Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
HBMO1 is a rectangle.HD = EM, OD = OE imply ΔHDO and ΔMEO are congruent.So <DHO = <EMO.Also <BHM = <BMH = ½ HO1M = 45 deg.Follows HBMO1 is a square.So <DOH = <EOM = 45 deg and H, O, M are collinear.Implies HBMO1 is a square and O is its center.Hence x = O1M = 2 OE =2r = b + c – a.
Referring to lines " Also <BHM = <BMH = ½ HO1M = 45 deg.Follows HBMO1 is a square.So <DOH = <EOM = 45 deg and H, O, M are collinear."Note that <DOH = <EOM = 45 deg if H, O, M are collinnear . Please provide explanation.Note that the solution never use given data " circle O, tangent to circle Q" . Please check it.
http://s25.postimg.org/59a5jt4qn/pro_1104.png This is a special case of general Sangaku problem andthe general solution is x= r+ 2.d.(s-a).(s-c)/(b.s)…………..(1)see sketch below for detailhttp://s25.postimg.org/vvmm8s8xr/General_Sangaku_problem.pngIn our case we have d= ½. b. replace it in (1) and simplify we getx= r+ (s-a).(s-c)/s ………(2)in right triangle ABC (s-a)= ½(-a+b+c) and (s-c)= ½(a+b-c)so (s-a).(s-c)= ¼.(b^2- (a-c)^2)replace b^2= a^2+c^2 in above and simplifywe get (s-a).(s-c)= ½. ac= area of triangle ABC= s.rso (s-a)(s-c)/s= r and x= 2.r= a+c-b
Draw OL // BC where L is on O1MApply Pythagoras to right Tr. O1OL(b/2 - x)^2 = (x-c/2)^2 + (a/2-x)^2 from which we have upon simplification using b^2=c^2 + a^2, x = a+c-b which is anyway = to 2r since BDOE is a square of side r and AF = c-r and FC = a-rSumith PeirisMoratuwaSri Lanka