20070207, 16:59  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×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 
20070208, 18:07  #2 
"Phil"
Sep 2002
Tracktown, U.S.A.
2137_{8} Posts 
30 degrees, but I am still looking for a solution that doesn't rely on trigonometry.

20070209, 08:39  #3 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
You are correct philmoore. This being a problem strictly on geometry , any other solution by trig, Coord etc is not permissble. Hint: Dont go angle chasing. Instead try a simple construction. Mally 
20070219, 15:32  #4 
"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 
20070224, 11:59  #5  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
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 

20070224, 12:14  #6  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Quote:
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 

20070225, 01:13  #7 
"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) = 4cos^{2}x  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. 
20070225, 11:13  #8  
"Lucan"
Dec 2006
England
6474_{10} Posts 
Quote:
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 

20070225, 13:47  #9  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
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 

20070225, 16:51  #10  
Bronze Medalist
Jan 2004
Mumbai,India
100000000100_{2} Posts 
Quote:
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' Glad you liked it Mally 

20070306, 12:28  #11  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
David Last fiddled with by davieddy on 20070306 at 12:30 Reason: typo 

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