easy little geometry problem

[futurebird]

Okay, I’m still thinking about my constrained version of the traveling salesmen problem. Here is an easy little geometry problem:

Triangle ABC has sides:

AB = p
BC = q
CA = r

Triangle ABD has sides

AB = p (duh)
BD = s
AD = t

Express the two possible lengths of DC in terms of p, q r, s and t

I’ve done it one way, but it looks like a big mess.

Comments

[never_the_less]

uf! I could give it a try but yeah, it's going to hve lots of tan/sin/cos and it's 1:50 am. mostly i just wanted to say hi, since i just added you. i find your journal pleasantly refreshing and earnest (in the best possible way), so i'm signing up. plus you are a runner!

[turingtest]

I think that forcing it to be done in terms of p,q,r,s,t will always force it to be messy, maybe like this:

CD = [ r2 + t2 - 2rt * cos( ÐCAB +/- ÐBAD ) ]1/2

where:

cos( CAB +/- BAD ) = [ cos( ÐCAB ) * cos( ÐBAD ) ] -/+ [ sin( ÐCAB ) * sin( ÐBAD ) ]

and:

cos( ÐCAB ) = (r2 + p2 - q2) / 2rp

cos( ÐBAD ) = (t2 + p2 - s2) / 2tp

sin( ÐCAB ) = [ 1 - cos2( ÐCAB ) ]1/2

sin( ÐBAD ) = [ 1 - cos2( ÐBAD ) ]1/2

Just Plug-n-Chug™ with the first equation above - not pretty!

While a nice trig exercise, an easier way to do this would probably be to set up a coordinate system and use linear algebra - all of this mess is just a clunky way to express simple vector addition.