1. $\widehat{BDC}=3x=\widehat{BCD}$.
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) $\Rightarrow \sin 3x=2\sin 2x\cos 3x\Leftrightarrow \sin x+\sin 3x-\sin 5x=0\Leftrightarrow$
$\Leftrightarrow \sin x(1-2\cos 4x)=0\Rightarrow \cos 4x=\frac{1}{2}\Rightarrow x=15$

2. 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) $\Rightarrow \sin 3x=2\sin 2x\cos 3x\Leftrightarrow \sin x+\sin 3x-\sin 5x=0\Leftrightarrow$
$\Leftrightarrow \sin x(1-2\cos 4x)=0\Rightarrow \cos 4x=\frac{1}{2}\Rightarrow x=15$

3. $\widehat{BDC}=\widehat{BCD}=3x\Rightarrow\triangle BDC$ 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?)

4. 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.

5. 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.

6. 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)

7. This is Langley's Problem!!

I know how to make it, but I'll wait if someone can show another solution!!

8. 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

9. Of course you can

10. Originally Posted by ThePerfectHacker
Famous problem.
Yes!
I didn't know that the problem was already posted.

11. 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.

12. 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.

13. 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.

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.

14. Okay, no problem

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

15. 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 $\Rightarrow \widehat{BDC}=4x\Rightarrow 4x+6x=180\Rightarrow x=18$

Page 3 of 4 First 1234 Last