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Blindfolded at the Great Wall of China (Posted on 2010-07-29) Difficulty: 4 of 5
You're blindfolded, disoriented, and standing one mile from the Great Wall of China. How far do you have to walk to guarantee that you will run into the wall?

Assume that the wall is infinitely long and straight.

No Solution Yet Submitted by David Shin    
Rating: 4.5000 (2 votes)

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Solution Not quite as far (Spoiler) | Comment 2 of 10 |
Taking a lead from Dej Marís posting, and assuming that the blindfold will not prevent complicated manoeuvres, I would follow a similar path, but make the first straight section longer than 1 mile before joining the unit circle. At first sight this would seem to add to the journey, but it actually allows the circular arc section to be shorter if the length is optimised. Here are the steps, starting from O.

OP        Straight line, more than a mile, in any direction from O.
PQ        Tangent at Q to the circle with radius 1 mile and centre at O.
            Let A denote the angle POQ.
QR        Circular arc subtending an angle 3*pi/2 - 2A at the centre, O.
RS        Straight line of length 1 mile, tangent to the circle at R.

Note that the arc length has been chosen so that PS will be a tangent to the circle, and therefore any line which meets the circle must also meet the path OPQRS. This is equivalent to saying that the path OPQRS must meet any line that is within 1 mile of O.

    Path length    =  OP + PQ + arc QR + RS

                   L    =  sec A + tan A + (3*pi/2 - 2A) + 1                    (1)

            dL/dA    =  sin A sec2 A  + sec2 A  - 2

For L to be a minimum length:     sin A sec2 A + sec2 A  - 2 = 0

which gives                                sin A + 1 - 2 cos2 A  = 0

                                                sin A + 1 - 2(1 - sin2 A) = 0

                                                2 sin2 A + sin A - 1 = 0

                                                (2 sin A - 1)(sin A + 1) = 0

sin A = 1/2 gives the only acute angled solution:  A = pi/6, and substituting in (1) then gives the minimum length as

                        Lmin  =  2/sqrt(3) + 1/sqrt(3) +3*pi/2 - 2(pi/6) + 1

                        Lmin  =  sqrt(3) + 7*pi/6 + 1        (~=  6.397242 miles)

  Posted by Harry on 2010-07-31 23:03:24
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