Simple Geom problem.

mfgoode · 2007-02-07, 16:59

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

philmoore · 2007-02-08, 18:07

30 degrees, but I am still looking for a solution that doesn't rely on trigonometry.

mfgoode · 2007-02-09, 08:39

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

davieddy · 2007-02-19, 15:32

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

davieddy · 2007-02-24, 11:59

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

mfgoode · 2007-02-24, 12:14

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

philmoore · 2007-02-25, 01:13

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.

davieddy · 2007-02-25, 11:13

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

davieddy · 2007-02-25, 13:47

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

mfgoode · 2007-02-25, 16:51

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'

Glad you liked it

Mally

davieddy · 2007-03-06, 12:28

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

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

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2007-02-08, 18:07	#2
philmoore "Phil" Sep 2002 Tracktown, U.S.A. 45F₁₆ 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

2007-02-25, 01:13	#7
philmoore "Phil" Sep 2002 Tracktown, U.S.A. 10001011111₂ 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.