.

Let .

In (1)

In (2)

(1) & (2)

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- Jul 13th 2007, 09:17 AMred_dog
.

Let .

In (1)

In (2)

(1) & (2)

- Jul 13th 2007, 09:20 AMKrizalid
- Jul 13th 2007, 11:18 AMred_dog
isosceles.

Let and the midpoint of .

Then

We have is cyclic .

Then

(What program do you use for drawing figures?) - Jul 13th 2007, 11:42 AMThePerfectHacker
- Jul 13th 2007, 11:57 AMKrizalid
- Jul 13th 2007, 12:36 PMred_dog
Here is the problem (it's not mine).

Let with and .

Let such that . Find .

(Krizalid, maybe you can draw the figure. Thank you) - Jul 13th 2007, 01:06 PMKrizalid
This is Langley's Problem!!

I know how to make it, but I'll wait if someone can show another solution!! - Jul 13th 2007, 01:09 PMThePerfectHacker
Famous problem.

Discussed Here.

(But I have an embarassing Trigonometry Solution!)

But there is a link on page #2 that provides another solution.

Can I post a problem please? Since I answered red_dog's solution.

See Here - Jul 13th 2007, 01:14 PMKrizalid
Of course you can :D:D

- Jul 13th 2007, 01:22 PMred_dog
- Jul 13th 2007, 01:23 PMThePerfectHacker
This one is really really hard. Not mine I found it in a book.

"Let be a convex polygon with and so that . Let be inside of the hexagon so that . Prove that . "

Hints (highlight below)

1) Reflect through BE

2)Use Ptolemy's Theorem. - Jul 13th 2007, 01:36 PMKrizalid
Sorry, I'm gonna have to ask if you can post another problem.

This is because some people wants to take part (and we dunno what their level is).

There're some hard problems but not so hard like this one.

Plus, this is starting, and the level of problems must up slowly.

In fact, I forgot to mention: Do not post Olympiad problems. - Jul 13th 2007, 02:33 PMThePerfectHacker
In that case you post a problem for me because I am leaving in like 5 minutes and not going to be here for a day. So that will delay this thread unless you want to wait.

Quote:

In fact, I forgot to mention: Do not post Olympiad problems.

- Jul 13th 2007, 02:43 PMKrizalid
Okay, no problem :):)

This problem looks like hard, but it's not when you find a good auxiliary construction.

http://img261.imageshack.us/img261/1...blem007ix3.gif - Jul 13th 2007, 10:24 PMred_dog
We construct the bisector of . Then is isosceles.

Applying bisector's theorem we have . (1)

Also we have (2)

(1) & (2) .

So the triangle is isosceles