Geometry Problem

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

Click the figure below to see the problem 716 details.

## Saturday, January 14, 2012

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

Labels:
90,
angle,
center,
degree,
intersecting circles,
perpendicular,
Problem,
radius

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Some clarifications and minor corrections of previous solution:

ReplyDeletehttp://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

Join BA and extend it.

ReplyDeleteDraw 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’

Very good solution Pravin !

ReplyDeletePeter Tran

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

ReplyDelete< 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