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[unparseable or potentially dangerous latex formula] has 9 digits, all distinct. In your head, find the missing digit.

Okay, that was pretty short. Here's another.

A quadrilateral has sides of 10, 12, 16, and 20. What is the largest possible area?

2^9=512

It doesn't have 9 digits.

O wait I feel stupid right now... didn't read subject

Oops, I had [unparseable or potentially dangerous latex formula], but didn't bracket in the 29, so I took out the 2...

[unparseable or potentially dangerous latex formula] has 9 digits, all distinct. In your head, find the missing digit.

hmm well the digits from 0-9 sum to 45 and its easy to see that 2^29 is congruent to 5 mod 9 so the missing digit must be 4

Okay, very well. How about the quadrilateral problem? And, if this is all too easy, maybe I'll give another that I like! I'm sorry (sorta).

A circle of radius r is determined by 3 points with integer coordinates. Show that at least 2 of these points are at least a distance of [unparseable or potentially dangerous latex formula] apart.

For the quadrilateral problem, the answer is about 194.4

yup.

i am just wondering do cyclic quads always have the largest area for a certain quad?

since in this problem the largest area is for a cyclic quad

that sounds like a good problem for you to solve stupidity!

why don't you explore it yourself, and report your results later ^_^

and by the way, i STILL haven't started my homework... i'm in trouble...

edit: ok i will see if i can prove that, it will give me something to do in school :)

edit: Oh, misread later posts...guess I'm not supposed to post the reason why, heh.

in the post above i meant to say where it has diagnols e,f not d,e

Kash gave Bretschneider's Formula. It's stronger than Heron's, but I wonder if there's a similar formula for pentagons. There's no hanging s to turn into (s-e). This page, I think, has some of the better quadrilateral formulas for area.

http://mathworld.wolfram.com/BrahmaguptasFormula.html

What about the general cyclic polygon? Is its area always determined by its side lengths only? There are two nice geometrical arguments I thought of that don't use any formulas.

Oh, and I thought this was cool. The quadrilateral talk reminded me of it.

http://staff.imsa.edu/~keyton/Talks%20a ... ystery.doc

To show that if a polygon is cyclic (able to be inscribed in a circle), you only need the side lengths to determine the area
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(1)Draw the radius to each of the vertices, i.e cut the polygon into pizza slices. You may rearrange the slices in any order to form any other polygon with the same side lengths. Note that you may have polygons with the same side lengths that are not cyclic.

(2) Cut through the polygon from one vertex to another vertex. Take one of the parts and flip it, then put it back. You can swap any two consecutive edges this way, so you can order the edges however you wish while preserving the area.




A circle of radius r is determined by 3 points with integer coordinates. Show that at least 2 of these points are at least a distance of [unparseable or potentially dangerous latex formula] apart.
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My solution wasn't this nice. This is the official Putnam solution. If R is the radius of the circumscribed circle, the area K is given by

[unparseable or potentially dangerous latex formula]

By Pick's theorem, the triangle area is never less than [unparseable or potentially dangerous latex formula].
[unparseable or potentially dangerous latex formula] where I is the number of lattice points in the interior and P is the number of points on the periphery.

[unparseable or potentially dangerous latex formula]

[unparseable or potentially dangerous latex formula]

Since the geometric mean has to be in the middle somewhere,

[unparseable or potentially dangerous latex formula]