# Geometry Marathon

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• Jul 13th 2007, 10:17 AM
red_dog
$\widehat{BDC}=3x=\widehat{BCD}$.
Let $AB=DC=a, \ BD=BC=b$.
In $\triangle ABD: \ \frac{a}{b}=\frac{\sin 3x}{\sin 2x}$ (1)
In $\triangle BDC: \ \frac{a}{b}=\frac{\sin 6x}{\sin 3x}=2\cos 3x$ (2)
(1) & (2) $\Rightarrow \sin 3x=2\sin 2x\cos 3x\Leftrightarrow \sin x+\sin 3x-\sin 5x=0\Leftrightarrow$
$\Leftrightarrow \sin x(1-2\cos 4x)=0\Rightarrow \cos 4x=\frac{1}{2}\Rightarrow x=15$
• Jul 13th 2007, 10:20 AM
Krizalid
Quote:

Originally Posted by red_dog
$\widehat{BDC}=3x=\widehat{BCD}$.
Let $AB=DC=a, \ BD=BC=b$.
In $\triangle ABD: \ \frac{a}{b}=\frac{\sin 3x}{\sin 2x}$ (1)
In $\triangle BDC: \ \frac{a}{b}=\frac{\sin 6x}{\sin 3x}=2\cos 3x$ (2)
(1) & (2) $\Rightarrow \sin 3x=2\sin 2x\cos 3x\Leftrightarrow \sin x+\sin 3x-\sin 5x=0\Leftrightarrow$
$\Leftrightarrow \sin x(1-2\cos 4x)=0\Rightarrow \cos 4x=\frac{1}{2}\Rightarrow x=15$

• Jul 13th 2007, 12:18 PM
red_dog
$\widehat{BDC}=\widehat{BCD}=3x\Rightarrow\triangle BDC$ isosceles.
Let $AB=DC=a, \ BE\perp DC, E\in (DC)$ and $F$ the midpoint of $AB$.
Then $FE=DE=EC=\frac{a}{2}\Rightarrow \widehat{DFC}=90$
$\widehat{FEC}=2x, \ \widehat{EFC}=\widehat{FCE}=x$
We have $\widehat{FBD}=\widehat{FCD}=x\Rightarrow FBCD$ is cyclic $\Rightarrow \widehat{DBC}=\widehat{DFC}=90$.
Then $3x=45\Rightarrow x=15$

(What program do you use for drawing figures?)
• Jul 13th 2007, 12:42 PM
ThePerfectHacker
Quote:

Originally Posted by red_dog
(What program do you use for drawing figures?)

I think it might be called "Asymptote" but I might be wrong.
• Jul 13th 2007, 12:57 PM
Krizalid
Quote:

Originally Posted by red_dog
$\widehat{BDC}=\widehat{BCD}=3x\Rightarrow\triangle BDC$ isosceles.
Let $AB=DC=a, \ BE\perp DC, E\in (DC)$ and $F$ the midpoint of $AB$.
Then $FE=DE=EC=\frac{a}{2}\Rightarrow \widehat{DFC}=90$
$\widehat{FEC}=2x, \ \widehat{EFC}=\widehat{FCE}=x$
We have $\widehat{FBD}=\widehat{FCD}=x\Rightarrow FBCD$ is cyclic $\Rightarrow \widehat{DBC}=\widehat{DFC}=90$.
Then $3x=45\Rightarrow x=15$

(What program do you use for drawing figures?)

Okay, that is correct.

Actually, drawing with Asymptote it's a lot of complex work, plus, CorelDraw's quality is the best.
• Jul 13th 2007, 01:36 PM
red_dog
Here is the problem (it's not mine).
Let $\triangle ABC$ with $\widehat{A}=\widehat{B}$ and $\widehat{C}=20$.
Let $M\in (AC), \ N\in (BC)$ such that $\widehat{ABM}=60, \ \widehat{BAN}=50$. Find $\widehat{BMN}$.

(Krizalid, maybe you can draw the figure. Thank you)
• Jul 13th 2007, 02:06 PM
Krizalid
This is Langley's Problem!!

I know how to make it, but I'll wait if someone can show another solution!!
• Jul 13th 2007, 02:09 PM
ThePerfectHacker
Famous problem.
Discussed Here.
(But I have an embarassing Trigonometry Solution!)

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

See Here
• Jul 13th 2007, 02:14 PM
Krizalid
Of course you can :D:D
• Jul 13th 2007, 02:22 PM
red_dog
Quote:

Originally Posted by ThePerfectHacker
Famous problem.

Yes!
I didn't know that the problem was already posted.
• Jul 13th 2007, 02:23 PM
ThePerfectHacker
This one is really really hard. Not mine I found it in a book.

"Let $ABCD EF$ be a convex polygon with $AB=BC=CD$ and $DE=ED=FA$ so that $\angle BCD = \angle EFA = \pi/3$. Let $G \mbox{ and }H$ be inside of the hexagon so that $\angle AGB = \angle DHE = 2\pi /3$. Prove that $AG+GB+GH+DH+HE \geq CF$. "

Hints (highlight below)

1) Reflect through BE

2)Use Ptolemy's Theorem.
• Jul 13th 2007, 02:36 PM
Krizalid
Quote:

Originally Posted by ThePerfectHacker
This one is really really hard. Not mine I found it in a book.

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, 03:33 PM
ThePerfectHacker
Quote:

Originally Posted by Krizalid
Sorry, I'm gonna have to ask if you can post another problem.

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.
This is an IMO problem. You are right, it is not right to post such hard problems.
• Jul 13th 2007, 03:43 PM
Krizalid
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, 11:24 PM
red_dog
We construct the bisector $CE, \ E\in BD$ of $\widehat{BCD}$. Then $\triangle BEC$ is isosceles.
Applying bisector's theorem we have $\frac{DE}{EB}=\frac{DC}{BC}\Rightarrow \frac{DE}{EC}=\frac{DC}{BC}$. (1)
Also we have $\widehat{DEC}=\widehat{DCB}=4x$ (2)
(1) & (2) $\Rightarrow \triangle DEC\sim\triangle DCB\Rightarrow \frac{DC}{BC}=\frac{DC}{BD}\Rightarrow BC=BD$.
So the triangle $BDC$ is isosceles $\Rightarrow \widehat{BDC}=4x\Rightarrow 4x+6x=180\Rightarrow x=18$
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