A games program was inspired by the exploits of "Indiana Jones", naturally modifications were made.

In one set of three events a pair of contestants had to retrieve a "gem".

Within the arena the pair would see a deep circular pit, the only aids allowed to enable their task, their 'prize' and any transformations made to the pit for each scenario.

Scenarios of the three distinct events:
In each case a pit of radius 6 feet has the "gem" dangling over its centre at just above head height as the contestants stand at the edge of the pit. Lightweight, anti-slip aluminium planks are presented as their only aids.
1. The top of a column, 1 foot in diameter, is horizontal to the pit opening and in the very centre of the pit. Two planks, length of 5 feet, are available.
2. The height of the column (in #1) is raised 2 feet above the pit opening. Just one 6 foot plank is available.
3. Only three planks are available and they are 8 feet long, there is no column!

[Bonus] Now it would seem that five planks of about 5 feet in length might also accomplish the task as at #3. How might those be arranged?
Note:
1. Jumping is disallowed and not necessary,
2. Measurements
- for a metric reader's perspectivity multiply each foot by 30.5 cm,
- planks are a little longer than one might expect (but that would be necessary in a real situation), and
3. Disregard plank thickness as it has no bearing within the spirit of the exercise.
(4. In a "real world" situation the partner would only be required for #3, thus two contestants.)

(In reply to re(3): weaving - good thought, but ... by brianjn)

<o:p>Since a weaving solution is allowed, I suggest the following:</o:p>

<o:p>Denote the planks AB, CD etc. Their respective midpoints are M1,M2,etc. The centre of the pit is O. AB is placed in a chord on the circumference of the pit touching it at points A and B. CD is placed in a chord touching B at C and with its other end D further along the circumference of the pit. M1 and M2 are therefore about 1/10 of the radius from the edge (in fact, about 0.55 feet). Lay EF across AB and CD such that OM3B is a straight line. What is the closest M3 can get to O while EF is still touching planks AB and CD? It might be easier to consider the situation where EF is placed tangentially to the edge of the pit, touching at M3, forming the right triangle M3(=B and C):M1:X, where M1XM3 is the right angle.</o:p>

<o:p>Using a =rcrdè, we obtain 5=6è: =5/6 radians as the value of the intercepted arc, and so 5/12 radians or 23.87 degrees as the angle XM3M1. Since we know by construction that line XM1 is 2.5. we can now compute length M1X to be about 1.10645, leaving a distance to O of 4.9, which is less than 5. EF is placed over AB and CD so that M3 is 4.9 from O. GH is then placed crossing EF perpendicularly at M3 with G at O and H projecting 0.1 feet further from O than M3. We call this projection the ’end bit’. Lastly, IJ is placed under AB, over the end bit and under CD so that M5 rests above some point on the end bit between M3 and H, thereby pinning GH in place.  A contestant then walks over AB, EF and IJ to collect the gem.
</o:p>

Posted by broll on 2010-04-15 05:00:12