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 Post subject: L'il quad fun!Posted: Tue, 7 Aug 2012 18:42:45 UTC
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Code:

C

D

A                      E                      B

Quadrilateral ABCD; E on AB: AE = BE = DE = CE = 100.
And AD = CD = 70.

Calculate BC. NO trigonometry allowed!

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 Post subject: Re: L'il quad fun!Posted: Wed, 8 Aug 2012 06:02:22 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6007
Location: 127.0.0.1, ::1 (avatar courtesy of UDN)
Denis wrote:
Code:

C

D

A                      E                      B

Quadrilateral ABCD; E on AB: AE = BE = DE = CE = 100.
And AD = CD = 70.

Calculate BC. NO trigonometry allowed!

Just to be clear --- what do you count as "trigonometry" here, e.g. does area of triangles count?

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 Post subject: Re: L'il quad fun!Posted: Wed, 8 Aug 2012 08:46:01 UTC
 S.O.S. Oldtimer

Joined: Fri, 27 Jul 2007 10:17:26 UTC
Posts: 278
Location: Chandler, AZ, USA
Spoiler:
151 ?

If so here's how:
Spoiler:
Find the length AC (use Pythagoras since AC DE).
AC BC also (AB is the diameter of a circle),
then use Pythagoras to get BC.

Law of cosines confirms my answer.

Great problem!

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 Post subject: Re: L'il quad fun!Posted: Wed, 8 Aug 2012 14:21:50 UTC
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Agree Alstat; in other words:
a = AE = BE = DE = CE = 100.
b = AD = CD = 70.

AC = bSQRT(4a^2 - b^2) / a

OK; we're not done!
Next, get solution using ONLY similar triangles! (no Pythagoras allowed!).

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 Post subject: Re: L'il quad fun!Posted: Fri, 10 Aug 2012 17:49:59 UTC
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Denis wrote:
Next, get solution using ONLY similar triangles! (no Pythagoras allowed!).

Exercise: deduce Pythagoras from similar triangles (and circle theorems if you don't know how to prove them from congruent triangles).

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 Post subject: Re: L'il quad fun!Posted: Fri, 10 Aug 2012 18:54:38 UTC
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outermeasure wrote:
Exercise: deduce Pythagoras from similar triangles (and circle theorems if you don't know how to prove them from congruent triangles).

Not sure what you mean, OM...
My way is to simply extend AD and BC so they meet at F.
Then triangle ABF is similar to triangle CDF.

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 Post subject: Re: L'il quad fun!Posted: Fri, 10 Aug 2012 20:49:48 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
Denis wrote:
outermeasure wrote:
Exercise: deduce Pythagoras from similar triangles (and circle theorems if you don't know how to prove them from congruent triangles).

Not sure what you mean, OM...
My way is to simply extend AD and BC so they meet at F.
Then triangle ABF is similar to triangle CDF.

He's saying "only similar triangles" may as well include Pythagoras' theorem since you can deduce it from similar triangles.

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 Post subject: Re: L'il quad fun!Posted: Fri, 10 Aug 2012 22:17:14 UTC
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CF / DF = AF / AB
CF / 70 = 140 / 200
CF = 49 : BC = 200 - 49 = 151

What's that got to do with deducing Pythagoras' theorem ?

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 Post subject: Re: L'il quad fun!Posted: Fri, 10 Aug 2012 22:23:45 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
Denis wrote:
CF / DF = AF / AB
CF / 70 = 140 / 200
CF = 49 : BC = 200 - 49 = 151

What's that got to do with deducing Pythagoras' theorem ?

Nothing, but saying "using only... and not..." when one could deduce the "and not..." doesn't make sense when one could implicitly use the "and not..." indirectly by proving it using the "using only..."

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 Post subject: Re: L'il quad fun!Posted: Sat, 11 Aug 2012 00:06:36 UTC
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Gotcha! Thanks.

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 Post subject: Re: L'il quad fun!Posted: Wed, 15 Aug 2012 22:27:27 UTC
 Senior Member

Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 129
Location: Hockeytown aka Detroit
Denis wrote:
Code:

C

D

A                      E                      B

Quadrilateral ABCD; E on AB: AE = BE = DE = CE = 100.
And AD = CD = 70.

Calculate BC. NO trigonometry allowed!

Kinda surprised noone mentioned this yet ._.

Spoiler:
You can win with Pythagorean and Ptomley's

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 Post subject: Re: L'il quad fun!Posted: Thu, 16 Aug 2012 04:19:50 UTC
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rdj5933mile5math64 wrote:
You can win with Pythagorean and Ptomley's

Why bother with Pythagoras in the statement? It follows from Ptolemy (and of course you can prove Ptolemy using similar triangles and angle subtended by chords).

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