- 15 Nov '11 15:53I think I instinctively chose the best position. I can't find a smaller solution than 7.5" and somehow the 45° angle makes sense to me.

If I put the phone fitting to the table sides and have the pencil in 45° I get sqrt(2)*(2+3.5) = 7.78

If I change the angle of the pencil in that position I get even more.

If you have a given fence length, then a square shape gives biggest area. I think it has to do with that. I'm lacking mathematical terms in English to explain it any better.

I'll be impressed if there is a solution which allows a shorter pencil length. - 15 Nov '11 17:31

It's definitely going to be one of those two options, and it's better to have the phone at a 45* angle than flag against the sides of the table.*Originally posted by crazyblue***I think I instinctively chose the best position. I can't find a smaller solution than 7.5" and somehow the 45° angle makes sense to me.**

If I put the phone fitting to the table sides and have the pencil in 45° I get sqrt(2)*(2+3.5) = 7.78

If I change the angle of the pencil in that position I get even more.

If you have a given fence length, then a squa ...[text shortened]... better.

I'll be impressed if there is a solution which allows a shorter pencil length.

I agree w/ the 7.5" answer. - 15 Nov '11 20:28I also agree with it. To prove that it is correct, though, requires two things I think. Showing that the shortest line passes through the x-axis and the y-axis at a 45-degree angle, and that that the part of that line in the 1st quadrant (x > 0, y > 0) is shortest when the line is on one of the long sides of the cell phone. Looks doable though.
- 16 Nov '11 04:16 / 1 editPosition the phone in the first quadrant so that two of its corners are at (a,0) and (0,a) and the line segment from (a,0) to (0,a) is one of the 3.5" sides. Now place the pencil along the opposite side of the (a,0), (0,a) side, such that it intersects the axes at (b,0) and (0,b). Call the pencil side the "outer" side and the opposite side the "inner" side.

Now imagine a line segment from O (the origin) to the outer side, intersecting the outer side at a 90 degree angle at a point P. Clearly it also intersects the inner side at a 90 degree angle as well at some other point Q. Note that the segment from the origin to P is the same length as the segment from P to (b,0) because the triangle defined by those three points is an isosceles right triangle.

We can find a as follows:

a**2 + a**2 = 3.5**2

a = sqrt(3.5**2/2)

Now we can find the length of segment OQ as follows by considering the triangle formed by O, Q, and (a,0). Note that the segment from Q to (a,0) is of length 3.5/2.

(OQ)**2 + (3.5/2)**2 = sqrt( 3.5**2/2 )**2 (that is, a**2 where a is given above)

(OQ)**2 + 3.5**2/4 = 3.5**2/2

(OQ)**2 = 3.5**2/4

OQ = 3.5/2

We know that the length of QP is 2 (since it's parallel to one of the 2-inch sides).

So OP = OQ + QP = 3.5/2 + 2 = 7.5/2

And since that equals half the length of the pencil, the pencil is 7.5" long.

I believe that the shortest hypotenuse for a right triangle is when the legs are equal, so this would be the optimal case, as other positions of the phone would make the legs of the triangle formed by the axes and pencil unequal. The case where the phone's sides run along the axes is a degenerate case of this positioning, and is suboptimal at any rate.

So: How did the above authors get 7.5" so easily? Did I miss a much simpler way of working it out? - 16 Nov '11 10:19 / 1 editI think some may have found it this way..

Let the cell phone be rectangle ABCD where A is on the y-axis and D on the x-axis. A pencil is placed along the long side BC, and the ends of it are at point on the y-axis and F on the x-axis. As the angles FCB and BCF are equal, both 45 degrees, the lines CD and and DF are of equal length, as are EB and AB. The length of the pencil is

EF = EB + BC + CF = AB + BC + CD = 2 AB + BC

so the pencil length equals one long side of the cell phone plus two short sides.

3.5" + 2 x 2.0" = 3.5" + 4.0" = 7.5". - 16 Nov '11 12:00Yeah... I missed seeing that there were a couple of isosceles right triangles with sides along the 2" side of the phone. Meant to update my post last night after realizing that but had trouble accessing the site.

Not sure we've proven satisfactorily that this is an optimal solution, though. I agree that intuitively it seems so... but intuition isn't proof.

Ditto for the truck and desert problem from the other day. No proof that it's optimal. Anyone out there motivated to show that? - 16 Nov '11 14:04Would be nice if the shortest line in the first quadrant that goes through any point (a,b) where a > 0, b > 0 would always make a 45-degree angle with the axises. However.. that is so only for the points on the line y = x ( >0 ) . The equation it makes is

(d(k))^2 = (a - bk)^2 + (bk - a)^2/k^2 = u^2 (1 + k^(-2)) where u = bk - a.

Setting d'(k) = 0 and keeping a and b around as parameters. And then finding the possible values of a and b as the cell phone turns up to 45 degrees from a position where one long side is on the y-axis.. no more is needed as the rest is covered by symmetry.

This starts resembling the basketball problem too.. looks basically clear enough at first glance, but soon enough it generates really unpleasant equations. - 20 Nov '11 20:01 / 1 editassuming the pencil cannot be broken...

Calling the table corner point A, and the ends of the pencil x,y surely the shortest line will form an isosceles right-angled triangle axy. This makes angles Axy and yxA each 45 degrees.

placing the phone inside, with corners hijk, where hi = jk = 2, ij = kh = 3.5 we have formed two more equilateral triangles, xhi, and yij, for each we know one angle (45) and one length, and can work out the hypotenuse with the sine rule.

The pencil length is the sum of the two hypotenuses

= 2/sin(45) + 3.5/sin(45) = sqrt(2).(2+ 3.5) = 5.5 * sqrt(2)