## Saturday, June 8, 2013

### Problem 885: Intersecting Circles, Three Tangent Lines, Midpoint

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 885.

1. build :
EM, FE, FH
FHC=FCE=EMC
FEC=FHC => FHC~FCE => HC/CE=FC/FE => HC=CE*FC/FE
EFC=ECM => FEC~CEM => CM/FC=CE/FE => CM=CE*FC/FE
HC=CM => M midPoint of HM

2. Let HF and EM intersect at P.
Apply inversion on C with any inversion radius.
3 tangents are invariants.
Circle G becomes line through E'F' and parallel to H'M'
Similar on other two circles.
Triangle P'H'M' is similar to C'E'F'.
Remainings are straight forward.
QED

3. This might seem a bit strange but here goes. I am a quilter and my guild has presented us with a challenge to create at 20" block using a specific set of requirements. We were given a color, a number and a noun. Mine is three green circles. When I googled that I came up with your proof as a possibility. What is the solution so that I can use that as part of my block? Thanks!

1. To Jennifer
See above two solutions of problem 885
http://gogeometry.blogspot.com/2013/06/problem-885-intersecting-circles.html
Thanks

4. AG и BC перпендикулярно FC, АС и GB перпендикулярно CE, то ACBG параллелограмм. Значит <GAC=<GBC=<GBE. <HAC=2<HFC=2<ECM=<EBM. <HAG=(<HAC=<EBM)+(<GAC=<GBE)=<GBM, HA=AC=GB, и AG=BC=BM, значит что треугольники HAG и GBM конгруэнтный. GH=GM и GC перпендикулярно HM, то есть треугольники GCH и GCM конгруэнтный.

5. AG and BC are perpendicular to FC, AC and GB perpendicular to CE, then ACBG parallelogram. So <GAC = <GBC = <GBE. <HAC = 2 <HFC = 2 <ECM = <EBM. <HAG = (<HAC = <EBM) + (<GAC = <GBE) = <GBM, HA = AC = GB, and AG = BC = BM, means that the triangles HAG and GBM congruent. GH = GM and GC perpendicular to HM, so that triangles GCH and GCM congruent.

6. < FHC = < FCE =< CME , < FCH = < FEC and < CFE = < ECM

So easily Triangles HFC, EFC and ECM are similar

Hence CE/EM = CF/CM….(1) from similar triangles EFC and ECM

And CE/CF = EM/CH…(2) from similar triangles ECM and HFC

Comparing (1) and (2) CM = CH

Sumith Peiris
Moratuwa
Sri Lanka

7. everyone starts with equality of angles fhc,fce and so on. it must be trivial but I can't see the proof for that. illuminate me, please ?

8. everyone begins with equality of angles fhc, fce and so on. I must be missing something trivial but what is the proof for that?

1. < FHC = < FCE because EC tangent to circle A at C

EC is the tangent to circle C at C so < CME = < FCE