Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1019.
Algebra solution :Let C’ is the point of sysmetry of C over AB => AC’=ACLet α=∠(BAD) , AB= 2.R and AH= 2pWe have AD= 2R.cos(α) , DF=2R.cos(α) ^2 +2pSince AD=DF => 2.R.cos(α)= 2p+2Rcos(α)^2 => R.cos(α)^2-R.cos(α)+p=0Let α1 and α2 are solutions of this quadratic equationSummation of solutions= 1 => cos(α1)+cos(α2)=1=> AD/AB+AC’/AB=1Or AB=AD+AC
Consider: R=DA, r = CA, 2a = AB(where O is the center)Join B and D ->Triangle BDA is right triangleJoin B and C ->Triangle BCA is right triangleDraw a radius DF (DF is perenpedicular to MN)Draw a radius CE (CE is perenpedicular to MN)Draw an tangent to circle O threw A in that way that it cut DF at L and CE at M thus MEFL is rectangle.Triangle DAL is similiar to BDA and so MAC to BCA (because they has 90 angle of MEFL and becasue of the tangent : R/2a = DL/R (triangles DAL and BAD) - >DL = R^2/2a THUS LF = R - R^2/2a AC/BA = MC/AC -> r/2a = MC/r (triangles CAM and BCA) - >MC = r^2/2a THUS ME = r - r^2/2aSince MEFL is rectangle we get that ME = LF:r - r^2/2a = R - R^2/2a after short develop we get2ar - r^2 = 2aR - R^2R^2 - r^2 = (R - r)(R+r) = 2a(R - r) thus 2a = R+r - > AB = DA + AC
Choose midpoint of AH to be origin and BH as y-axis.If a=AH/2 and r=AB/2, parabola 4ay=x^2 cuts circle x^2+(y-(a+r))^2=r^2 at C and D==>4ay+y^2-2y(a+r)+(a+r)^2=r^2,y^2-2y(r-a)+a^2+2ar=0,CE+DF=(sum of y-coordinates of C and D)+2a=2(r-a)+2a=2r=AB
Let AC = a and AD = b and AB = c and AH = xUsing Pythagoras EH^2 = a^2-(a-x)^2 = a^2(c^2-a^2)/c^2 since EH.AB = BC.CASo (a-x)^2 = a^4/c^2Hence a-x = a^2 /c......(1)Similarly b-x = b^2/c.....(2)(1) - (2) a-b = (a^2-b^2)/cDividing both sides by a-b,c = a+bSumith Peiris MoratuwaSri Lanka
Note that E,A,D are collinear as well as F,A,C though this fact was not used in the above proof.
Another problem from the above.,,,Prove that 1/AH = 1/AC + 1/AD
Also EH/ BC + FH/BD = 1