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#1 2006-12-07 18:57:40

Toast
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Registered: 2006-10-08
Posts: 1,321

Hard(ish) Geometry Problem 3

Um, this problem I found in the deductive proofs chapter is I think really hard, not quite sure how and where to start...dunno

Thx...

(APE and CPB are similar triangles)

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#2 2006-12-09 02:55:55

krassi_holmz
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Registered: 2005-12-02
Posts: 1,905

Re: Hard(ish) Geometry Problem 3

Here:

Last edited by krassi_holmz (2006-12-09 03:44:52)


IPBLE:  Increasing Performance By Lowering Expectations.

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#3 2006-12-09 03:03:25

Toast
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Registered: 2006-10-08
Posts: 1,321

Re: Hard(ish) Geometry Problem 3

Hey that's the lambda from counter-strike, isn't it?

Last edited by Toast (2006-12-09 03:08:25)

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#4 2006-12-09 03:45:43

krassi_holmz
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Registered: 2005-12-02
Posts: 1,905

Re: Hard(ish) Geometry Problem 3

Sorry for delaymant now it's edited.


IPBLE:  Increasing Performance By Lowering Expectations.

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#5 2006-12-09 03:53:39

Toast
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Registered: 2006-10-08
Posts: 1,321

Re: Hard(ish) Geometry Problem 3

thanks krassi_holmz, ill try to understand that  wink... hehe

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#6 2006-12-09 05:49:32

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Hard(ish) Geometry Problem 3

There must be a mistake.  We are told that AE/ED = m/n, but we already know that E bisects AD.  Thus, AE=ED, so AE/ED = 1.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#7 2006-12-09 18:03:27

Toast
Real Member
Registered: 2006-10-08
Posts: 1,321

Re: Hard(ish) Geometry Problem 3

Yeah, if we know the ratio then what is the use of pronumerals? Sadly I won't know the answer until I go back to school in a little under 2 months...

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#8 2006-12-09 18:55:16

krassi_holmz
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Registered: 2005-12-02
Posts: 1,905

Re: Hard(ish) Geometry Problem 3

First I thought the same, but the proof works for every m,n.
If E bisects AD, then E is (the center of the bisectors) in ΔABD, Let BD intersects with AC in K. Then BK=KD and AK is a bisector of the triangle. Thus AP=2PK. Whe we disect ΔBCD in the same way, we have: KC=3PK, SO AC/AP=(AP+PK+KC)/AP=6PK/2PK=3=2+1=2+n/m.


IPBLE:  Increasing Performance By Lowering Expectations.

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