Geometry Problem

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

Click the figure below to see the complete problem 660.

## Thursday, August 25, 2011

### Problem 660: Triangle, Circle, Diameter, Perpendicular Lines

Labels:
circle,
diameter,
perpendicular,
triangle

Subscribe to:
Post Comments (Atom)

Here’s a proof by brute force. Draw a line // DF from A meeting EF & GD in M & N resply. Now by similar triangles, AM=GE*AF/FG while AN =AG*DF/FG. Now let A be the origin, AC the x-axis and AM, the y-axis so that we can write eqns. for EF & GD as follows:x/AF+y*FG/(GE*AF)= 1 and x/AG+y*FG/(AG*DF)=1 from which we can eliminate y to obtain the x-co-ordinate of H the intersection of EF & GD as x= (GE*AF–AG*DF)/(GE-DF.Now let /_DAC= α=90–C and /_CAE=β=180–A. Hence we can say that:x=[b^2 sinβcosβ(cosα)^2-b^2 sinαcosα(cosβ)^2]/(sinβcosβ-sinαcosα)=2bcosβcosαsin(β-α)/(sin(2β)–sin(2α))

ReplyDelete=bsinC*cosA/sin(B) using standard trigonometric formulae.Further, c/sinC = b/sinB by the sine rule hence, x = c*cosA. In other words, the x-co-ordinate of H is the same as that of B or HB is perpendicular to CA, our x-axis.

Ajit

http://img220.imageshack.us/img220/3329/problem660.png

ReplyDeleteHere is the geometry solution.

Connect EC and AD.

EG cutt DC at M and DF cut AE at N

Let H’ and B’ are the projections of H and B over AC

Note that (ADF)=(ACD) and (GCE)=(FNA) (see picture)

Triangles AFD similar to triangle GEC and AF/FD=GM/GC ( 1)

Triangles AFN similar to triangle EGC and AF/FN=GE/GC ( 2)

Divide expressions (1) to (2) we have FN/FD=GM/GE (3)

From (3) we also get FD/GE=ND/ME=k

Triangle HFD similar to triangle HEM with ratio of similarity = k so H’F= k/(1-k) x FG

Triangle BND similar to triangle BEM with ratio of similarity = k so B’F= k/(1-k) x FG

H’ coincide with B’ and HB perpendicular to AC

Peter Tran

Extend DA, CE to meet at X. Join BX.

ReplyDeleteExtend CA to meet BX at Y.

AD ⊥ BC and CE ⊥ AB

Follows X is the orthocentre of ∆ABC.

So BYX ⊥ AC

Let Z be the midpoint of AC.

Clearly D, Y, E and Z are concyclic (∵they lie on the ninepoint circle of ∆ABC)

So ∠DYZ = ∠DEZ

= ∠EDZ (∵DZ = EZ)

= ∠EYZ

Same as ∠DYF = ∠EYG

Follows ∆DYF ∼ ∆EYG

Hence YF/YG = DF/GE = HD/HG (∵FD ∥ GE)

Follows HY ∥ DF or HY ⊥ AC

Already BY ⊥ AC

Hence HB ⊥ AC

http://s1092.photobucket.com/albums/i418/vprasad_nalluri/?action=view¤t=GoGeom660.jpg

ReplyDeleteHallo Peter Tran, can you exlain why

ReplyDeleteFrom (3) we also get FD/GE=ND/ME=k. Thank's

To Anonymous

ReplyDeleteSee details below.

From (3) FN/FD=GM/GE

we have FN/GM=FD/GE=(FN+FD)/(GM+GE)=ND/ME (Properties of ratio)

Peter Tran

http://hiphotos.baidu.com/gzyoushi/pic/item/18fa9d6d3f5046964216944a.jpg

ReplyDeleteCHINESE SORRY