Here is alittle problem I was given last year on the AIME, solve it and I will believe that you guys are indeed smart mathematically, this was a non calculator question


[quote]
A tripod has three legs each of length 5 feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is 4 feet from the ground In setting up the tripod, the lower 1 foot of one leg breaks off. Let h be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then h can be written in the form [img]http://www.artofproblemsolving.com/Wiki/images/math/7/1/9/7198d85ac08e4b084111df94b262f6d1.png[/img] where m and n are positive integers and n is not divisible by the square of any prime. Find [img]http://www.artofproblemsolving.com/Wiki/images/math/0/5/9/059eb58a507bd548a19329f44fac8a21.png[/img] (The notation [img]http://www.artofproblemsolving.com/Wiki/images/math/2/f/c/2fc4d67260ce7b740b9dedf6da69b7dc.png[/img] denotes the greatest integer that is less than or equal to x.)[/quote]

Sounds like it is a homework problem. Reason being is that you registered and signed up today to get this answered and have never been seen on the forums until now.

Nah, I got it last year, when i took the AIME, i just found this forum

[quote author=CornMuffin link=topic=16257.msg164177#msg164177 date=1170797415]
solve it and I will believe that you guys are indeed smart mathematically
[/quote]

If you don't already believe that Yoni is one of the most clever people you'll ever meet just by browsing this forum..

Bleh, icky. Non-formal discussion:
Let S be the top of the tripod and A, B, C be the points on the floor (when the tripod is not broken).
Let T be the breaking point on leg SA.
That is, AT = 1, TS = 4.
Let M be the midpoint of the triangle ABC (which is an equilateral triangle, of course). This is the point directly under S.
Also, let D be the midpoint of BC.

Goal 1: Find BC.

In the triangle SMB, angle <SMB is a right angle:
SM^2 + MB^2 = SB^2
4^2 + MB^2 = 5^2
--> MB = 3

We go back to triangle ABC, and we know AM = BM = CM = 3.
Since this triangle is equilateral, AM, BM and CM are angle bisectors.
Therefore (sine theorem):
BM / sin(<BCM) = BC / sin (<BCM)
3 / sin(30 degrees) * sin (120 degrees) = BC
BC = 3sqrt(2) = sqrt(18)

End of goal 1.

Goal 2: Find SD.

In the triangle SBC, angle <SDC is a right angle.
Pythagoras:
SD^2 = SC^2 - DC^2 = SC^2 - BC^2 / 4
SD^2 = 5^2 - 18/4 = 41/2
SD = sqrt(41/2)

End of goal 2.

Goal 3: Find sin(<SDA).

Since AD is a height in ABC, we know that M is on AD.
So we look at the triangle SDA.
SM is perpendicular to plane ABC, so angle <SMA is right.

sin(<DAS) = SM / SA = 4 / 5
SD / sin(<DAS) = AS / sin(<SDA)
sqrt(41/2) / (4/5) = 5 / sin(<SDA)
sin(<SDA) = (25/4) / sqrt(41/2)

End of goal 3.

This is getting tiresome. I'll leave you with directions.
Find TD by using the cosine theorem.
Find sin(<TDA) using the sine theorem.
Find sin(<SDT) using the sine of sum-of-angles identities (we know <SDA and <TDA).
Then, by the sine definition, sin(<SDT) will give you h' / 5, where h' is the height you seek.

I don't get the point of the question, unfortunately, since I didn't reach the answer. Oh well.

ya, i really dont like that question lol...

but the answer is 183, i wasn't able to come up with the answer, i just looked at the answer key :)

Here's the practical answer: It's a tripod with one of the legs a foot shorter than the rest, it won't stand up, it'll fall over. go get a replacement.

Did you take the AMC this year?  I found it to be overbearingly hard.  Last year my school sent 11 students to take the AIME, but this year, we only had two break 100.  Perhaps the national average will be below that 100 marker this year, or perhaps my school simply graduated a large number of smart seniors :-/