Hi All,
This is one of the questions that I liked most of all in my teaching year. By the way, vast majority didn't able to solve it. So let's see who will be able to solve that?

Ernest Friedman-Hill wrote:Damn, you're right (confirmed numerically; I can't see a tractable way to do the calculus.)

I just did the calculus. Lots of chain rule and square roots.

First, the width of the figure is 6, right?

Let's say x and y meet at a point that is z from the left side.

x = sqrt(4 + z^2)
y = sqrt(36 + (6-z)^2)

d = x + y
Take the first derivative of d, set it to 0 and solve for z. (Not intuitively difficult... just a lot of pushing numbers around.)

z = 1.5 and (6-z) = 4.5

Fancy that... two 3-4-5 right triangles: x = 2.5 and y = 7.5, so x+y = 10.

Now that I see the answer, I want to generalize the solution and say that min(x+y) happens when the two line segments make similar triangles. (2, 1.5, 2.5) and (6, 4.5, 7.5). Maybe it you shift the point where they meet one direction or there other, then one is getting longer faster than the other is shrinking. I'm not positive of that though. Now that I write that, it sounds pretty good.

Mike Simmons
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Ryan McGuire wrote:Now that I see the answer, I want to generalize the solution and say that min(x+y) happens when the two line segments make similar triangles. (2, 1.5, 2.5) and (6, 4.5, 7.5). Maybe it you shift the point where they meet one direction or there other, then one is getting longer faster than the other is shrinking. I'm not positive of that though. Now that I write that, it sounds pretty good.

Yup. The physical analogy I mentioned is to imagine x + y as a length of elastic that wraps around a frictionless roller that moves freely along the bottom of the diagram. Being elastic, it contracts so as to minimize its length, subject to the other constraints of the problem. In order for the roller to be at equilibrium, the lines would have to form equal angles with the base line on either side. Once you see that the two right triangles are similar, the math is pretty simple, just setting up a couple ratios.

Mike Simmons wrote: ...that moves freely along the bottom of the diagram.

Yes, it would have more suitable to say it previously...

But, there is an easy way to solve it by drawing symmetrical line... (So i can show you if you want)

You can check it in this way, too.
Regards,

Mike Simmons
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Rizvan Asgarov wrote:

Mike Simmons wrote: ...that moves freely along the bottom of the diagram.

Yes, it would have more suitable to say it previously...

Nah. Figuring that out was part of the problem setup, no? We need to pay attention both to what is specified, and what isn't, in order to solve the problem.

Rizvan Asgarov wrote:But, there is an easy way to solve it by drawing symmetrical line... (So i can show you if you want)

Well, sure. That's pretty much equivalent to what Ryan and I were saying, observing that the two right triangles on the bottom (the ones sharing a side each with the baseline) are similar. Or more specifically, that the angle of segment 'x' with the baseline is the same as that of segment 'y' with the baseline. Yes, it's simple once you realize that. The question is, how do you realize it? And how do you justify it? That's where I brought in the physics analogy, and Ryan talked about what happens if you shift the meeting point to one side or the other.

Out of curiosity, what subject, and what course, were you teaching when you used this problem? What sort of mathematical background did the students have?

Hi again,
How are you? Hope that everything is going 'OK'...

Mike Simmons wrote: The question is, how do you realize it?

And this is the solution by drawing symmetrical line:

And x+y is min. because of right line (as connecting 2 points)

Mike Simmons wrote: Out of curiosity, what subject, and what course, were you teaching when you used this problem? What sort of mathematical background did the students have?

It has been questioned me by one of my schoolmate and i can't remember - maybe this is an geometry olympiad question (for the 10-11th classes). But i could not solving it completely. By the way, this story is very long. We have graduated from Turkish high school where we were accepted to this school from the seventh form by examination. These schools have been founded by "Cag Oyretim" Company. By the way, i want to note that these schools being under the company of "Cag Oyretim" has earned carried out activity 285 medals (65 golds, 99 silvers and 121 bronzes) from international Olympiads during 15 years. Nowadays (3 May), Our students got a gold and 2 bronze medals from the 44th International Mendeleyev Olympiad in Azerbaijan (on chemistry subject)

Thank you for your interest...
Best Wishes,
Rizvan

Mike Simmons
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Ah, I see. Nice. A more purely geometric argument than mine was, though ultimately equivalent. Thanks for elucidating.

Ryan McGuire
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Rizvan Asgarov wrote:Hi All,
This is one of the questions that I liked most of all in my teaching year. By the way, vast majority didn't able to solve it. So let's see who will be able to solve that?

Good luck,
Rizvan

For those reading this question years after it was solved, let me try to describe the now-missing diagram:

Draw a quadrilateral with points at A=(0,2), B=(0,0), C=(w,0), D=(w, 6), where w is some unspecified width. Make w approximately equal to something like 5.

Obviously AB has a length of 2, and CD has a length of 6. The length of AD is given as 2*sqrt(13). Angles ABC and BCD are right angles.

Now pick a point E somewhere along BC and draw line segments AE and DE. AE has a length of x, and DE has a length of y.

Find the position of point E along BC such that (x+y) is a minimum.

A man has worked the whole day at some distance from the camp where he's staying.
Now he's tired and decides to walk back to the camp. But since he is very thirsty as well,
he decides to walk to the nearby river first, drink some water and then walk to camp.
This river is a special one: it is as straight as a ruler!
Man and camp are, of course, on the same side of the river.
Now, since he is very tired, he wants to walk as little as possible.
To what point of the river should he walk?