Thursday, October 29, 2015

Geometry Problem 1159: Triangle, Circumcircle, Sagitta, Chord, Arc, 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 1159: Triangle, Circumcircle, Sagitta, Chord, Arc, Metric Relations.

Posted by Antonio Gutierrez at 8:28 AM
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3 comments:

  1. c.t.e.oOctober 29, 2015 at 2:20 PM

    Join B to O Fröm pythagora theorem r2 = 3/2 sq2R (tr BEO and BDE) The same r3 = 1/2sq10R
    From similar tr BDE and BHF we get BH = 3sq5

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  2. Peter TranOctober 29, 2015 at 5:10 PM

    http://s29.postimg.org/8hhggawgn/pro_1159.png

    Let R= radius of circle O
    We have ∡ (BHD)= ∡ (BHF)=90 => D, H, F are collinear
    DB^2= DẸ 2R = 18.R
    BF^2= FG. 2R= 10.R
    DB/ BF= sqrt(18/10)= sqrt(1.8)
    Observe that F is the midpoint of arc BC so
    ∡ (CBF)= ∡ (BDF)= angle A/2
    So triangle BDH simillar to FBG
    DB/BF= BH/FG = sqrt(1.8)
    BH= 5 x sqrt(1.8)= 3.sqrt(5)

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  3. Sumith PeirisOctober 29, 2015 at 9:36 PM

    Tr.s BDE and BHF are similar so

    BD/BF = 9/BH ....(1)

    Tr. s BDH and BGF are similar so

    BD/BF = BH/5 ....(2)

    From (1) and (2),

    9/BH = BH /5

    Hence BH = 3sqrt5

    Sumith Peiris
    Moratuwa
    Sri Lanka

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