A Beauty Cold and Austere: Solution

By Ben Monreal and Justin Werfel

Answer: ST ELSEWHERE

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 *a _{8}* = H and

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 *a _{6}* = 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
*a _{11}* = 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 *a _{10}* = 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 *i _{max}* you
choose, you’ll wind up drawing

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 *u _{min}* and

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 *a _{1}* = S and

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 *u _{max}* = 19, so

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:

*a*= 19 = S from di Suvero beam nearer signature_{1}*a*= 20 = T from di Suvero beam further from signature_{2}*a*= 5 = E from one end of Venet spiral_{3}*a*= 12 = L from other end of Venet spiral_{4}*a*= 19 = S from antenna length of Wiley_{5}*a*= 5 = E from selecting the right Melchert ring_{6}*a*= 23 = W from Lewitt line counting_{7}*a*= 8 = H from the x-coordinate and . . ._{8}*a*= 5 = E from the y-coordinate of the pool reflecting in the surface of the Kapoor_{9}*a*= 18 = R from locating the gap in the Graham yin-yang_{10}*a*= 5 = E from finding the treeless link in the Qiang chain_{11}

and the answer is **ST ELSEWHERE**.