Geometry Problem

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

Click the figure below to see the complete problem 829.

## Friday, November 30, 2012

### Problem 829: Triangle, Circumcircle, Parallel, Chord, Side, Metric Relations

Labels:
chord,
circumcircle,
metric relations,
parallel,
triangle

Subscribe to:
Post Comments (Atom)

x=5sqrt(5/2)

ReplyDeleteDenote BF, BG by p & q resply. Use Stewart's Theorem to obtain: 125+x^2=6(p^2+5), 50+4x^2=6(q^2+8)and p/(p+5/p) =q/(q+8/q).

ReplyDeleteSolve simultaneously to get x=5√(2.5) --- same as Atharva

Since DE//AC so Arc(AD)=Arc(EC) => angle (ABF)=angle(CBG)

ReplyDeleteSo in triangle ABC ,line BF and BG are isogonal to each other.

Per Steiner’s theorem for isogonal lines BF and BG

(AF/FC).(AG/GC)=AB^2/BC^2

So 1/5 . 4/2 =25/x^2

So x=5.sqrt(5/2)

Problem 8Since ACED is an isosceles trapezoid, AE = CD = y (say).

ReplyDeleteLet BF = p, DF = q, BG = u and GE = v

From similar triangles,

∆ABF & ∆DCF ; y/5 = q/1 = 5/p…..(1) and

∆BCG & ∆AEG; y/x = v/2 = 4/u ….(2)

Now since AC//DE ; p/q = u/v and substituting for p, q, u & v from (1) & (2) we have

(25/y)/(y/5) = (4x/y)/(2y/x) or x2 = 125/2 which gives us x = 5√(5/2)

Sumith Peiris

Moratuwa

Sri Lanka