First problem! I don't know how many I'll go to. Depends on when I run out of interesting problems
A hexagon with sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle.
Find the radius of the circle.
Day two:
By the way, this one's easier.
How many zeros are at the end of 100000! ?
Day three:
We all know it's impossible to construct a circle with the same area as a square within the Archimedean constraints (used only straight-edge and compass, and must be done in a finite number of constructions).
Show how to construct a circle with the same area as a square if you remove one of those constraints.
Have fun.
And by the way, do you know about those nifty "Mira" things? They're basically red sheets of translucent reflective plastic. You can use them for constructions, and you can actually trisect an angle with one. They won't help you with this problem, though.
Day four:
This is from an old math competition. I don't remember which, exactly, but J. Riley wrote it.
If w, x, y, and z are finite, not all zero, and satisfy the equations
w = bx + cy + dz x = aw + cy + dz y = aw + bx + dz z = aw + bx + cy
Umm... not enough info? What are the interior angles of the hexagon? (or maybe it's not necessary, I dunno )
Edit: after further thought and some MS Paint diagrams, I've decided to demand a diagram for this question, so that we know exactly what this hexagon looks like. At the moment we don't even know which sides have which lengths, for one thing.
Edit: actually it depends on what you mean by inscribed. If we had a normal hexagon inscribed by the circle then every vertex would touch the circle, but if this particular hexagon was inside a circle only 2 vertices will touch, so find the largest radius you need.
actually he just giving us all a riddle.. he not asking for help, hence the title of the thread.. if he do neeed help, he could have just googled the question .. and the answer/explanation should be at first result.
On December 15 2005 19:59 Ender wrote: The hexagon is inside an eclipse, not a circle.
Edit: actually it depends on what you mean by inscribed. If we had a normal hexagon inscribed by the circle then every vertex would touch the circle, but if this particular hexagon was inside a circle only 2 vertices will touch, so find the largest radius you need.
First of all, an eclipse is a car, not a shape. Secondly, he said it's a circle and is asking for radius, something that an ellipse (I'm assuming that's what you ment) does not have.
That having been said, I have no idea how to do this problem. But I'd be interested to know now.
if that's true, then in this example the circle depends on the distance of the opposing sides with length 2. but this distance depends on the angle at the connections the 7 and the 11 side share. if you choose it to be 180°, you'd get a rectangle 2x18. from that point you could decrease the angles, resulting in "coffins" with different heights, thus hexagons with different "outcircles". or is the angle at the 7-11 connections somehow determined?
I also get 7, but I used MatLab's Fzero command on @(r)r-1/(cos(-asin(7/(2*r)) - asin(11/(2*r)))) because I've long forgotten how to simplify that kind of thing.
what about a hexagon with the side sequence 2-7-11-2-7-11 with the angles 180,180,360(or 0),180,180. that would result in a single line with the length 20, thus resulting in an outcircle with the radius 10. or is this construction forbidden by a property of hexagons?
EPT
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![[image loading]](http://img206.imageshack.us/img206/1292/hexagon3kx.png)
