Geometry

A Curious Geometric Limit

Inspired by Alexander Bogomolny (@CutTheKnotMath) and his curious geometric limit, I offer this trigonometric approach.

Consider the following diagram in which we have circle X of radius 3 centered at (3,0) and circle O of radius k centered at the origin. Call the intersection of the these circles in the first quadrant B. Let A be the intersection of circle O with the y-axis, and extend line AB until it intersects the x-axis at E. Our goal is to show that $\lim_{kto}OE=12$ .

Consider the diagram below.

Let $\angle{OXB} = \theta$ . Some simple angle chasing, using properties of isosceles triangles and right triangles, gives us that $\angle{OEA} = \frac{\theta}{4}$ .

Since AO = k by definition, we have $tan(\frac{\theta}{4}) = \frac{k}{OE}$ , and so OE = $\frac{k}{tan(\frac{\theta}{4}) }$ .

Since OB = k is the base of isosceles triangle OXB, we can drop an altitude from X to OB and find that $sin(\frac{\theta}{2}) = \frac{\frac{k}{2}}{BX} = \frac{\frac{k}{2}}{3}$ . This gives us $k = 6sin(\frac{\theta}{2})$ .

So OE = $\frac{k}{tan(\frac{\theta}{4}) }$ = $\frac{6sin(\frac{\theta}{2})}{tan(\frac{\theta}{4}) }$ . By the double angle formula for sine, we have

$sin(\frac{\theta}{2}) = 2sin(\frac{\theta}{4})cos(\frac{\theta}{4})$ ,

Making this substitution, and by writing tangent as the quotient of sine and cosine, we get

OE = $\frac{6sin(\frac{\theta}{2})}{tan(\frac{\theta}{4}) } = \frac{12sin(\frac{\theta}{4}) cos((\frac{\theta}{4}))}{\frac {sin(\frac{\theta}{4})}{cos(\frac{\theta}{4})} } = 12 {(cos(\frac{\theta}{4}))}^2$ .

Finally, since $\theta\to 0$ as $k \to 0$ , we have

$\lim_{k\to 0}OE = \lim_{\theta \to 0}12 {(cos(\frac{\theta}{4}))}^2 = 12\lim_{\theta \to 0} {(cos(\frac{\theta}{4}))}^2=12$ .

By patrick honner, 15 yearsMay 15, 2011 ago

One-Cut Challenge: Quadrilaterals

Here’s another introduction to the one-cut challenge, this time with quadrilaterals. This is another activity for students of all ages from my Fun with Folding series, and you can try this out even if you haven’t fully explored the one-cut challenge with triangles.

Recall that in the one-cut challenge, your task is to produce the given shape as a cut-out using only straight folds and asingle straight cut.

Let’s start with a square. One simple approach is to fold the square in half, then fold the top and bottom halves up along the vertical side.

This same approach also works with rectangles.

A second approach with the square involves fewer folds. First, fold along the main diagonal, then go from there.

In addition to being a more efficient solution, and raising the general question “What is the minimum number of folds needed to produce the shape?”, this approach produces some confusing results when applied to a rectangle!

Another easy quadrilateral to work with is the isosceles trapezoid.

But watch out: an arbitrary quadrilateral poses quite a challenge!

Happy folding!

Have more Fun With Folding!

By patrick honner, 15 yearsMay 10, 2011 ago

Challenge Geometry Resources Teaching

One-Cut Challenge: Triangles

Here’s an introduction to the one-cut challenge using triangles, from my Fun with Folding series, suitable for students of all ages (including teachers!). This is a rich, compelling problem that touches on a lot of sophisticated ideas in geometry, but is simple enough to start playing around with right away.

The one-cut challenge is as follows: given a shape made up of connected straight line segments (i.e. a polygonal chain), can you produce the shape as a cut-out using only straight folds and a single straight cut?

A good place to start is with an equilateral triangle. This is a fairly easy problem to solve, given the inherent symmetry in the figure. Fold across any line of symmetry to produce a new figure that looks like two line segments meeting at an angle. Fold those together along their vertex, and cut!

The next step is trying this with an isosceles triangle, whose single line of symmetry still allows this approach to work.

Now the kicker: try it out on a scalene triangle! No more lines of symmetry, and all of the sudden this is a pretty challenging problem!

Happy folding!

Have more Fun With Folding!

By patrick honner, 15 yearsMay 9, 2011 ago

Appreciation Geometry Photography

Math Photo: Triangles on the Beach

While on the beach one day, I found a triangular rock. So I started on some beach relief.

Then I iterated.

And I iterated again.

And I continued to iterate, until the whole beach was one big Sierpinski Triangle!

By patrick honner, 15 yearsMay 8, 2011 ago

Challenge Geometry

2011 AIME A #8: Triangular Tables

I was inspired to have some more fun with folding by a question from this year’s American Invitational Mathematics Examination (AIME) that turned triangles into tables and asked “How high can the table go?”. (You can find the question here).

Investigating the problem seemed like more fun than solving it, so I cut out a triangle from some foam board and scored lines near the vertices.

Then I folded the corners and made the following table with an irregular hexagonal top!

I made a few, to see what kinds of heights I could get.

There are so many fun questions to explore here! What comes to mind?

A Curious Geometric Limit

One-Cut Challenge: Quadrilaterals

One-Cut Challenge: Triangles

Math Photo: Triangles on the Beach

2011 AIME A #8: Triangular Tables

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