1. The Superman Problem

I am having a difficult time understanding a problem...
The problem, in essence is the Pythagorean Theorem, and I see how it IS Pythagorean Theorem, but I want to go deeper than the answer.

The problem is this:
Superman starts at home, and decides to exercise, by running. He runs exactly one mile north, decides it isn't enough, and runs exactly one mile east. He then runs two miles south, two miles west, three miles north, three miles east, four miles south, four miles west, and so on until he has run a distance of 11,001 miles, at which point he rests, and flies directly home. What is the distance he had to fly to get home?

Now, the problem itself calls upon the 11,001 miles distance. I have figured this out, but my process was lengthy, as I actually got a large paper and drew it out, then calculated. That was a huge pain in the butt, and I'd rather have a set equation, or a simple process (i.e. no more than 5 minutes), but I don't quite comprehend how to simplify it.

So, my questions, in essence, are:
1. In the exact same problem, what would be the distance he had to fly home for x miles?
2. Why is this?

Thank you in advance for any time spent on this!

2. Re: The Superman Problem

Possible method:
-start at origin (0,0); after 1N,1E: at (1,1) ; 2S,2W: (-1,-1) ; 3N,3E: (2,2) ; 4S,4W: (-2,-2) ....
-calculate position (x,y) that Clarke Kent ends at
-distance = SQRT(x^2 + y^2)

Take 79 miles travelled: position = (-4,-4) after 72, so (-4,3) after 79.
So distance = SQRT[(-4)^2 + 3^2] = 5.

Looks promising...I'll try that tomorrow; you try too!...g'nite!

3. Re: The Superman Problem

Thank you, but how exactly would I determine the point of any given distance? When getting into hundreds of thousands of miles run, then things actually get complex, and counting where exactly 500,000 miles is on the outward spiral would be nothing short of a Herculean Task... I admit, a graph with a house as the point of origin is genius, but how do I determine where he has to start from to fly home? (i.e. the (1,1) and the (-3, -4) on the graph to plug into the equation)

4. Re: The Superman Problem

billnyethphyicsguy,

If you scrol to the bottom of this front page forum you will find two more sister sites, one is a physics forum, should you not post there?

but how do I determine where he has to start from to fly home? (i.e. the (1,1) and the (-3, -4) on the graph to plug into the equation)

Should you not consider the point that Superman stopped at last?

If he flys back, as the crow flys, can you not work it out using Vectors?

David

5. Re: The Superman Problem

Hello, billnyethephysicsguy!

Superman starts at home and decides to exercise by running.
He runs exactly 1e mile north, decides it isn't enough, and runs exactly 1 mile east.
He then runs 2 miles south, 2 miles west, 3 miles north, 3 miles east,
4 miles south, 4 miles west, and so on until he has run a distance of 11,001 miles,
at which point he rests and flies directly home.
What is the distance he had to fly to get home?

My approach is similar to Wilmer's.

Superman starts at the origin (0,0).

After 4 stages of his journey, he is at (-1, -1).

After 8 stages, he is at (-2, -2).

After 12 stages, he is at (-3, -3).

In general, after $\displaystyle 4n$ stages, he is at $\displaystyle (\text{-}n,\,\text{-}n)$

We have: .$\displaystyle (1+1) + (2+2) + (3+3) + \hdots + (m+m) + r \;=\;11,\!001$

. . We find that: .$\displaystyle m = 104,\:r = 81$

That is, he makes 208 stages of his run, then runs another 81 miles.

After $\displaystyle 4n = 208$ stages, he is at $\displaystyle (\text{-}52,\,\text{-}52)$

Then he runs 81 miles north to $\displaystyle (\text{-}52,\,29)$

His distance is: .$\displaystyle \sqrt{29^2+52^2} \:=\:\sqrt{3545} \:\approx\:59.54$ miles.

6. Re: The Superman Problem

Finding distance from home after running 11,001 miles is NOT the answer here (just an example);
the question is: what is distance from home after running k miles?

Soroban's m=104 is correct; however, its "calculation" is (use "sum of evens" formula):
m(m + 1) = k
m^2 + m - k = 0
m^2 + m - 11001 = 0
solve to get m = ~104.38; take integer portion: 104

So this would be 1st step in calculating distance from home.

Similarly with my example k = 79:
m^2 + m - 79 = 0
m = ~8.41 = 8
So distance ran at that point is 8(9) = 72 : at (-4,-4)
I used 79 because it results in a 3-4-5 right triangle, 5 being distance from home.
Something "simple" to experiment with; no need to concern ourselves with "huge" mileages:
what applies to a simple case will apply to 10^(infinity) miles!

Perhaps Soroban (fresly rested!) can wrap that up?!
As is, I see the main problem as being: where is SuperMan? (x,y) or (x,-y) or (-x,y) or (-x,-y) ?
At (x,x), he's headed South; at (x,-y), West; at (-x,-y), North; at (-x,y), still North (1 mile before turning East).

7. Re: The Superman Problem

Got it...with some help...

"distance from home" after running k miles =

SQRT{j^2 - [j - k MOD(2j)]^2} where j = INT{[INT(SQRT(k)) - 1] / 2} + 1

if k = 11001, then distance = 59.539902586.... or 59.54 as per Sir Soroban...