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A spider is sitting in the middle of one of the smallest walls in a room and a fly is resting beside the window. What is the shortest distance the spider would have to crawl to catch the fly? Buildings in Ah Canul Group dating back to 500-600 AD. (I am unable to move these photos around. They are out of order. I'm going to have to wing it and post the information separately.). Exit Path is a multiplayer and uniplayer guantlet-style racing game through perilous traps and platforms. Desktop Solitaire free download - Palm Desktop, 123 Free Solitaire, Desktop Themes, and many more programs.

Exit Pathspiter GamesOn Sunday, July 5th, we drove south of Mérida on the Ruta Puuc, an area dotted with ruins, cenotes and underground caverns. We found Oxkintok by accident. It was a surprisingly impressive discovery. As we drove the final four kilometers on the gravel road we were surrounded by partial ruins and unrestored pyramids. Thirty groups of buildings have been discovered at the site. Of those, three main groups have been restored in an 8 sq. km. area. The Ah May, Ah Canul and Ah Zib Groups each from separate eras of life in Oxkintok.Exit Pathspiter Games

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Archaelogists have become excited about Oxkintok in the past several years. This site is located halfway between Celestún on the Gulf Coast and Dzibilchaltún north of Mérida. The three centers are connected by sacbes, or raised white roads. In fact, the number of large constructions at Oxkintok and the abundance of sacbes proves there was a major political, economical and cultural exchange; the Maya had contact with the center of México as well as other Meso-American groups.
Exit pathspiter games freeThey've found inscriptions in the architecture here dating back to the 'Early Oxkintok' phase (300-500 AD) that tell of the Pre-Classic history of life in 300 BC. One of the large structures looks much like the pyramids of Teotihuacán outside of México City, a major Aztec site.

You sent in a large number of solutions to this problem but many of them only considered the spider moving horizontally or vertically but not diagonally across the sides of the room. I have included below a net of the room with the path of the spider as it takes the shortest route over the ceiling. I hope it helps you to see what was going on. However the spider could walkaround the walls to get to the fly or along the floor - risking a human foot!

Andrei of Tudor Vianu National College calculated the distance the spider must travel in the original problem and has worked out when it is best to go via the floor instead of the ceiling.

First, I observed that the shortest distance can be found on the net of the box.
In the figures below, the distances on the box are written, if the spider travels by the ceiling (figure 1) or by the lateral wall (figure 2).
In figure 1 - I marked the position of the fly (F), and the spider (S).
The next step is to calculate the distance travelled by the spider for each case.
Applying Pythagoras Theorem in Figure 1, I obtain:

$ FS = 7.40355 m $

$ FS^2 = 7.5^2 + 0.25^2$

The shortest distance is $7.40355$ m, i.e. the first situation. This is because the dimension 'wide' is larger than the dimension 'high'and the spider starts from the middle of the wall.

Now I analyze the situation when the fly goes down the wall. In the first case, with the fly fixed, it was situated in the upper middle of the face, so it was better for the spider to go on the top of the box. When the fly arrives at the middle (height) of the box, it is the same for the spider to go over the top or over the bottom of the box. When the fly goes still further to the bottom, thespider should go on the bottom of the box.


However in some cases it is quicker to go via the side wall instead of either the floor or the ceiling. Gillian sent us in the following diagrams to help explain how she worked out when this is the case:
My first diagram shows the measurements when the spider is crawling along the ceiling. She has to go along $(8.75 - h)$ metres and down $1.5$ metres, so the shortest distance she can travel is the square root of $(8.75-h)^2 + 1.5^2$. As the height, $h$, of the fly decreases, the distance increases.

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If she crawls along the floor, the shortest distance she can travel is the square root of $(6.25+h)^2 + 1.5^2$. As h descreases, the distance decreases. These two results could be seen just by common sense.
My second diagram shows the measurements when the spider is crawling along the wall closest to the fly. This way, she goes along $7.5$ metres and up $(h - 1.25)$ metres. The shortest distance she can crawl is the square root of $7.5^2 + (h-1.25)^2$.
At $h = 1.25$ metres, it is quicker for her to crawl along the wall than either the floor or the ceiling. (Distance $= 7.5$m, as opposed to $7.65$m if she were going along the floor or ceiling.)

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To see when it's quicker to go along the wall than the ceiling, I solved the inequality
$(8.75 - h)^2 + 1.5^2 geq 7.5^2 + (h - 1.25)^2$

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$(8.75 - h)^2 - (h - 1.25)^2 geq 7.5^2 - 1.5^2$
$7.5times (10 - 2h) geq 54$
$h leq 1.4$
When the fly is less than $1.4$m from the ground, it is quicker to go by the wall than by the ceiling. Similarly, when it is higher than $1.1$m, it is quicker to go by the wall than the floor.