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Denis
 Post subject: L'il quad fun!
PostPosted: 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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outermeasure
 Post subject: Re: L'il quad fun!
PostPosted: Wed, 8 Aug 2012 06:02:22 UTC 
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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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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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alstat
 Post subject: Re: L'il quad fun!
PostPosted: Wed, 8 Aug 2012 08:46:01 UTC 
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Is the answer
Spoiler:
151 ?


If so here's how:
Spoiler:
Find the length AC (use Pythagoras since AC \perp DE).
AC \perp 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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Denis
 Post subject: Re: L'il quad fun!
PostPosted: 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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outermeasure
 Post subject: Re: L'il quad fun!
PostPosted: 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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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Denis
 Post subject: Re: L'il quad fun!
PostPosted: 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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Shadow
 Post subject: Re: L'il quad fun!
PostPosted: Fri, 10 Aug 2012 20:49:48 UTC 
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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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Denis
 Post subject: Re: L'il quad fun!
PostPosted: 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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Shadow
 Post subject: Re: L'il quad fun!
PostPosted: Fri, 10 Aug 2012 22:23:45 UTC 
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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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Denis
 Post subject: Re: L'il quad fun!
PostPosted: Sat, 11 Aug 2012 00:06:36 UTC 
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Gotcha! Thanks.

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rdj5933mile5math64
 Post subject: Re: L'il quad fun!
PostPosted: Wed, 15 Aug 2012 22:27:27 UTC 
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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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outermeasure
 Post subject: Re: L'il quad fun!
PostPosted: 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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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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