请输入您要查询的字词:

 

单词 ProofOfMorleysTheorem
释义

proof of Morley’s theorem


The scheme of this proof, due to A. Letac,is to use the sines law to get formulas for the segmentsAR, AQ, BP, BR, CQ, and CP,and then to apply the cosines law to the triangles ARQ, BPR, and CQP,getting RQ, PR, and QP.

To simplify some formulas, let us denote the angle π/3, or 60 degrees,by σ.Denote the angles at A, B, and C by 3a, 3b, and 3c respectively,and let R be the circumradiusMathworldPlanetmath of ABC. We haveBC=2Rsin(3a). Applying the sines law to the triangle BPC,

BP/sin(c)=BC/sin(π-b-c)=2Rsin(3a)/sin(b+c)(1)
=2Rsin(3a)/sin(σ-a)(2)

so

BP=2Rsin(3a)sin(c)/sin(σ-a).

Combining that with the identity

sin(3a)=4sin(a)sin(σ+a)sin(σ-a)

we get

BP=8Rsin(a)sin(c)sin(σ+a).

Similarly,

BR=8Rsin(c)sin(a)sin(σ+c).

Using the cosines law now,

PR2=BP2+BR2-2BPBRcos(b)
=64sin2(a)sin2(c)[sin2(σ+a)+sin2(σ+c)-2sin(σ+a)sin(σ+c)cos(b)].

But we have

(σ+a)+(σ+c)+b=π.

whence the cosines law can be applied to those three angles, getting

sin2(b)=sin2(σ+a)+sin2(σ+c)-2sin(σ+a)sin(σ+c)cos(b)

whence

PR=8Rsin(a)sin(b)sin(c).

Since this expression is symmetricMathworldPlanetmath in a, b, and c, we deduce

PR=RQ=QP

as claimed.

Remarks: It is not hard to show that the triangles RYP, PZQ,and QXR are isoscoles.

By the sines law we have

ARsinb=BRsina  BPsinc=CPsinb  CQsina=AQsinc

whence

ARBPCQ=AQBRCP.

This implies that if we identify the various verticeswith complex numbersMathworldPlanetmathPlanetmath, then

(P-C)(Q-A)(R-B)=-1+i32(P-B)(Q-C)(R-A)

provided that the triangle ABC has positive orientation, i.e.

(C-AB-A)>0.

I found Letac’s proof athttp://www.cut-the-knot.org/triangle/Morley/index.shtmlcut-the-knot.org,with the reference Sphinx, 9 (1939) 46.Several shorter and prettier proofs of Morley’s theorem can also beseen at cut-the-knot.

随便看

 

数学辞典收录了18232条数学词条,基本涵盖了常用数学知识及数学英语单词词组的翻译及用法,是数学学习的有利工具。

 

Copyright © 2000-2023 Newdu.com.com All Rights Reserved
更新时间:2025/5/4 5:02:12