Geometry

Math Photo: Riemann Building

Something compelled me to photograph this building as I was strolling around one steamy afternoon.

And then I realized why: it reminded me of a 3-dimensional Riemann sum!

I did not attempt to estimate its volume.

By patrick honner, 14 years7 years ago

Alexander Bogomolny (@CutTheKnotMath) recently posted an interesting interactive diagram entitled “Simultaneous Diameters in Concurrent Circles“. The diagram demonstrates that if three circles intersect in a single point and any two centers are collinear with common chords, then the third center must also be collinear with a common chord.

Inspired by this demonstration, I offer a proof of this fact from a different perspective.

Claim: If two circles centered at A and B intersect at X and Y, and AY intersects circle B at M and BY intersects circle A at N, then the center of the unique circle that passes through M, N, and Y lies on the line through XY.

Proof

We start with the following diagram.

Since NY is a chord of circle A, we know A lies on the perpendicular bisector of NY. The same argument goes for B and chord MY. Construct the perpendicular bisectors and call their point of intersection C. (Note that as long as X and Y are distinct, these perpendicular bisectors can not be parallel).

Notice immediately that C is the circumcenter of triangle NYM, as it is the intersection of the perpendicular bisectors of NY and MY. Thus, C is the center of the circle that passes through N, Y, and M.

Notice also that AM is perpendicular to BC and BN is perpendicular to AC; thus Y is the orthocenter of triangle ABC! (The orthocenter is the point of concurrency of the altitudes of a triangle). Therefore Y lies on the altitude to AB that passes through C. That is to say, CY is perpendicular to AB. But XY is also perpendicular to AB (common chords are perpendicular to lines through centers), so X, Y, and C must be collinear.

Therefore, C is both the center of the circle through N, Y, and M, and C lies on line XY.

Thanks to @CutTheKnotMath for the inspiration! It’s fun to see the orthocenter make an appearance here, as this point of concurrency seems to have fewer applications than its more famous relatives centroid, circumcenter, and incenter. An even more exciting role is played by the orthocenter in this problem.

By patrick honner, 14 years7 years ago

Geometry Photography

Math Photo: Triangular Tower Top

Another scene from ever-under-development Brooklyn.

By patrick honner, 14 years8 years ago

Challenge 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, 14 years8 years ago

Challenge Geometry Teaching

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!