Geometric problem
in [Math]
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From: José Carlos Santos on 21 Jul 2010 10:18 Hi all, Let P, P_1 and P_2 be three non-collinear points of the plane such that the angle P_1-P-P_2 is acute. Let Q_1 be the orthogonal projection of P_1 on the line defined by P and P_2 and let Q_2 be the orthogonal projection of P_2 on the line defined by P and P_1. Problem: to prove that Q_1 belongs to the line segment whose endpoints are P and P_2 or that Q_2 belongs to the line segment whose endpoints are P and P_1. Analytically, this is easy. One can assume wlog that P = (0,0). If _v_ is the vector PP_1 and _w_ is the vector PP_2, then what I want to prove is that <v,w>/||w|| <= ||w|| or <v,w>/||v|| <= ||v||; in other words, I want to prove that <v,w> <= ||w||^2 or <v,w> <= ||v||^2. This is an easy consequence of the Cauchy-Schwarz inequality. Can anyone provide a geometric proof? Best regards, Jose Carlos Santos
From: Rob Johnson on 21 Jul 2010 12:25 In article <8aodtcFv16U1(a)mid.individual.net>, Jose Carlos Santos wrote: >Let P, P_1 and P_2 be three non-collinear points of the plane such that >the angle P_1-P-P_2 is acute. Let Q_1 be the orthogonal projection of >P_1 on the line defined by P and P_2 and let Q_2 be the orthogonal >projection of P_2 on the line defined by P and P_1. Problem: to prove >that Q_1 belongs to the line segment whose endpoints are P and P_2 or >that Q_2 belongs to the line segment whose endpoints are P and P_1. > >Analytically, this is easy. One can assume wlog that P = (0,0). If _v_ >is the vector PP_1 and _w_ is the vector PP_2, then what I want to >prove is that <v,w>/||w|| <= ||w|| or <v,w>/||v|| <= ||v||; in other >words, I want to prove that <v,w> <= ||w||^2 or <v,w> <= ||v||^2. This >is an easy consequence of the Cauchy-Schwarz inequality. > >Can anyone provide a geometric proof? WLOG, assume segment(P1,P) is no longer than segment(P2,P). Then, since the hypotenuse is the longest side of a right triangle, we have that segment(Q1,P) is no longer than segment(P1,P) which by assumption is no longer than segment(P2,P). Thus, Q1 is on segment(P2,P). Does that suffice? Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: José Carlos Santos on 21 Jul 2010 19:34 On 21-07-2010 17:25, Rob Johnson wrote: >> Let P, P_1 and P_2 be three non-collinear points of the plane such that >> the angle P_1-P-P_2 is acute. Let Q_1 be the orthogonal projection of >> P_1 on the line defined by P and P_2 and let Q_2 be the orthogonal >> projection of P_2 on the line defined by P and P_1. Problem: to prove >> that Q_1 belongs to the line segment whose endpoints are P and P_2 or >> that Q_2 belongs to the line segment whose endpoints are P and P_1. >> >> Analytically, this is easy. One can assume wlog that P = (0,0). If _v_ >> is the vector PP_1 and _w_ is the vector PP_2, then what I want to >> prove is that<v,w>/||w||<= ||w|| or<v,w>/||v||<= ||v||; in other >> words, I want to prove that<v,w> <= ||w||^2 or<v,w> <= ||v||^2. This >> is an easy consequence of the Cauchy-Schwarz inequality. >> >> Can anyone provide a geometric proof? > > WLOG, assume segment(P1,P) is no longer than segment(P2,P). Then, > since the hypotenuse is the longest side of a right triangle, we > have that segment(Q1,P) is no longer than segment(P1,P) which by > assumption is no longer than segment(P2,P). Thus, Q1 is on > segment(P2,P). > > Does that suffice? Indeed! Thanks a lot. Best regards, Jose Carlos Santos
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