mersenneforum.org Simple Geom problem.
 Register FAQ Search Today's Posts Mark Forums Read

 2007-02-07, 16:59 #1 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22×33×19 Posts Simple Geom problem. Here is a simple problem from the Maths Tripos. ABC is an isoceles triangle, whose vertex angle at A is 20* . The point D is on AC such that DBC is 60*. E is on AB such that ECB =50. Find BDE Mally
 2007-02-08, 18:07 #2 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 21378 Posts 30 degrees, but I am still looking for a solution that doesn't rely on trigonometry.
 2007-02-09, 08:39 #3 mfgoode Bronze Medalist     Jan 2004 Mumbai,India 22·33·19 Posts You are correct philmoore. This being a problem strictly on geometry , any other solution by trig, Co-ord etc is not permissble. Hint: Dont go angle chasing. Instead try a simple construction. Mally
 2007-02-19, 15:32 #4 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts I'm pretty sure I've seen this problem before. If the the answer is 30 degrees then if C' ia the reflection of C in AB then C'ED is a straight line. Does this get us anywhere? Love, David
2007-02-24, 11:59   #5
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by davieddy I'm pretty sure I've seen this problem before. If the the answer is 30 degrees then if C' ia the reflection of C in AB then C'ED is a straight line. Does this get us anywhere? Love, David
Why don't you respond to this Mally?
It is a good problem and you posed it.
It might also keep you out of trouble meddling
in posts you don't understand.

David

2007-02-24, 12:14   #6
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts

Quote:
 Originally Posted by davieddy Why don't you respond to this Mally? It is a good problem and you posed it. It might also keep you out of trouble meddling in posts you don't understand. David
Well David I did and gave a crucial hint but somehow my post did not register.

Well here it is again.

Keeping the same notation/lettering/terminology simply mark E' on AC so that E'BC = 20*. Therefore all 3 triangles EBC , BE'C and DE'B are all isosceles.
Hence BEE' is equilateral and triangle EE'D is isosceles.
Now take it from there David and you"ll get home

Mally

 2007-02-25, 01:13 #7 philmoore     "Phil" Sep 2002 Tracktown, U.S.A. 3×373 Posts Very nice problem and solution, although I don't think that the construction is at all obvious, and I would not consider it "simple"! I constructed the point F on AB such that DF and DE had the same length, hoping that I could then prove that DF was parallel to CE. I was able to prove it, but only by using the triple angle formula cos(3x) = 4cos2x - 3cos x where x was 10 degrees, so cos(3x) was 1/2. So messy, I was really looking forward to seeing a simpler proof. I have an interesting diagram inspired by Davieddy's comment, but I will have to wait until Monday to upload it.
2007-02-25, 11:13   #8
davieddy

"Lucan"
Dec 2006
England

647410 Posts

Quote:
 Originally Posted by mfgoode Keeping the same notation/lettering/terminology simply mark E' on AC so that E'BC = 20*. Therefore all 3 triangles EBC , BE'C and DE'B are all isosceles. Hence BEE' is equilateral and triangle EE'D is isosceles. Now take it from there David and you"ll get home Mally
Thankyou Mally.
Even if I had seen that solution before, I'm not surprized I'd
forgotten it! As philmoore says, not that obvious or simple.

Furthermore how do you spot all the isosceles triangles if you
don't go "angle chasing" as you advised us against?

David

2007-02-25, 13:47   #9
davieddy

"Lucan"
Dec 2006
England

2×3×13×83 Posts

Quote:
 Originally Posted by philmoore I have an interesting diagram inspired by Davieddy's comment, but I will have to wait until Monday to upload it.
Would it be a nonagon or 18-agon with some coincidences on the diagonals?

This problem suggests that choosing a multiple of 36 for the number
of degrees in a circle was inspired. When does it date from?

David

2007-02-25, 16:51   #10
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

1000000001002 Posts

Quote:
 Originally Posted by philmoore Very nice problem and solution, although I don't think that the construction is at all obvious, and I would not consider it "simple"! .
Well philmoore I warned you that it is from the Maths Tripos, which is by all standards, a stiff examination of Cambridge Univ. England.

I agree its not obvious or simple. It requires, I would say, to think of a few steps ahead and look out for short cuts to the solution.

Hence I said - not to go angle chasing but rely on a construction first, and after that naturally work out the angles.

As the saying goes 'One man's meat is another man's poison'

Mally

2007-03-06, 12:28   #11
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by philmoore I have an interesting diagram inspired by Davieddy's comment, but I will have to wait until Monday to upload it.
I still wait with interest, Phil

David

Last fiddled with by davieddy on 2007-03-06 at 12:30 Reason: typo

 Similar Threads Thread Thread Starter Forum Replies Last Post Unregistered Information & Answers 3 2012-11-26 02:55 paul0 Factoring 5 2011-11-02 23:21 rainchill Software 8 2007-08-28 08:46 mfgoode Puzzles 82 2007-05-02 08:13 akruppa Puzzles 28 2006-02-04 03:40

All times are UTC. The time now is 06:39.

Fri Jan 28 06:39:48 UTC 2022 up 189 days, 1:08, 2 users, load averages: 1.85, 1.45, 1.45