Plate notation

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In geometry, the Conway triangle notation, named after John Horton Conway, allows trigonometric functions of a triangle to be managed algebraically. Given a reference triangle whose sides are a, b and c and whose corresponding internal angles are A, B, and C then the Conway triangle notation is simply represented as follows:

S=bcsin⁡A=acsin⁡B=absin⁡C

where S = 2 × area of reference triangle and

Sφ=Scot⁡φ.

in particular

SA=Scot⁡A=bccos⁡A=b2+c2−a22
SB=Scot⁡B=accos⁡B=a2+c2−b22
SC=Scot⁡C=abcos⁡C=a2+b2−c22
Sω=Scot⁡ω=a2+b2+c22      where ω is the Brocard angle.
Sπ3=Scot⁡π3=S33
S2φ=Sφ2−S22SφSφ2=Sφ+Sφ2+S2    for values of   φ  where   0<φ<π
Sϑ+φ=SϑSφ−S2Sϑ+SφSϑ−φ=SϑSφ+S2Sφ−Sϑ

Hence:

sin⁡A=Sbc=SSA2+S2cos⁡A=SAbc=SASA2+S2tan⁡A=SSA

Some important identities:

∑cyclicSA=SA+SB+SC=Sω
S2=b2c2−SA2=a2c2−SB2=a2b2−SC2
SBSC=S2−a2SASASC=S2−b2SBSASB=S2−c2SC
SASBSC=S2(Sω−4R2)Sω=s2−r2−4rR

where R is the circumradius and abc = 2SR and where r is the incenter,   s=a+b+c2   and   a+b+c=Sr

Some useful trigonometric conversions:

sin⁡Asin⁡Bsin⁡C=S4R2cos⁡Acos⁡Bcos⁡C=Sω−4R24R2
∑cyclicsin⁡A=S2Rr=sR∑cycliccos⁡A=r+RR∑cyclictan⁡A=SSω−4R2=tan⁡Atan⁡Btan⁡C


Some useful formulas:

∑cyclica2SA=a2SA+b2SB+c2SC=2S2∑cyclica4=2(Sω2−S2)
∑cyclicSA2=Sω2−2S2∑cyclicSBSC=S2∑cyclicb2c2=Sω2+S2

Some examples using Conway triangle notation:

Let D be the distance between two points P and Q whose trilinear coordinates are pa : pb : pc and qa : qb : qc. Let Kp = apa + bpb + cpc and let Kq = aqa + bqb + cqc. Then D is given by the formula:

D2=∑cyclica2SA(paKp−qaKq)2

Using this formula it is possible to determine OH, the distance between the circumcenter and the orthocenter as follows:

For the circumcenter pa = aSA and for the orthocenter qa = SBSC/a

Kp=∑cyclica2SA=2S2Kq=∑cyclicSBSC=S2

Hence:

D2=∑cyclica2SA(aSA2S2−SBSCaS2)2=14S4∑cyclica4SA3−SASBSCS4∑cyclica2SA+SASBSCS4∑cyclicSBSC=14S4∑cyclica2SA2(S2−SBSC)−2(Sω−4R2)+(Sω−4R2)=14S2∑cyclica2SA2−SASBSCS4∑cyclica2SA−(Sω−4R2)=14S2∑cyclica2(b2c2−S2)−12(Sω−4R2)−(Sω−4R2)=3a2b2c24S2−14∑cyclica2−32(Sω−4R2)=3R2−12Sω−32Sω+6R2=9R2−2Sω.

This gives:

OH=9R2−2Sω.

References

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