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Pyramid Maneuver (Posted on 2012-12-06) Difficulty: 3 of 5
A walking track is constructed over the surface of a pyramidal structure having a square base of side 100 feet and base angles of the inclined faces equal to 60 degrees each. The track was laid as the shortest path from one corner of the base to its opposite one.

Is it feasible for a traveller who cannot negotiate a slope stiffer than 1 in 2 to undertake and finish walking uphill along the track?

No Solution Yet Submitted by K Sengupta    
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Solution No trig soluiton | Comment 2 of 4 |
First make a 1/50 scale model, so the base is a 2 unit square.  This won't change the slope.

Call the base ABCD and the vertex X.  If we make the path from A to C by way of segment XB we will need the point on XB that the path crosses.

Call the center of the square Y and the midpoint of AB M.  Triangle XYM is right with angle M= 60 degrees with MY=1 so MX = 2 and XY = √3.  Triangle XMB is a right triangle with MX = 2 and MB = 1 so XB = √5.  Triangle YMB is an isosceles right triangle with MY = 1 so YB = √2.

Consider now isosceles triangle AXB.  The path sought is the altitude from A to XB, call the point it hits XB by H.  Using the area formula twice we have .5(AB)(XM) = .5(XB)(AH) solving gives AH = 4/√5.  Solving right triangle AHB gives HB = 2/√5.

We now know where on XB the path goes and just need the rise and run to find the slope.  Drop a segment straight down from H and it will be perpendicular to YB at a new point, P.  The slope of the path sought is HP/AP.

Triangles HPB and XYB are similar so solving XB/HB = XY/HP = YB/PB gives PB = 2√2/5 and HP = 2√3/5
Subtracting YB - PB = YP = 3√2/5

Right triangle AYP has legs AY = √2 and YP = 3√2/5 so AP = 2√17/5.

Finally the sought slope is HP/AP= (2√3/5)/(2√17/5) = √(3/17)
Cross multiplying to check whether this is less than 1/2 gives √12 < √17 so it is.

The traveler can negotiate this slope.

Edited on December 10, 2012, 1:48 pm
  Posted by Jer on 2012-12-07 11:44:59

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