Thursday, November 25, 2010

Problem 544: Right Triangle, Altitude, Median, Equal angles, Measure

Geometry Problem
Click the figure below to see the complete problem 544 about Right Triangle, Altitude, Median, Equal angles, Measure.

Problem 544: Right Triangle, Altitude, Median, Equal angles, Measure.
Go to Complete Problem 544

Posted by Antonio Gutierrez at 1:50 PM
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8 comments:

  1. PravinNovember 26, 2010 at 5:43 AM

    Denote BC = 2y.
    So BE = EC = y
    From triangle ABC:
    tan C = x/2y
    In triangle ABE:
    angle BAE = C and
    tan C = y/x
    Follows x^2 = 2 y^2,
    tan C = y/x = 1/sqrt 2
    Note angle ABD
    = complement of angle DBC= C
    Follows angle AFD = 2C
    In triangle AFD:
    cos 2C = 1/AF,
    AF = [1 + (tan C)^2]/ [1 - (tan C)^2]
    = [1 + (1/2)]/ [1 - (1/2)]
    AF = 3
    Also cos 2C = 1/3,
    1 - 2 (sin C)^2 = 1/3,
    sin C = 1/sqrt3
    Now by Pythagoras
    AD^2 = 9 - 1 = 8,
    AD = 2 sqrt2
    In triangle ABD
    sin <ABD = sin C = AD/AB,
    1/sqrt3 = 2(sqrt2) / x
    x = 2 sqrt (2/3)

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  2. PravinNovember 26, 2010 at 9:44 PM

    The last step should read x = 2 sqrt6
    Alternatively,
    ABE ~ CBA => x^2 = BC.BE = (1/2) BC^2
    => BC = x sqrt2 and BE = x/ sqrt2
    AE^2 = x^2 + (x^2)/2 = 3x^2 /2, AE = x sqrt(3/2)

    ADB ~ EBA => x^2 = AE.BD = [x sqrt(3/2)].BD
    => BD = x sqrt(2/3)
    AD^2 = x^2 - BD^2 = x^2 - (2x^2/3) = x^2/3
    AD = x/sqrt3

    Finally AF^2 - AD^2 =1^2,
    => 3x^2/8 - x^2/3 = 1, x^2 = 24 , x = 2 sqrt6

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  3. Peter TranNovember 27, 2010 at 12:31 AM

    Problem 544

    Draw line NFM // BC ( N on AB , M on AC)
    Let BE=EC= a
    1. F is the midpoint of NM. (properties of // lines)
    FN=FM=a/2
    Triangle(ABE) similar to triangle (CBA) (case AA)
    So BE/AB=AB/BC or a/x= x/(2.a) and x=a.SQRT(2)
    2. AC=a.sqrt(6) and AM= a/2.sqrt(6) ( Pythagoras theorem)
    3. Triangle (FDM) similar to triangle(ANM)
    So FD/AN=FM/AM or 1/AN=a/(a.sqrt(6))
    AN=sqrt(6) and AB=2.sqrt(6)

    Peter Tran

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  4. Peter TranNovember 27, 2010 at 12:41 AM

    See link below for the sketch of the problem 544

    http://img440.imageshack.us/img440/2296/problem544.png

    Peter Tran

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  5. c .t . e. oNovember 27, 2010 at 1:48 PM

    Draw EP//BD => EP = 2 => BD = 4 => BF = AF =3 =>
    AD = 2√2 => x² = 4² + (2√2)²
    x = 2√6

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  6. Ivan BazarovMarch 16, 2014 at 5:06 PM

    x^2=BE*(BC=2BE). F is circumcenter of AEB with radius R. <BDE=<FEB=90-a so BE^2=R(R+1).Then the original equation becomes x^2=2R(R+1)
    But from Pythagoras in triangle ABE x^2=(2R)^2-(BE^2=R(R+1))=3R^2-R
    Now we have x^2=2R^2+2R=3R^2-R. Solving for R gets us R^2-3R=0
    R is not 0 so with R=3, x^2=24, x=2sqrt(6).

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  7. APOSTOLIS MANOLOUDISJune 22, 2016 at 12:10 PM

    Problem 544
    Fetch AM=//BE then B, F and M are collinear (<ACB=<ABD=α=<BAE ).Then AF=BF=FM=y.
    Is triangle ADM similar triangle BDC so AD/DC=DM/DB=AM/BC=1/2.Therefore DC=2.AD,
    BD=2.DM or y+1=2(y-1) or y=3,then AD^2=3^2-1=8, but AC=3.AD.
    Then x^2=AB^2=AD.AC=3AD^2=3.8=24, or x=√24=2.√6.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  8. Sumith PeirisJune 23, 2016 at 5:18 AM

    Draw EG // BD, G on AC


    < ABD = á hence F is the centre of right triangle ABE. Let AF=FE=BF=y


    From the mid point theorem, EG = 2 X DF = 4 hence y+1 = 2 X EG = 4 so y =3


    Applying Pythagoras to right triangle ABE, x2 + a2 = 4y2 = 36…..(1)


    Since AB is tangential at A to circle AEC, x2 = a2/2 …..(2)


    From (1) and (2) ; 3x2 = 36


    x = √24 = 2√6


    Sumith Peiris
    Moratuwa
    Sri Lanka

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