If your proof doesn't use trig. then it's okay.

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- Jun 22nd 2007, 07:36 AMKrizalid
If your proof doesn't use trig. then it's okay.

- Jun 22nd 2007, 08:12 AMOReilly
- Jun 23rd 2007, 07:52 PMblurghmy proof
correction, my proof does use similar triangles. i found a mistake in my proof and had to fix it. :D (hehe, oops)

um, i haven't got the hang of Latex, so i hope you don't mind if i give you a link to a pdf file instead. (you'll have to download it, sorry!)

although i think and am more fluent in English, i always solve math problems in Vietnamese, so my proof is basically a literal translation of what i would've written in Vietnamese, so I'm sorry if it doesn't make sense in some parts. All symbols are what the Vietnamese use in our math, if there is anything different from what most people would write please tell me.

once again, sorry for making you download a pdf file.

:rolleyes:

my proof - Jun 23rd 2007, 09:31 PMIlaggoodly
impressive proof!

- Jun 24th 2007, 09:33 AMKrizalid
Nice proof.

You can shorten it using similar triangles.

In your sketch we have that $\displaystyle \overline{AD}=\overline{CD}=\overline{ND}$, let's denote this segments "k". On the other hand we have that $\displaystyle \measuredangle~CBN=\measuredangle~BCN=45^\circ$.

Nevertheless, $\displaystyle \triangle{ACB}\sim\triangle{BCD}\implies\overline{ BC}=k\sqrt2$. Finally, $\displaystyle \overline{BN}=\overline{CN}=k\implies\triangle{CDN }~\text{is equilateral}$, then $\displaystyle \measuredangle~NDC=2x=60^\circ~\therefore~x=30^\ci rc$ and the proof is completed.

Post your problem now :D - Jun 24th 2007, 07:55 PMblurgh
sweet! i like your approach!

ok here's one of my favourite geometry problems. well actually i've changed it a bit, now it's a bit harder! if someone wants a hint i'll give them the original version... maybe. ;)

Let there be a rectangle that has a length that is twice the size of it's width, as shown in the diagram below.

http://i79.photobucket.com/albums/j1.../60degrees.png

Prove x=60 degrees. - Jul 12th 2007, 02:43 PMOReilly
Has anyone tried to solve this?

I have tried many ideas, but failed. - Jul 12th 2007, 02:48 PMKrizalid
http://img487.imageshack.us/img487/5449/sdfgd5.png

Prove that $\displaystyle \overline{AB}=\overline{AC}$ :D:D - Jul 12th 2007, 02:50 PMOReilly
- Jul 12th 2007, 04:14 PMKrizalid
I found another proof:

http://img293.imageshack.us/img293/4...degreesnx6.png

- Take a point $\displaystyle F$ on $\displaystyle \overline{CD}$ such that $\displaystyle \measuredangle~CBF=60^\circ\implies\overline{BF}=2 m~\therefore~\overline{AB}=\overline{BF}~\blacktri angleleft$
- From $\displaystyle \blacktriangleleft$ we have that $\displaystyle \measuredangle~FAB=75^\circ\implies\triangle{ADF}\ cong\triangle{BCE}~\therefore~\overline{AF}=\overl ine{BE}$
- Nevertheless, $\displaystyle \overline{AB}\parallel\overline{EF}\implies\text{q uadrilateral }ABEF~\text{isosceles trapezium}$ (and cyclic).
- Finally $\displaystyle x=\measuredangle~EAD=\measuredangle~FAD+\measureda ngle~EAF=15^\circ+45^\circ=60^\circ$ and the proof is completed $\displaystyle \blacksquare$

- Jul 12th 2007, 04:49 PMOReilly
Nice proof.

- Jul 13th 2007, 03:22 AMblurgh
very nice proof! i like it alot! my proof involved creating an equadrilateral triangle with one side being BE and overlaps triangle ABE. once again, very sweet approach.

:cool:

well, knock us out with another question. *shudders* i'm scared of what your hard questions are like... scared and intrigued. - Jul 13th 2007, 05:15 AMDivideBy0
- Jul 13th 2007, 07:13 AMOReilly
- Jul 13th 2007, 08:40 AMKrizalid
Here's my problem :D:D:D

http://img63.imageshack.us/img63/4651/problem006cp7.gif

Not so hard but it's a beautiful problem :)