<p>I think I posted a response with a link to a graph but can’t remember where it is hidden. The above link should, however, point you to the full answer. </p>
<p>**A file of soldiers marching in a straight line one behind the other is one kilometer long. An inspecting officer starts at the rear, moves forward at a constant speed until he reaches the front, then turns aroudn and travels at the same speed until he reaches the last man in the year. by this time, the column, marching at a constant speed, has moved one kilometer forward so that the last soldier is now in the position the front soldier was when the whole movement started. How far did the inspecting officer travel? **</p>
<p>This was also posted in 2003 on the older CC forum.</p>
<p>I don’t think the answer can uniquely be determined. Suppose the officer marches k times as fast, where k > 1 (actually, k > 2). Then the officer marches k km in total, because the soldiers marched 1 km total. But k is arbitrary (at least I believe it is…too lazy to do algebra right now).</p>
<p>Okay, so suppose the officer travels k times as fast. Suppose the soldiers travel x km when the officer arrives. Then the officer travels kx km, and </p>
<p>kx = 1+x</p>
<p>Coming back, the soldiers travel 1-x km in the same time the officer travels x km. Therefore,</p>
<p>x = k(1-x) → x = k - kx</p>
<p>So we have the system of equations</p>
<p>kx = 1+x
k-kx = x</p>
<p>Adding, we get k = 1 + 2x. Substitute to get</p>
<p>(1+2x)x = 1+x → 2x^2 = 1 → x = sqrt(2)/2. Substitute into k = 1 + 2x to get k = 1 + sqrt(2)</p>
<p>Yay! Yes, k would therefore be the distance the officer walked. Nice problem.</p>
<p>Here’s a fun one:</p>
<p>You are standing in the middle of a circle. A dog runs along the circumference of the circle and it wishes to chase/eat you. The dog runs four times as fast as you do. You can only run along a 2-dimensional plane, no bridges or tunnels or anything. Given that the dog can use whatever strategy it wishes to chase you, how do you escape the circle?</p>
<p>The dog runs the entire circumference of 3.14 x 2R. If you place yourself at a point that is opposite to the dog on a line crossing the center but about 3/4 of the distance of center to the circumference, you can make it. From the center to this point you can outrun the dog to keep him in the opposite position. </p>
<p>The dog needs 3.14 to run to the escape point and you need only 3.</p>
<p>I think I get your solution, but note that the dog can use whichever strategy it chooses. So you have to make sure your solution can counteract all dog strategies.</p>
<p>@DrSteve thanks for bringing that up. Apparently the most recent problem on that thread (product of n consecutive integers is always divisible by n!) has a really easy one-line solution.</p>
<p>Here’s one from the time DS was doing AMCs and ARMLs:
You have a rectangle made by m parallel streets and n perpendicular avenues. You can hence go from one corner to the diagonally opposite one in m+n blocks. How many different m+n block long paths can you do this in?</p>
<p>The dog can only run along the circle, and cannot enter. What the dog can do is to run or stand still. Even when the dog runs, he cannot preclude the person in the circle to reach a point that will allow that person to only have a 3/4 of the radius to cover to the edge of the circle. </p>
<p>After a while, even if the dog tries to keep the distance shorter than a full 1/2 circle, he will give up, because the person cannot be precluded to find that “sweet” spot as it takes him less time to run within an area that is 1/4 from the center to the edge of the circle. You can imagine the person running in an extending spiral and the dog running the full circle. The dog cannot keep up with the person in the middle. The person, on the other hand, can reach the point that is only 3/4 removed from the circle edge and keep the dog at the most opposite point, and thus force the dog to cover that full 1/2 circle. The best the dog can do is in 3.14x time. The person only needs a 3x time. </p>
<p>The theory is easy to show with a graph of the original circle plus a smaller one with a radius of 1/4 the orginal. A lot harder to explain with words. It is, however, all based on forcing the dog to have to run 1/2 of the circle to reach the escape point of the person. Something that is easily achieved by the person, as long as he maximized his “bonus” headstart which is 1/4 of the radius.</p>
<p>xiggi, your strategy is absolutely right. At any circle within a radius of r/4, if you run along the circumference, your angular velocity is higher than the max that Fido can manage. So if you stagger up to just under the r/4 mark, you can always run along the circumference there, outrunning Fido to get to the opposite side. At this point you head directly along the radius to the outer circle - the remaining 3/4 r. Fido will have to do pi r to get to you, but since 3 < pi, you reach the edge first even if he runs 4 times faster.</p>
<p>A more complex problem would be finding the optimal path.</p>
<p>@Dad<em>of</em>3; Your parallel streets and avenues problem is a great old puzzle from combinatorics. I’ll post a hint and also a variation…</p>
<p>The hint first: suppose you are starting at the southwest corner of the grid. You have to take a certain number of steps to the north and also a certain number of steps to the east. Counting the ways is like counting the ways to arrange a string of letters where the string consists of only n’s and e’s.</p>
<p>Also, did you mean to say that there are m+n and n+1 streets? Because if the grid is m by n, the shortest path would be (n-1) + (m-1). For example, if you are on first street and the town has streets that number up to 5th street, your longest walk is 4 blocks, not 5. [I used this problem in a sample lesson I taught nearly 20 years ago and made exactly that mistake as I posed the question…]</p>
<p>And for anyone who wants a variation: add diagonal lines to the grid so that at every corner it is possible to walk north, east or northeast. Now how many paths are there?</p>
<p>BTW, I think this problem may have been discussed in the Scientific American book of puzzles and brain teasers. Don’t remember the exact name of those books but I think that they are still in print. Anyone who is enjoying this thread would love those books.</p>
<p>I was not precise - meant m blocks by n blocks. Had drilled into my son all the errors we make with boundary conditions and here I am doing the same.</p>
<p>@xiggi, Sorry I think I mis-interpreted your solution. Yes, if I run around the circle of radius r/4 - a small number, then my angular velocity is faster than the dog’s.</p>
<p>It’s really easy using trig. Let angle CDE = θ. By law of sines on triangles DEC and DEB respectively,</p>
<p>(sin θ)/EC = (sin 10)/DE</p>
<p>(sin (θ+30))/BE = (sin 20)/DE</p>
<p>Dividing second equation by first,</p>
<p>(EC sin (θ+30))/(BE sin θ) = (sin 20)/(sin 10)</p>
<p>And we know that, from triangle EBC, that (sin 80)/BE = (sin 60)/EC, so EC/BE = (sin 60)/(sin 80). Plug this into the previous equation and solve for θ.</p>
<p>Gee, I did this about nine years ago, and I might be rusty. This is what I think I did. I would not be surprised that there might be a shorter way, but it is what it is.</p>
<p>1000/y = 1000/(x-y) + 1000/(x+y)</p>
<p>Replace 1000 by 1 (could have done that from the start)
1/y = 1/(x-y) + 1/(x+y)
And play a bit around for
1/y = x+y/(x-y)(x+y) + x-y/(x+y)(x-y)
1/y = x+y+x-y/(x-y)(x+y)
1/y = 2x/(x-y)(x+y)
1/y = 2x/x^2 - y^2
2xy= x^2 - y^2</p>
<p>Oh great, I got the same thing again! :)</p>
<p>From here, we follow the above
Simplify to
2xy = x^2 - y^2
Divide both sides by y^2 we get
2(x/y) = (x/y)^2 - 1</p>