Triangle Puzzle

Originally posted by richjohnson
As you may remember from mathematics classes, the answer is not as imporant as the method used to arrive at it. The answer alone will never get you full credit.

Amen. The process by which you come to an answer, even the wrong one, informs a person more about how (or how not) to approach a problem than an unexplained answer.

I guess I think of these puzzles (and this forum) as a place to exchange thought processes rather than a place to give answers. Of course, all of us like to be the one who solves it, but it's more important to me to be able to solve it in the end, even if it is after one, two, or twenty other people have figured it out.

Nemesio

Originally posted by THUDandBLUNDER
An isosceles triangle ABC has a point D on AB and a point E on AC.

Given:
Angle BAC = 20 degrees
Angle EBC = 50 degrees
Angle DCB = 60 degrees.
AB = AC

Find angle CDE.

ghostintheshell has correctly guessed./_CDE=30 deg.but he hasn't given proof. HERE is an elegant proof.
construction:- draw DP parralel to BC,EQ parralel to CB, point P on AC,point Q on AB...Also draw equilateral triangle RBC on base BC.
Obviously triangles BQR,CRE, & ABC are each similar isosceles triangles with vertex angle 20 deg. Triangles QDR & EPR are each isosceles with vertex angle /_DQR=/_PER=100 deg.
Thus, DQ=QR=ER=EP.
Now CD/DP=AD/DP( isosceles Tr DAC )
=CE/ER ( similarity of Tr's CER & ADP )
=CE/EP.
HEnce DE is internal bisector of /_CDP.
hence required /_CDE=30 deg is immediately inferred.
It was a really elegant problem.

Originally posted by ranjan sinha
ghostintheshell has correctly guessed./_CDE=30 deg.but he hasn't given proof. HERE is an elegant proof.
construction:- draw DP parralel to BC,EQ parralel to CB, point P on AC,point Q on AB...Also draw equilateral triangle RBC on base BC.
Obviously triangles BQR,CRE, & ABC are each similar isosceles triangles with vertex angle 20 deg. Triangles ...[text shortened]... /_CDP.
hence required /_CDE=30 deg is immediately inferred.
It was a really elegant problem.

why this squable? Between nemesis, thunderbolt, ghostinthemachine , all are all wrong. The poser's solution is also correct, though he has used trigonometry for his proof. But his secretiveness and hiediousness is unbecoming. Ranjan's geometrical proof is more appealing.
THUD & BLUNDER rightly comes to
_______sin(x+20)/sin x =sin 80/sin 40.
This gives cot x =(2cos40-cos20)/sin20.
A little bit of trigonometrical manipulation
gives cot^2(x)= (cot x)^2
={4 cos^2(40)-sin^2(20)-2cos20}/sin^2(20).A few more steps yields
{cot x}^2 = 3 ( exactly)
Hence x=30 deg exactly.
using

Originally posted by rspoddar82
The poser's solution is also correct, though he has used trigonometry for his proof. But his secretiveness and hiediousness is unbecoming.

Hiediousness? LOL Yes, although it is common knowledge that everybody shuns me and are instantly repelled by the sight of me, you have no right to call me 'secretive' as I posted a link to a PDF containing eight different solutions.

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Let angle BAC = 2a
Let angle DCB = b
Let angle EBC = c

Find angle CDE in terms of a, b, and c

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PS 'The word 'hideous' has nothing to do with hiding. Duh.

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Oh god, I thought this was over. I sent my "evidence" to nemesio.
Whom, I'm assuming has worked it out for himself.

For the millionth time NO ONE GUESSED. I solved the problem geometrically and I provided the correct answer on page 1 of this post.

Sorry that I wasn't proactive in providing in advance the "empiracle proof" that only the math geeks would be screaming for. When I posted my answer I expected at least someone to acknowledge that I had the correct answer and if they were interested then ask how I did it. I never expected that I'd have to provide and defend my method this badly. What I expected was for someone to say "great job bro!", and acknowledge the correct answer.

Originally posted by GhostintheShell
Oh god, I thought this was over. I sent my "evidence" to nemesio.
Whom, I'm assuming has worked it out for himself.

For the millionth time NO ONE GUESSED. I solved the problem geometrically and I provided the correct answer on page 1 of this post.

Sorry that I wasn't proactive in providing in advance the "empiracle proof" that only the ...[text shortened]... pected was for someone to say "great job bro!", and acknowledge the correct answer.

Great job, bro! You da man!

But can you finish Part 2 below before your mummy comes to collect you from school?

Let angle BAC = 2a
Let angle DCB = b
Let angle EBC = c

Find angle CDE in terms of a, b, and c

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I'm 33 years old, and no I'm not solving anymore of your puzzles if this is the way you reward the people who solve them, jerk.

4+5 answer is 100deg. duh

Originally posted by richjohnson
As you may remember from mathematics classes, the answer is not as imporant as the method used to arrive at it.

How did you arrive at this conclusion? 😀

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Originally posted by GhostintheShell
I'm 33 years old, and no I'm not solving anymore of your puzzles if this is the way you reward the people who solve them, jerk.

You mean we can't play with your ball anymore?
:'(

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Given side AB=AC and the angle BAC=20deg. the top angle of the isosli triangle angle ACB must = angle ABC and therefore both equal 80deg. since 80+80+20=180.

By drawing a line through the 2 midpoints on AB and AC you create a line parallel to the base, AB and AC are the transverals that cut the parallel lines. So knowing angle ABC=80deg. its supplementary angle is 100 deg. also = to the angle opposite it CDE

hey ghost,...Geometrical proof does not mean drawing a neat diagram using ruler . protractor etc. and then measuring the angle by protractor in the dagram. This is not proof, this is only measurement.

Why don't you read the previous posts. I did draw it out, in order to reason through the puzzle the taking angles I knew, and the geometric principles I could apply to arrive at the exact answer.

I never measured it, and I never guessed it either. I solved the puzzle using only simple additive and subtractive geometry. By mirroring several lines and extending angles you can easily solve this problem using no trig, no protractor, with an elegant and simple method.

If your small mind can't come up with the answer to the problem without breaking out a scientific calculator to do the trig and a bit of simple algebra then that's not my problem.

And yes, I'm taking my ball (brain) away and refusing to play with you.

Were you able to solve it on your own as elegantly as I did or did you break out the calculator?

Originally posted by GhostintheShell
Were you able to solve it on your own as elegantly as I did or did you break out the calculator?

Nobody who knows that sin30 = 1/2 would need a calculator.



Originally posted by GhostintheShell
I'm 33 years old...

So don't forget, big boys don't cry!

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