points $x_1, x_2, x_3, x_4$ corresponding to $n$. We project all four points, giving us $x_2, x_3, x_4,$ and a new point $x_5$. By 1, cross ratios are invariant so the cross ratio of these four new points is $\frac{-1}{3}$. By 2, we know that this cross ratio corresponds only to points in the $\frac{1}{n}$ configuration described above. So $x_5$ has distance $\frac{1}{n+4}$ from the bottom right corner.

I'm open to critiques here. There is probably some way to condense this, or at least improve readability.


The rectangular construction produces actually a perspective row. It is like looking to a row of trees under perspective view. For geometric explanation view Perspective Fields part 1, pages 20-24 at: http://www.chrisvantienhoven.nl/index.php/component/docman/cat_view/1-downloads.html?Itemid=&orderby=dmdate_published&ascdesc=DESC It also works when taking a parallelogram instead of a rectangle.

  • $\begingroup$ a small correction: it should be Perspective Fields part 2, pages 20-24. $\endgroup$ – Chris Aug 8 '12 at 6:23

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