Saturday, January 14, 2012

Problem 716. Intersecting Circles, Center, Radius, Perpendicular, 90 Degrees

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

Click the figure below to see the problem 716 details.

1. Some clarifications and minor corrections of previous solution:
http://img849.imageshack.us/img849/782/problem716.png

Draw diameter FK of circle O’ . Connect KA and KB
KA perpen to FA , KB perpen to FB
From G &H draw GA’//KA and HB’//KB ( see picture) . GA’ cut FK at N
Note that GA’ perpen to FA , HB’ perpen to FB

1. Triangle FAG congruence to FBH (case AA)
Let FG/FA=FH/FB= k and m(HFB)=m(GFA)=x
So FA’/FA= FG/FA . cos(x)= k.cos(x)= FN/FK
And FB’/FB=FH/FB. cos(x)= k.cos(x) =FA’/FA=FN/FK
So GA’ and HB’ will intersect at point N on FK (properties of // lines and congruence triangles)

2. In triangle GHN, HA’ and GB’ are altitudes and F will be the orthocenter of triangle GHN so O’F perpen To GH
With the same reason as part 1, OC will perpen to ED and L will be the orthocenter of triangle C’MF’
And ML will perpen to C’F’
Peter Tran

2. Join BA and extend it.
Draw tangent at F to circle (O’) to meet BA extended at X.
Let tangent XF be extended to meet C’F’ at Y.
∠YFB = ∠FAB (angle in the alternate segment).
= ∠HAB = ∠HGB (angles in the same segment).
∴ XY ∥ MH.
In turn this implies F’O’F ⊥ MH.
By a similar argument C’OC ⊥ MD.
∴ L is the orthocenter of ∆ MC’F’ and
Hence ML ⊥ C’F’

3. Very good solution Pravin !
Peter Tran

4. Let CC’ meet circle O at P and DE at Q. Let F’F meet HG at R.

< EDA = < CBA = < CPA, hence APDQ is concyclic and so < PQF’ = < PAC = 90.

Similarly we can show that < C’RF = 90

Hence L is the orthocenter of Triangle MC’F’ and so ML is perpendicular to C’F’

Sumith Peiris
Moratuwa
Sri Lanka