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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,732

This thread started on puzzles and games, but bobbym said it was transferred here. Cannot find it so, in case you missed the first problem, I'll repeat it now.

The diagram shows a triangle ABC

with ABC = 80 and ACB = 80

D lies on AC so that DBC = 60

and E lies on AB so that ECB = 50.

To find (by Euclidean geometry) x = EDB

Then bobbym gave me a new puzzle to try.

Given: A triangle ABC with A = 20 and B = C = 80.

E lies on AB so that AE = BC.

Find angle AEC.

It's taken me a while but I've finally got a solution that I think stands up OK. As soon as I've figured how to hide it, I'll post it.

*Last edited by bob bundy (2010-07-11 19:53:52)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,260

Hi bob bundy;

Put it betwen the hide tags.

[ and then "hide" and then ] put your proof here. Then [ , then "\hide" then ]

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,732

Oh ****!!!. That was supposed to be the start diagra, not the solution diagram. And, no matter how many times I edit it out, it's just stuck there now. Sorry.

Here's the diagram I meant to post:

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,260

Hi bob bundy;

I moved my question to here where everyone could see it.

http://www.mathisfunforum.com/viewtopic … 26#p146326

post # 136

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,732

I thought I had a solution but then it evaporated when I realised I'd assumed something that looked right but minus any proof!

But now I think I've cracked it. It's a long proof so apologies. If you've got a quicker way I'd be glad to see it.

` `

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,260

Hi bob bundy;

Looks good from here! Good work!

If you've got a quicker way I'd be glad to see it.

I am looking at a short proof right now, but I do not undertsand it yet. As soon as I can at least follow it, I will post it for you. Please be patient, If I can't get it I will post it anyway because you might follow it and then you can explain it to me.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,732

It would really make my day to explain to you so go ahead and make me happy!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,260

Working to get the picture and post over here!

**In mathematics, you don't understand things. You just get used to them.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,260

Draw isosceles triangle AEQ with AQE = 20°. Since. AE = BC, the latter is equal to ΔABC.

AQ = AC.

CAQ = EAQ - EAC = 80° - 20° = 60°

ΔACQ is equilateral. So:

ΔCQE, EQ = CQ and CQE = 60° - 20° = 40°

CQ = EQ = AQ.

CEQ = (180° - 40°) / 2 = 70°

AEC = 80° + 70° = 150°.

**In mathematics, you don't understand things. You just get used to them.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,732

Hi

That is brief! I think a few added details plus diagram below will make it clear .

Draw perpendicular bisector of AE and place Q on this line so that AQE = 20

This triangle will be isosceles so the bottom angles are both 80.

The base AE = BC so AQE and ABC are congruent triangles.

Hence AB = AC = AQ = EQ

So centre A, radius AC also goes through Q (and B)

QAE = 80 => QAC = 60 making AQC an equilateral triangle.

AQC - AQE = 60 - 20 = 40.

Then in the isosceles triangle QEC the angle QEC = 70.

Add QEA = 80 and you're there.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,260

Hi bob bundy;

Thanks, for providing more. I started to understand the proof when I discovered it was harder to draw a decent diagram than to understand it.

**In mathematics, you don't understand things. You just get used to them.**

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**Mark-nz****Member**- Registered: 2012-10-07
- Posts: 3

Can we assume that if x>1 that ln(x) >x -1 (??)

therefore x > exp(x - 1) = exp(x) / e

x .1 means for all x there must exist p>0 such that x=1+p and substituting this for x we see that

1 + p > exp(1+p)/e = exp(p).e/e = exp(p)

Our assumption produces then 1+p > exp(p) and this is false if p=2,3,4,.......

So assumption is false, proving the result.

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**Mark-nz****Member**- Registered: 2012-10-07
- Posts: 3

Oops..x>1 in line three please...typo!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,260

Hi Mark-nz;

Welcome to the forum. Please start a new thread in "Help Me" for your question.

**In mathematics, you don't understand things. You just get used to them.**

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