tag:blogger.com,1999:blog-6933544261975483399.post2953498880406029133..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1292 Square, Circle, Right Triangle, Arc, Inradius, Radius, Incircle, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-86893770234768802202016-12-11T21:32:36.825-08:002016-12-11T21:32:36.825-08:00Thank you for your explanation.
Per your response,...Thank you for your explanation.<br />Per your response, I understand that in your solution, you suppose that<br />. r= incircle radius of triangle QAD not triangle AED per problem statement<br /> . r1= incircle radius of triangle QN’M’ not as per problem statement<br />This special case only happen when E coincide to Q.. in this case triangle AED become AQD and r1= √2. r<br />What happen for general case when E do not coincide to Q ? how about ratio of r1/r in this case ?<br />Please provide more detail <br />See below for the sketch as per your solution<br />https://goo.gl/photos/8FNGp86bfDVgp2jPA<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-32589299861219529222016-12-11T01:53:00.982-08:002016-12-11T01:53:00.982-08:00Question posed from Peter Tran about problem 1292 ...Question posed from Peter Tran about problem 1292 regard last 4 lines of my solution.<br />I suppose to insert circle with radius r1 a r in triangles QN’M’ and QAD the base of which is F and H on N’M’ and AD.<br />AD2= 12+ 12 AD=√2 ; area∆ AQD= (1x1)/2 =1/2 ; p=1+AD/2 = 1+√2/2 = (2+√2)/2; <br />r= area∆AQD/p= 1/2:(2+√2)/2 = 1/(2+√2)<br />N’F=M’F=QF=1 ; QM’=QN’=√2 ; area ∆N’M’Q= (√2x√2)/2 =1 ; p’ of ∆QM’N’=(√2+√2+2)/2=√2+1<br />r1= area∆N’M’Q/p’ = 1/(√2+1); <br />r1/r = 1/(√2+1/:1/(√2+2) = √2 = α<br />This rapport is independent of position of E on circle AQD end is independent of the dimensions of r1 and r<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65734026498100340812016-12-10T20:52:06.904-08:002016-12-10T20:52:06.904-08:00Refer to last 4 lines of your solution.
how do you...Refer to last 4 lines of your solution.<br />how do you get r= 1/(2+√2) and r1= 1/(1+√2) ?<br />Is r and r1 depend on value of AE and ED ? Please explain<br /><br />Peter<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-37436990645425691212016-12-10T15:53:21.789-08:002016-12-10T15:53:21.789-08:00This is a special case of general Sangaku problem ...This is a special case of general Sangaku problem and the general solution is <br />x= r+ 2.delta..(s-d).(s-a)/(e.s)…………..(1)<br />see sketch below for detail<br />http://s25.postimg.org/vvmm8s8xr/General_Sangaku_problem.png<br />In our case we have 2.delta= e.(sqrt(2)-1)<br />replace it in (1) and simplify we get<br />x= r+ (sqrt(2)-1) .(s-a).(s-d)/s ………(2)<br />in right triangle AED (s-a)= ½(-d+e-a) and (s-d)= ½(a+e-d)<br />so (s-a).(s-d)= ¼.(e^2- (d-a)^2)<br />replace e^2= a^2+d^2 in above and simplify<br />we get (s-a).(s-d)= ½. ad= area of triangle ABC= s.r<br />so (s-a)(s-d)/s= r<br />and x= r+(sqrt(2)-1).r= sqrt(2) r<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60311324045163843872016-12-10T15:52:32.992-08:002016-12-10T15:52:32.992-08:00https://goo.gl/photos/jk5hvVT2HwxfZPv9A
Let e=AD,...https://goo.gl/photos/jk5hvVT2HwxfZPv9A<br /><br />Let e=AD, d=AE, a=ED and x= r1<br />Observe that AQED is cyclic and AD and QN are diameters<br />Q, O, M are collinear and EO meet circumcircle of AED at N<br />∠ (QEN)= 90 degrees<br />And QA=QD=AN=ND=e /sqrt(2)<br />We have r=1/2(a+d-e)<br />Apply Ptolemy’s theorem in qua. AQED we have<br />QE.AD + AQ.ED=AE.QD => QE= (d-a)/sqrt(2)<br />Apply Pythagoras’s theorem in tri. QO1E we have<br />(e/sqrt(2)-x)^2=(x.sqrt(2))^2+(d-a)^2/2<br />Replace d^2+a^2=e^2 in above and simplifying we have <br /> .x^2+sqrt(2).e.x –a.d= 0<br />Determinant delta= 2(a+d)^2 after replace e^2=a^2+d^2<br />And positive solution x= sqrt(2).(a+d-e)/2 = sqrt(2).r <br /><br /><br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40323131358232589562016-12-10T14:38:36.586-08:002016-12-10T14:38:36.586-08:00Solution of problem 1292
We call AE=b; ED=c; AD=a;...Solution of problem 1292<br />We call AE=b; ED=c; AD=a; F is the perpendicular to AD from Q that meet circle ABCD<br />K is the point of tangent of circle O1 parallel to AD that meet extension of ED and EA at M and N <br />H is the point of cross AD and QF; AH is the radius of circle AQD; G is the point that cross MN with QF;<br />r1/r= ME/ED=EN/EA=MN/AD= α<br />radius r = area ∆ AED/p in which p= (a+b+c)/2<br />radius r1= area ∆ EMN/p' in which p’= (a+b+c) α/2 and so <br />r1/r= α for all the position in which E is on the circle AQD.<br />We need now to calculate α.<br />If we consider the right triangle AQD and put AQ=QD=QF= 1 = radius of circle ABCD;<br />we have AD= √2; r= 1/(2+√2)<br />M’ and N’ the point in which the tangent to F cross the extension of QD and QA<br />FM’=FN’=1; M’N’=2; QM’=QN’= √2<br />r1= 1/(1+√2) ; r1/r= (2+√2)/(1+√2) ; (r1)2/(r)2= (2+4+4√2)/(1+2+2√2)=2 and so r1/r= α =√2 <br />Anonymousnoreply@blogger.com