Geometry Marathon!

mathwizarddude

Post next problem after you solve one - full solution please (problems preferrably not from the resource section already with well-written solutions since everyone has access to it plus there is no point in reposting them when there are already solutions there).

Problem 1: In concave hexagon $ABCDEF$ , $m\angle A = m\angle B = m\angle C = 90^\circ$ , $m\angle D = 100^\circ$ , and $m\angle F = 80^\circ$ . Also, $CD = FA$ , $AB = 7$ , $BC = 10$ , and $EF + DE = 12$ . Compute the area of the hexagon.

**Posted:** Fri Nov 06, 2009 10:50 pm

winternights1

I haven't solved the problem, but I think I did make progress on it.

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edit: Oops. Most of this is wrong, because I read angle F = 80 as angle E = 80... so I was essentially working a different problem maybe

. Sorry about that Blush

**Posted:** Sat Nov 07, 2009 7:38 pm

krishkoushik · P versus NP Offline Joined: 20 Aug 2009 Posts: 32

SOLUTION TO PROBLEM 1:

extend AF to X such that $\angle AXD=90^\circ$

area of hexagon=area of rectangle-area of FEDX

$FE=a, ED=12-a$

$FX=b, XD=b-3$

$FD^2=EF^2+ED^2=FX^2+XD^2$

$a^2+(12-a)^2=b^2+(b-3)^2$

$144-2a(12-a)=9+2b(b-3)$

$135=4(\frac{a(12-a)}{2}+\frac{b(b-3)}{2}$

$area(FEDX)=33.75$

$area(ABCDEF)=70-33.75=36.25$

PROBLEM 2
Prove that the area of triangle formed by the lengths of medians of a given triangle is three fourth the area of the given triangle.

**Posted:** Sat Nov 07, 2009 10:54 pm

grn_trtle

I'm feeling extremely uncreative tonight, so here's a solution with analytical geometry

Solution

Edit: Oh, someone else feel free to post another problem. Sorry, I don't really know of any good ones that haven't been posted here already. If I find something I'll put it up.

**Posted:** Sun Nov 08, 2009 12:49 am

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$\arctan\alpha + \arctan\beta = \text{arg}\{(1+\alpha i)(1+\beta i)\} = \arctan\left( \frac{\alpha+\beta}{1-\alpha\beta} \righ...$

mathwizarddude

Is there a synthetic solution to the median problem?
Problem 3:Two square pieces of paper are made subject to the constraint that the sum of their areas is 2 square units. These pieces of paper are to be put inside a rectangle such that the sides of the sqauares are parallel to the sides of the rectangle. What is the least possible value for the area of a rectangle if we want to be sure tht this rectangle will enclose these two pieces of paper? Prove that your answer is the minimum.

luisgeometria

**Posted:** Sun Nov 08, 2009 3:49 pm

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Platon said: "Nobody untrained in geometry may enter my house"

Poincare

New problem?

**Posted:** Mon Nov 09, 2009 3:23 pm

_________________
Please visit this website: freerice.com and play for just 5 minutes and help end hunger.

mathwizarddude

Problem 3:Two square pieces of paper are made subject to the constraint that the sum of their areas is 2 square units. These pieces of paper are to be put inside a rectangle such that the sides of the sqauares are parallel to the sides of the rectangle. What is the least possible value for the area of a rectangle if we want to be sure that this rectangle will enclose these two pieces of paper? Prove that your answer is the minimum.
Problem 4:Prove that of all triangles inscribed in a given triangle, the one with least perimeter connects the feet of the given triangle.

**Posted:** Mon Nov 09, 2009 4:19 pm