tag:blogger.com,1999:blog-6933544261975483399.post6805341956014917255..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 331. Square, Point on the Inscribed Circle, Tangency PointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-64715307030709373182010-11-27T04:04:45.442-08:002010-11-27T04:04:45.442-08:00It is well known that :
If Z is the midpoint of a ...It is well known that :<br />If Z is the midpoint of a segment XY with XZ = ZY = a, and P is any point not collinear with X,Z,Y then<br />PX^2 + PY^2 = 2[PZ^2 + a^2] or equivalently,<br />2PZ^2 = PX^2 + PY^2 - 2a^2<br />In the present problem,<br />let 2k be the length of each side of the square.<br />For convenience, denote the distances of P from A,B,C,D by a,b,c,d respectively<br />2PH^2 = a^2 + d^2 - 2k^2,<br />2PE^2 = a^2 + b^2 - 2k^2,<br />2(PH^2 - PE^2) = d^2 -b^2 = PD^2 - PB^2 etcPravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-76429813666864698872009-09-01T20:30:45.642-07:002009-09-01T20:30:45.642-07:00Isn't the distance formula in analytical geome...Isn't the distance formula in analytical geometry nothing but simple Pythagoras?<br />VihaanVihaanhttps://www.blogger.com/profile/14122321143712979648noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-80119985692544325602009-08-16T11:42:52.044-07:002009-08-16T11:42:52.044-07:00Assume that P is at x distance from line BC and at...Assume that P is at x distance from line BC and at y distance from DC and side of square be a. Now if you just apply simple pythagoras and just simlply put the values you will get PH^2 - PE^2 = a(y-x) and also PD^2 - PB^2 = 2a(y-x).shaileshhttps://www.blogger.com/profile/13003995866362418349noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33154224911570645692009-07-31T00:43:33.807-07:002009-07-31T00:43:33.807-07:00Let O be (0,0) and the side of the square=2a then ...Let O be (0,0) and the side of the square=2a then A:(-a,-a),B:(-a,a),C:(a,a) & D:(a,-a) while E:(-a.0),F:(0,a),G:(a,0) & H:(0,-a) and the radius of the inscribed circle=a. Now using the distance formula it can be shown that PH^2-PE^2=2a(y-x) while PD^2-PB^2=4a(y-x) from which we deduce that: (PD^2-PB^2)/2 = PH^2-PE^2<br />Vihaan: vihaanup@gmail.comVihaanhttps://www.blogger.com/profile/14122321143712979648noreply@blogger.com