Monday, May 19, 2008

Elearn Geometry Problem 41



See complete Problem 41
Sangaku, Mickey Mouse, Circles, Tangent. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 comments:

  1. Using Pythagoras, BC^2 = d^2 + (c - b)^2
    Let angle CAB = z
    Using cosine rule: BC^2 = (a+c)^2 + (a+b)^2 - 2(a+c)(a+b)cos(z)
    Or d^2 + (c - b)^2 = (a+c)^2 + ((a+b)^2 - 2(a+c)(a+b)cos(z)
    From where, d^2+ b^2+c^2 - 2bc = a^2+c^2+2ac + a^2+b^2+2ab - 2(a+c)(a+b) cos(z)
    Or d^2=2a^2+2ac+2ab+2ac-2(a+c)(a+b)cos(z)
    Or d^2=2(a+b)(a+c)-2(a+c)(a+b)cos(z) or d^2
    = 2(a+b)(a+c)(1-cos(z) --------(i)
    From the triangle in circle A we can write,
    x^2 =a^2 +a^2-2a^2*cos(z)
    =2a^2(1-cos(z)
    Or(1-cos(z)) = x^2/2a^2-------------(ii)
    From (i) and (ii), we get:
    d^2 = 2(a+b)(a+c)x^2/2a^2
    = (a+b)(a+c)x^2/a^2
    Or x^2 = a^2*d^2/[(a+b)(a+c)]
    QED
    Ajit:ajitathle@gmail.com

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  2. bae deok rak(bdr@korea.com)October 7, 2010 at 10:05 PM

    By cosine rule,
    d^2 +(b-c)^2 =BC^2
    =(a+b)^2+(a+c)^2-2(a+b)(a+c)cos(ang(BAC))
    =(a+b)^2+(a+c)^2-2(a+b)(a+c)(a^2+a^2-x^2)/2aa
    =2a^2+b^2+c^2+2ab+2ac-2(a^2+ab+ac+bc)
    +(a+b)(a+c)(x^2/a^2)
    <=>
    d^2=(a+b)(a+c)(x^2/a^2).

    We get, x^2 =(a^2 d^2)/(a+b)(a+c)

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  3. From Moscow: I have pure synthetic solution but not sure if correct. Please check. Dropping perpendiculars from A, F, and G onto DE to get A' F' G', we have 1)A'F'/A'D=a/(a+b)and 2)A'G'/A'E=a/(a+c). Looking at pentagon ABCDE, We note that angle A + angle B + angle C + 90 + 90= 540, so angles A, B, and C add to 360. We see that angle FDE is half of B and angle GED is half of C. Therefore FD and GE meet at point P such that angle FPG is half of angle A so P is on circle A. But angle BDF = 90-angle FDE = BFD = AFP = APF, so AP is perpendicular to DE. x/d=PF/PE=PG/PD, so x^2/d^2=(PF*PG)/(PE*PD)=(A'F'*A'G')/(A'D*A'E)=a^2/((a+b)(a+c)). The last two equality statements we get from 1) and 2) and we are done.

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    Replies
    1. To Ivan from Moscow, Problem 41. Your great pure synthetic solution is correct. To complete your solution, what is the reason for the step: x/d=PF/PE=PG/PD. Thanks.

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