First, the title:
“Mathematics, rightly viewed, possesses not only truth, but supreme beauty—a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show.” –Bertrand Russell
The expressions given all define parametric curves corresponding to the shapes of sculptures and art in the MIT Public Art Collection.
Type or copy the equations into your favorite plotting routines. Mathematica will import MathML without much trouble, or you can copy-paste into Matlab, Python, C, etc., with only light editing. Each set of equations is a 2D or 3D vector (xy or xyz) which you can plot over a parameter range. If there’s one parameter, u, you’ll get a line. If there are two (u,v) you’ll get a surface. When there’s an integer parameter like i, you should draw a separate line for each value of i. Look at the plots.
(There are some you can’t recognize yet; you don’t know all the plotting parameters. We’ll get back to these later.)
You should recognize some familiar MIT public-art works, and you can click through the MIT Public Art Collection on the web to identify some more obscure ones.
Now, look at the list of “parameter values you like.” When you made a plot over a parameter range, you were probably thinking about the limits. But any specific value of the parameter, between the limits, identifies a particular place along the curve. In this puzzle, you’ll find that every “parameter value you like” is a number between 1 and 26. Convert these to letters with A = 1, etc., to get the answer ST ELSEWHERE.
Below we show all of the equations, and parametric plots with the appropriate readout-value highlighted.
(This solution file is available as a Mathematica notebook.)
The equation should generate a rectangular segment of a 2D parabola, resembling the reflective Anish Kapoor work in the Stata east lobby. Near the easternmost door of the lobby is a cluster of fire-pulls and whatnot as described. If you stand with your head in this cluster and look at the Kapoor, you’ll see a reflecting pool—i.e., the reflection of a pool in one of the large-format photographs that line the Stata first-floor street. The reflection is in the right center of the rectangle, i.e., the coordinate represented by u = 8, v = 5, so a8 = H and a9 = E.
Here is where you have to stand.
And here is what the Kapoor looks like when you stand there. The blue patch is the reflecting pool.
There is an electrical outlet under the next-to-last ring, which is the one plotted by the i = 5 curve. Therefore a6 = E.
The ninth ring, counting west to east, is uniquely the treeless ring flanked by treeful ones. That’s the ring you plot when i = 5, so a11 = E.
This parameterization plots a solid yin-yang about the shape and height of Graham’s pavilion inside Simmons Hall. The real thing has a gap in the glass around u = 18, so a10 = R.
This parameterization is a pseudorandom-length collection of radial lines. (The pseudorandomness comes from the large numbers in the trig function arguments.) And it’s the first time you're being asked to plot something over an unknown range. “Try different values of imax and see what it looks like” is a fine approach, especially if you’ve realized that the parameter should fall between 1 and 26. But you can also inspect the expression. Whatever value of imax you choose, you’ll wind up drawing imax + 7 lines. Choosing imax = 23 will get you 30 lines, the number in the real Sol Lewitt wall drawing, so a7 = W.
You’re given the complete parameterization and endpoints of one of the two spirally elements of Venet’s sculpture. Go ahead and plot that one and try to recognize it. To get the other element to be the right length, fiddle with the endpoints umin and umax until you have something that looks like the sculpture. Setting umin too low will make the second spiral crash into the first one; setting umax too high will give it a slinky tail stretching far to one side. To get the real thing you have to plot from u = 5 to u = 12, so a3 = E and a4 = L.
Here we’ve parameterized a collection of straight lines (fairly accurately!) and a twiddly curve (less accurately, but you’ve done more than enough typing by now) that look like Mark di Suvero’s red welded beam assembly. Di Suvero’s signature is plasma-cut into the foot of the eastern leg of the triangle, something you may only be able to identify by going there and searching. That’s given by the third line parameterized here. Two beams meet that one: the nearer one about 3/4 of the way along, which you can find by trial and error is u = 19. and the further one at the endpoint which is given as u = 20. (In the example plot below those are shown in cyan and red segments respectively.) Therefore a1 = S and a2 = T.
The antennae of this butterfly, part of a collection visible on the back of Building 44, have curls which develop very rapidly with u. The antennas are obtained at the correct length by plotting with umax = 19, so a5 = S. It’s a fairly steep function. Below, I provide examples of the butterfly you’ll draw if you incorrectly guess umax = 18 or umax = 20.
If you just typed all that math into plotting software by hand . . . good for you! You are glad Louise Nevelson’s Transparent Horizons wasn’t here, aren’t you?
Put all of the extracted answers together to get:
and the answer is ST ELSEWHERE.