tag:blogger.com,1999:blog-6933544261975483399.post5847590890275794725..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 371: Square, Inscribed circle, Triangle, AreaAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-6933544261975483399.post-29112268274164518792020-11-24T11:24:10.909-08:002020-11-24T11:24:10.909-08:00See graph here.
∠CDF = ∠DAG ⇒ ∆DMG and ∆ADG are s...See graph <a href="https://drive.google.com/file/d/1zvOx8LHRs8pcbhpH53MdLVu7xx5tU9DP/view?usp=sharing" rel="nofollow"> here</a>.<br />∠CDF = ∠DAG ⇒ ∆DMG and ∆ADG are similar ⇒ ∆GMN is rectangle in M<br />∆ENG rectangle in N and ∠ENF = 45° ⇒ ∆GMN is isoscele in M and MN = MG = DG/sqrt(5) <br />Hence S1 = 1/2.MG.MN = DG∧2/10 = S/40 QED<br />Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35481957922279476722019-01-31T05:15:04.869-08:002019-01-31T05:15:04.869-08:00Join O to G. FD meet OG at P. ∆AMD ~ ∆MGD => ˂ ...Join O to G. FD meet OG at P. ∆AMD ~ ∆MGD => ˂ M = 90°<br />From similarity if MG = x => MP = x/2, MD = 2x<br />=> S(PGD) = 1/2(xˑ5x/2) = 1/16 S<br />=> 1/2 (x²) = 1/40 Sc.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4365489444129222192019-01-24T16:31:36.463-08:002019-01-24T16:31:36.463-08:00For ease of calculations, let the length of the sq...For ease of calculations, let the length of the square be 2a units<br />Since ADG congruent to DCF & AD perpendicular to DC => AG perpendicular to FD => NMG is a right triangle --------(1)<br />Let m(MDG)=α and connect EG,FH and EN<br />similarly m(AGE)=α and m(HFD)=α (since AGE,HFD,DAG & CDF are congruent triangles)<br />since m(HEG)=45 and m(HEN)=m(HFN)=m(HFD)=α => m(NEG)=45-α ---------(2)<br />m(ENG)=90 (angle in semi-circle) ---------(3)<br />From (2) & (3) m(EGN)=45+α<br />=> m(MGN) = 45 (as m(AGE)=α)<br />Hence (1) is an isosceles right triangle and hence MG=MN ----------(4)<br /><br />Denote I as point of intersection of EN and AG & observe that triangles GEI and DNG are similar<br />Since EG=2.DG<br />=> GI=2.DN<br />Since m(ENF)=m(INM)=45 and NM perpendicular to IG=> M is midpoint of IG and MG=MN=ND<br />Apply pythogrous to MGD => MG=MN=a/sqrt(5)<br />Area of MGN=S1=a^2/10= 4a^2/40=S/40<br />=>S=40.S1 Q.E.D<br />Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57328953465728849212019-01-22T08:55:36.385-08:002019-01-22T08:55:36.385-08:002nd Pure Geometry Solution, much easier
< NGD ...2nd Pure Geometry Solution, much easier<br /><br />< NGD = < NFG = < MCG since FMGC is concyclic as a result of congruent triangles AGD & FCD<br /><br />So CM // GN & N is the mid point of DM<br /><br />So S1 = S(MGD)/2 = S(FCD)/10 since Tr.s MGD & FCD are similar & FD^2 = 5.GD^2<br /><br />But S(FCD) = S/4 and the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39514437421531991022016-10-15T07:48:51.262-07:002016-10-15T07:48:51.262-07:00Pure Geometry solution
Let AB = a and MG = p
Tr....Pure Geometry solution<br /><br />Let AB = a and MG = p<br /><br />Tr.s FCD and AGD are congruent SAS<br /><br />Hence < GDM = < GAD and so < GMD = 90.<br /><br />So p = AG/GD^2 = a /(2sqrt5)<br /><br />< CGF = 45 hence < MNG = 45<br /><br />So S1 = p^2/2 = a^2/(2X(2sqrt)^2) = a^2/40<br /><br />Hence S = 40S1<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83641929614540292582009-11-13T20:39:33.386-08:002009-11-13T20:39:33.386-08:00http://i026.radikal.ru/0911/b5/5341e5c9858f.jpg
S...http://i026.radikal.ru/0911/b5/5341e5c9858f.jpg<br /><br />Solution without analitycal geometry. Fales theorem - main idea.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38174058085429089162009-10-26T20:38:43.095-07:002009-10-26T20:38:43.095-07:00How about thia for a plane geometry solution?
Let ...How about thia for a plane geometry solution?<br />Let the square side=2a. Tr. GAD & Tr. FDC are congruent and hence angle GAD = ang. FDC. But ang. FDC + ang. ADF = 90 and hence ang. GAD + ang. ADF = 90 and, therefore, ang. AMD = 90. Now AG=V(4a^2+a^2)=V5a=FD where V = square root. Further, AD*DG/AG =MD which gives us MD=2a/V5. Moreover, ND*DF=DG^2 or ND =a^2/(V5a)=a/V5 or N bisects MD which makes Tr.GMN=Tr.GMD/2 -----(1).<br />Now Tr. GMD and Tr. AFC are similar and hence Tr.GMD/a^2=(2a/V5)^2/4a^2 or Tr. GMD =4a^2/20 which makes Tr. GMN= 4a^2/40 by (1) above or S1=S/40, as required<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26848767221870574202009-10-25T06:12:59.503-07:002009-10-25T06:12:59.503-07:00Some ideas:
Right triangles
Congruent triangles
Si...Some ideas:<br />Right triangles<br />Congruent triangles<br />Similar triangles<br />Angles in a circle<br />Area of a triangle and squareAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-30728638153845514862009-10-24T20:35:20.159-07:002009-10-24T20:35:20.159-07:00Let O be (0,0) and G be (a,0). This makes F:(0,a),...Let O be (0,0) and G be (a,0). This makes F:(0,a), A:(-a,-a) and G:(a,-a). Determine equations of FD and AG as 2x + y = a & x - 2y = a. Their intersection is M: (3a/5, -a/5). The circle at O is x^2 + y^2 = a ^2 and its intersection with FD gives us N as (4a/5,-3a/5). <br />Area of a triangle with vertices at (x1,y1), (x2,y2) and (x3,y3) is given by: (1/2)[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]. Plugging in the values, we get area of tr. GMN as (1/2)[a(-a/5+3a/5)+(3a/5)(-3a/5)+(4a/5)(a/5)]= a^2/10 = 4a^2/40.<br />In other wods, S1 = S/40<br />Antonio, how may one solve this without using analytical geometry?<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com