Geometry Marathon

**Krizalid** · June 22nd 2007, 07:36 AM

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

**OReilly** · June 22nd 2007, 08:12 AM

Originally Posted by blurgh

i've figured out the solution but the proof doesn't use similar triangles.

Post your solution.

**blurgh** · June 23rd 2007, 07:52 PM

correction, my proof does use similar triangles. i found a mistake in my proof and had to fix it.

(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.

my proof

**Ilaggoodly** · June 23rd 2007, 09:31 PM

impressive proof!

**Krizalid** · June 24th 2007, 09:33 AM

Nice proof.

You can shorten it using similar triangles.

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

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

Post your problem now

**blurgh** · June 24th 2007, 07:55 PM

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.

Prove x=60 degrees.

**OReilly** · July 12th 2007, 02:43 PM

Has anyone tried to solve this?

I have tried many ideas, but failed.

**Krizalid** · July 12th 2007, 02:48 PM

Prove that $\overline{AB}=\overline{AC}$

**OReilly** · July 12th 2007, 02:50 PM

Originally Posted by Krizalid

Prove that $\overline{AB}=\overline{AC}$

I had good idea, and tried many variants of it, but like I said I failed.

If anyone wants I will post it, maybe someone will get new idea.

**Krizalid** · July 12th 2007, 04:14 PM

I found another proof:

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

**OReilly** · July 12th 2007, 04:49 PM

Nice proof.

**blurgh** · July 13th 2007, 03:22 AM

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.

well, knock us out with another question. *shudders* i'm scared of what your hard questions are like... scared and intrigued.

**DivideBy0** · July 13th 2007, 05:15 AM

Originally Posted by blurgh

correction, my proof does use similar triangles. i found a mistake in my proof and had to fix it.

(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.

my proof

Wow, how were you able to make it so neat? A special program?

**OReilly** · July 13th 2007, 07:13 AM

Originally Posted by blurgh

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.

well, knock us out with another question. *shudders* i'm scared of what your hard questions are like... scared and intrigued.

Can you post your proof?

**Krizalid** · July 13th 2007, 08:40 AM

Here's my problem

Not so hard but it's a beautiful problem

Thread: Geometry Marathon

LinkBack

Thread Tools

Display

my proof

Similar Math Help Forum Discussions

Secondary school's competition - marathon.

Modern Geometry: Taxicab Geometry Problems

DE Marathon

Small limit marathon

Limit Marathon

Search Tags