# A Great Calculus Problem… that is Powerfully Related to Geometry

I’m sitting in a building at Exeter, digesting lunch and waiting for the next session to begin. I’m at what has so far been a really valuable math teacher conference. What impresses me most, besides the amazing and neverending supply of food that they offer, is the population of teachers that come. Many of the teachers I’m talking with have 20+ years of experience in the classroom.

Over the next few days I’m going to use this blog to talk about some of the tidbits of interesting problems I’ve been presented with, to good resources or programs that I was introduced to, to neat ways to present topics in class, to ideas that I’ve been inspired to think about.

I’m going to start with a nice calculus problem — probably good for a AP Calculus BC class, but there is definitely a way I could show this problem to my non-AP class.

Here’s the problem. You’re in a museum and you’re looking at a painting which is hung above eye level. (There is a specific painting which is hung high in the entrance room at the Brooklyn Museum that I think of with this problem.) You are standing some distance away from it. The question is: what is the largest angle ($\alpha$) that you can get as you walk forwards and backwards? (See diagram below for setup.)

So to be clear, as you move the eyeball forward and backwards along the dashed blue line, what’s the largest angle you can create? Of course if you walk right up to the painting, or far away, the angle is going to decrease to 0. If you can’t see that, look at the diagrams below.

So of course there has to be some perfect distance that will give you the maximal angle. You see where this is going…

Find that maximum angle! (Use the variables in the diagram below.)

Of course this doesn’t have to be a painting. It could be, as the speaker pointed out, an overhead view of a hockey rink, with the painting being the goal, and the eyeball being the player with a puck. Where does the hockey player have the maximum angle to shoot and make it into the goal?

I want you to have the fun of solving it, but the solution I came up with was:

$\alpha=\tan^{-1}(\frac{P}{2\sqrt{Y(P+Y)}})$

However, there is something pretty amazing about this problem, something that is powerfully seen with geometry software like geogebra or geometer’s sketchpad. Check out the sheet I made and see what happens as I bring the person close to the picture and look for the optimal angle? When you look at this, try to see if there is a geometry connection to our solution for the largest angle…

Vodpod videos no longer available.

more about “Geogebra_Picture_Problem“, posted with vodpod

Do you see the geometry connection? The optimal angle exists when the circle created by the top of the picture, the bottom of the picture, and the eyeball is tangent to the line of sight. Now my charge to you — which would be my charge to my students — is to (a) explain in words why this is true and (b) use geometry to calculate this optimal angle. You know, this work is an exercise for the reader. I mean, I’m not going to do everything for you. Sheesh.

1. I remember this problem from Math 1 at RPI. We figured out the best row to sit in the large lecture hall.

2. Matt E says:

I went to that conference in 2000! It was excellent.

I think if you find X in terms of Y1 = Y and Y2 = Y+P you find something else interesting!

3. Guillaume Filteau says:

Dear samjshah,

I tried to solve the problem, however I apparently made a mistake among the way, as the answer I computed does not agree with yours.

First, I build a function f(X) to get angle alpha from horizontal distance X.

Angle alpha is equal to the angle at the eyeball position for the whole triangle, minus the angle of the smaller triangle XY. We can therefore write

f(X) = atan((P+Y)/X) – atan(Y/X)

We need to find the maximum of this function. This is a calculus problem. We therefore find the derivative f'(x):

f'(X) = -(Y+P)/(X^2+(Y+P)^2) + Y/(X^2+ Y^2)

Setting this derivative to 0 to find the maximum, we get two roots:

(Y^2+Y*P)^(1/2),
-(Y^2+Y*P)^(1/2)

Assuming only the first root is of interest, we can therefore evaluate f(X) at X = (Y^2+Y*P)^(1/2).

This gives the maximum angle,

alpha_max = f((Y^2+Y*P)^(1/2))
alpha_max = atan((Y+P)/(Y^2+Y*P)^(1/2))-atan(Y/(Y^2+Y*P)^(1/2))

Best,
Guillaume

1. Dear Guillaume,

You may have already settled this by now but actually your solution and samjshah’s agree perfectly! The identity

tan (a-b) = (tan a – tan b) / (1 + tan a tan b)

allows you to simplify your solution into the form samjshah gives.

All the best, Ben

1. It is not really the same. I think there is a typo on samsjah solution.
The numerator should be Y instead of P, that what I was trying to say. But there was some error on my latex.

4. Usually posed as the best seat in the movie theater. (although a bit less accurately. The floor in the theater will slope back. In fact, that gets hard, if you do it right)

Jonathan

5. I found $\alpha=\tan^{-1}\left(\frac{Y}{2\sqrt{(P+Y)Y}\right)$ which is equivalent to Guillaume solution above.

6. Great problem, samjshah! Thanks for calling our attention to the geometry connection. Sometimes I feel like I could spend a whole lifetime just thinking about circles and triangles, and seeing this connection makes me feel that way.

7. Hello Sam! (Hope I got derived your name correctly from ‘samjshah’…)

How wonderful to come across this lovely problem (I’d searched Google for “great calculus problem”), and to come across a fellow passionate mathematics teacher in New York City!

The geometry connection you pointed out here really ties together the theorems about arc length and angle measure in a way I’d never seen before. Let me see if I can explain what I can mean in text…

Imagine an arbitrary circle is drawn, and a line drawn through it, subtending a (minor) arc of length L. Now let us consider a point p on the opposite side of the line from the arc. Extend rays from this point, subtending the arc, and forming an angle t.

From the standard theorems, we know:

• if p is on the circle, then t = L/2
• if p is inside the circle, then t = (L+?)/2 for some ?
• if p is outside the circle, then t = (L-?)/2 for some ?

Hence p is largest inside the circle, smaller on the circle, and smallest outside the circle. What an elegant little rule of thumb! (Also it might be interesting to consider the equivalence relation on the half-plane given by p ≈ q iff t(p) = t(q)…)

* * *

In any case, your ‘hint’ above made me wonder, just how does one construct the circle through the top and bottom of the painting, tangent to the line of sight? I decided to explore a strategy similar to how one constructs a circle through three arbitrary non-collinear points: namely, by drawing perpendicular bisectors.

First, I drew the perpendicular bisector through the painting (parallel to the line of sight). All points on this line are equidistant from the top and bottom of the painting. But the distance from this line to the line of sight is known: P/2 + Y. Thus we should construct a circle centered at either the top or bottom of the painting with radius P/2 + Y, and find where it intersects the perpendicular bisector. This point of intersection is the center of a circle of radius P/2 + Y, which goes through both the top and bottom of the painting, and which is tangent to the line of sight.

If the circle is drawn in this way, one sees a right triangle formed by the center of the circle and half of the painting. The angle at the center of the circle is Alpha; the opposite side to this angle has length P/2, and the hypotenuse has radius length P/2 + Y. Thus:

Alpha = arcsin( (P/2) / (P/2 + Y) ) ,

which is probably a spot simpler than your formula using the arctan. (All of a sudden it’s clear to mean where the terminology ‘arcsin’ comes from!)

Using the Pythagorean theorem, one easily computes that X, the distance from the eye to the wall, equals sqrt(PY + Y^2). With this value known, our two formulas for Alpha are easily seen to be equivalent. Of course, one might consider it part of the problem to find X!

* * *

Thanks for all the mental stimulation, two and a half years after your original post. But then again, math is timeless, right? :)

+j

8. (Apologies — posting this so I can get follow-up comments via email… should there be any…)