Friday, October 23, 2015

Geometry Problem 1156: Square, Tangent Line, 90 Degrees, Metric Relations

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1156: Square, Tangent Line, 90 Degrees, Metric Relations.

Posted by Antonio Gutierrez at 9:47 AM
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8 comments:

  1. Peter TranOctober 23, 2015 at 12:37 PM

    Let u= ∠(BCE) and t= tan(u)
    We have ET=x.tan(u)=3 ..... (1)
    DG= x/cos(90-2.u)-x= x/sin(2u)-x
    And ET/DG=3= tan(u)/(1/sin(2u)-1)
    Replace sin(2u)=2t/(t^2+1) in above
    We get 2.t^2/(t-1)^2=3 so t= √ (6)/( √ (6)+2)
    Replace value of t in (1) we get x= 3+√ (6)

    ReplyDelete
  2. AjitOctober 23, 2015 at 11:15 PM

    Let the intersection of EG & AD be called F and let FT=FD=y. From similar triangles FEA & FDG we can say that (x-3)/1=(x-y)/y and right triangle FEA gives us, (3+y)²=(x-3)²+(x-y)². Solving these two equations, we obtain the only admissible value of x as (3 +√6)

    ReplyDelete
  3. Sumith PeirisOctober 24, 2015 at 2:19 AM

    Purely Geometry solution

    Let TG = p

    Then from Pythagoras p^2 + x^2 = (x+1)^2

    So p^2 = 2x+1. (1)

    Now write the area of Tr. CEG in 2 ways

    1/2 x(p+3) = 1/2 x (x+1)

    So p = x - 2. and from (1)

    (X-2)^2 = 2x + 1 so x^2 -6x + 3 = 0

    Solving the quadratic

    x = 3 + or - sqrt 6

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sumith PeirisOctober 24, 2015 at 9:21 AM

    Alternatively since 2 altitudes of the Tr. ECG are equal the corresponding sides can easily be proved to be equal

    Hence x+1 = p+3 and so on

    ReplyDelete
  5. Peter TranOctober 24, 2015 at 4:08 PM

    Note that x=3+√ (6) is only acceptable solution of this problem.
    Peter Tran

    ReplyDelete
  6. PravinOctober 25, 2015 at 9:14 AM

    EC bisects angle BET (BEC)
    So angles GCE = CEB = CEG
    Thus GE = GC = 1 + x,
    GT = GE - 3 = x - 2
    Let GC be extended to meet circle at D'
    So GT^2 = GD. GD' = 1.(1 + 2x) = 1 + 2x.
    Follows (x- 2)^2 = 2x + 1, x^2 - 6x + 3 = 0 etc

    ReplyDelete
  7. Loukas SidiropoulosApril 24, 2016 at 9:29 AM

    I use the point F like the user Ajit did. My solution here:
    https://www.facebook.com/photo.php?fbid=10206223938546060&set=a.10205987640598759.1073741831.1492805539&type=3&theater

    ReplyDelete
  8. Stan FulgerApril 25, 2016 at 12:03 AM

    Excellent observation of Sumith Peiris, GC=GE; now we can apply the power of G w.r.t. circle, GT^2=GD(GD+2CD), so (x-2)^2=2x+1, a.s.o.

    ReplyDelete

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