The Riddler – Can You Solve the Puzzle of the Picky Eater

Took anoth­er one week hia­tus from The Rid­dler, but am back to this one’s!

The Rid­dler is a week­ly puz­zle series from Oliv­er Roed­er at FiveThir­tyEight that usu­al­ly involves some com­bi­na­tion of math, log­ic and probability.

This weeks FiveThir­tyEight Rid­dler, Can You Solve the Puz­zle of the Picky Eater is:

Every morn­ing, before head­ing to work, you make a sand­wich for lunch using per­fect­ly square bread. But you hate the crust. You hate the crust so much that you’ll only eat the por­tion of the sand­wich that is clos­er to its cen­ter than to its edges so that you don’t run the risk of acci­den­tal­ly bit­ing down on that charred, stiff perime­ter. How much of the sand­wich will you eat?

Extra cred­it: What if the bread were anoth­er shape — tri­an­gu­lar, hexag­o­nal, octag­o­nal, etc.? What’s the most effi­cient bread shape for a crust-hater like you?

So let’s start with a 2 x 2 square, as pic­ture below

To sim­pli­fy things we’re only going to deal with the first quad­rant, specif­i­cal­ly half of the first quad­rant (see shad­ed por­tion. To find the area clos­er of the shad­ed tri­an­gle that’s clos­er to the cen­ter than the edge, we need to find a func­tion f(x) that rep­re­sents the all the points that are equidis­tant from the edge to the cen­ter. The dis­tance of that equa­tion from the cen­ter is defined by:  sqrt(x² + f(x)²)

The dis­tance from that equa­tion to the edge in the shad­ed por­tion is defined by: 1‑x. Set the two equal to each oth­er, and solv­ing for f(x) pro­duces: f(x) = sqrt(1–2x). So our plot now looks like:

The shad­ed area can be split up into two dis­tinct areas. The area under g(x) from 0 to the inter­sec­tion of g(x)/f(x) (which we’ll call c). And the area under f(x) from c to 0.5.

g(x) is defined by that angle theta, which in the case of a square shape eval­u­ates to: g(x) = x. We set g(x) = f(x), and solv­ing for x will pro­vide c. So we need to solve x in:r cotangent(θ)*x = sqrt(1–2x). That yields x = c  = 2^1/2 — 1.

To solve for the area under­neath our curve we eval­u­ate: [Inte­gral g(x) from 0 to c] + [ [Inte­gral f(x) from c to 0.5]. Since g(x) is a tri­an­gle with a base and height of c and g© respec­tive­ly , the [Inte­gral g(x) from 0 to c] term is 0.5*c²*cotangent(θ). We’ve sim­pli­fied our equa­tion now to: [0.5*c²*cotangent(θ)] + [ [Inte­gral f(x) from c to 0,5]. Using c  = 2^1/2 — 1 we solve and get the area of the shad­ed region to be: 0.1094. The base and height of the tri­an­gle in the first quad­rant (shad­ed region in the 2nd pic­ture) is 1 and g(1) respec­tive­ly. So the area of that tri­an­gle com­putes to 1*g(1)*0.5  = 0.5. So the pro­por­tion of the sand­wich eat­en is: 0.1094/0.5 = .2190. 

The answer to this week’s Rid­dler is “You will eat 21.9% of the sandwich”.

Extra Cred­it

So now for the extra cred­it, we look at oth­er shapes. Let’s take for exam­ple a hexa­gon, as shown in the graph below.

Just as before we’re only con­cerned with the first quad­rant, specif­i­cal­ly the shad­ed tri­an­gle in the first quadrant.

You’ll see that a lot of the math we did still applies. f(x) remains sqrt(1–2x). The only vari­able that’s changed is theta. Chang­ing theta changes the val­ue of g(x), which in turn changes the val­ue of area of the shad­ed tri­an­gle and val­ue of c, but since those are all depen­dent on theta, the only term we need to be con­cerned with is theta. Theta in the case of the square was 45o, and in the case of the hexa­gon is 60o. The gen­er­al equa­tion for theta, in a reg­u­lar poly­gon with n sides can be defined by θ = [(n‑2)*(π/2)]/[n*2].  So in the gen­er­al case of a poly­gon with n sides our steps are:

1. Find Theta via  θ = [(n‑2)*(π/2)]/[n*2]
2. g(x) is now defined by g(x) = cotan­gent(θ)*x
3. f(x) remains f(x) = (1–2x)^1/2 as per the equa­tion solved earlier
4. Solve f(x) = g(x) for the inter­sec­tion point of the two curves which we’ll define as c
5. Inte­grate g(x) from 0 to c, which can be sim­pli­fied to  [0.5*c²*cotangent(θ)]
6. Inte­grate f(x) from c to 0.5
7. Add up the results of steps 5 & 6 to find the area of the sand­wich eat­en for that sub-sec­tion in 1st quadrant
8. Cal­cu­late the area of the entire sub-sec­tion in the 1st quad­rant, defined by  1*g(1)*0.5
9. Divide the result of Step 7 by Step 8 to find the pro­por­tion of sand­wich eat­en by the picky eater.

Rather than doing this all by hand, I wrote up a script in R for n in 3:50, which can be found here on my Github. That gen­er­at­ed the fol­low­ing results:

We can see as the num­ber of edges (n) gets larg­er, the pro­por­tion of sand­wich eat­en approach­es 0.25. A poly­gon with infi­nite edges is a circle.

For a cir­cle with radius r, the area that is equidis­tant to the cen­ter as it is to the edge is defined by the con­cen­tric cir­cle with radius r/2. In that case the pro­por­tion eat­en is: [(r/2)²*π]/[r²*π] which sim­pli­fies to 0.25. In the con­text of the graph this makes sense.

We con­clude that the most effi­cient shape is a cir­cle, with 25% of the sand­wich being eaten.