Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view the complete problem 989
Wednesday, March 5, 2014
Geometry Problem 989: Arbelos, Semicircles, Diameter, Perpendicular, 90 Degree, Common External Tangent, Rectangle, Midpoint of Arc
Subscribe to:
Post Comments (Atom)
http://s25.postimg.org/vsrf9vrnj/Prob_989.png
ReplyDeleteLet M, L, N, O are points shown on the sketch
1. We have MG⊥GH and NH⊥GH
Since LG=LB=LH=> L is the midpoint of GH
∠(GBH)= ½(∠GMB+∠BNH)= 90
And ∠ (MLN)=1/2(∠GLB)+ ∠BLH)=90
In right triangle MLN , BL^2=BM.BN
In right triangle ADC , DB^2=BA.BC= 4. BM.BN=> L is the midpoint of BD
And DGBH is a parallelogram with right angle ∠GBH => DGBH is a rectangle
2. Connect AG and CH
We have ∠ (AGB=∠ (BGD)= 90 => A,G,D are collinear
Triangles AMG and AOD are isosceles with common angle ∠GAM
So ∠ (AGM)= ∠ (ADO) => OD//MG => OD ⊥ EF
So D is the midpoint of arc EF
I have a comment on the second part.
ReplyDeleteDO || HN, and they are perpendicular to EF.
So DO bisects both chord EF and arc EF.
Thank you for your comment on my solution.
DeletePeter
Because <GAC=<FHC, AGHC is cyclic. Let AG and HC meet at D'. <D'AC+<D'CA=90, so D' must be on big circle. Furthermore, because D'G*D'A=D'H*D'C from cyclic property of AGHC, D' must also be on radical axis of 2 small circles, which is their common tangent. Therefore, D' coincides with D. It is easy to see that GDHB has 3 right angles from Thales theorem in each circle and must therefore be rectangle.
ReplyDelete<GAC=<FHC=arcDF/2+arcFC/2=arcDE+arcFC/2.
This means that arcDF=arcDE so D is midpoint of arcEF.
Draw a circle with diameter BD. It meets (AB) in X and (BC) in Y.
ReplyDelete<BXD=<BYD=90 deg by Thales' theorem.
<BXA=<BYC=90 deg by Thales' theorem.
So X lies on AD and Y lies on BD.
Thales again: <XDY = <ADC = 90 deg.
Now note that <BXY = 90 deg - <BDX = <BAX.
So XY is tangent to the (AB) at X and similarly tangent to (BC) at Y. So X=G and Y=H.
By similarity the tangent to the (AC) at D is parallel to GH. Let M be the midpoint of {AC), then MD is perpendicular to EF, hence bisects EF and bisects angle EMF, and D is midpoint of arc EF.
Let GH, BD cut at X. Let O be the centre of AC and let OD cut GH at Y.
ReplyDeleteLet AB = 2a, BC = 2b and let GH = 2t,
Now XG = XB = XH = t. Also GF^2 = (a-b)^2 + (a+b)^2 = 4ab = 4t^2 = BD^2 hence BX = FX = DX = GX = t. It follows that BFDG is a rectangle
If < DAO = p, < DOB = p = < BXH
Therefore OYXB is concyclic and so OY is perpendicular to EG
It follows that EY = YF and hence ED = DF which is the result we need
Sumith Peiris
Moratuwa
Sri Lanka
https://photos.app.goo.gl/Ps4iad2ZZNznvwP17
ReplyDelete