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.
Through Math for America, I am part of an on-going collaboration with the New York Times Learning Network. My latest contribution, a Test Yourself quiz-question, can be found here:
https://learning.blogs.nytimes.com/2011/06/08/test-yourself-math-june-8-2011/
This question is based on rising gas prices and their effects on family budgets and consumer behavior.
Packing things poses some surprisingly complicated mathematical problems. Like some of the nastiest problems in math, there isn’t necessarily a standard algorithm you can use to figure out how to pack things efficiently.
I tried to put five hard-boiled eggs in a cylindrical tupperware container.
But the top wouldn’t quite close.
Perturbed, I squeezed the eggs into the cylinder in a different manner. This took some doing, as they kept falling over.
But they fit! And amazingly, they didn’t even touch the bottom!
This is a nice chart from the Chronicle of Higher Education that compares average salaries by college major.
http://chronicle.com/article/Median-Earnings-by-Major-and/127604/
Each vertical bar represents a high-level category (like Social Science or Engineering) and by scrolling over each bar you see a detailed breakdown of salary by specific major (like Poly Sci or Architecture).
Not surprisingly, math majors do pretty well, and the more applied, the better!
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.