Sunday, August 2, 2009

Problem 333. Circle inscribed in a semicircle, Perpendicular

Proposed Problem
Click the figure below to see the complete problem 333 about Circle inscribed in a semicircle, Perpendicular to the common tangent.

 Problem 333. Circle inscribed in a semicircle, Perpendicular.
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Complete Problem 333
Level: High School, SAT Prep, College geometry


  1. If let OD=p, CD=r and OA=OB=R as in problem 311 and draw DG parallel to HE meeting CO in G. Then EH = DG and from triangle ODC we can say that: area of Tr.ODC = (1/2)DG*OC= (1/2)DC*OD or DG = DC*OD/OC In other words, DG = r*p/(r^2+p^2)^(1/2) whence, EH^2 = DG^2 = (rp)^2/(r^2 + p^2).
    Now from Tr. OCD (R-r)^2 = r^2 + p^2 or 2rR = R^2–p^2 or r = (R^2 - p^2)/2R. Hence we can say,
    EH^2= (p^2((R^2-p^2)/2R)^2)/[((R^2-p^2)/2R)^2+p^2] = (p^2(R^2-p^2)^2)/((R^2-p^2)^2+4R^2p^2)
    = p^2(R^2-p^2)^2/(R^2+p^2)^2
    But R-p = b & R+p = a and a+b=2R so p=(a-b)/2; hence EH^2 = ((a-b)(ab))^2/(a^2+b^2)^2.
    Now, using the result of Problem #311 we can say that DE^2 = 2(ab)^2/(a^2 + b^2) which gives us
    DH^2 = DE^2–EH^2 = 2(ab)^2/(a^2+b^2)-((a-b)(ab))^2/(a^2+b^2)^2 = [ab(a+b)/(a^2+b^2)]^2
    Or x = DH = a*b*(a+b)/(a^2 + b^2). QED.

  2. u can just say that X=EG=EO-OG , while OG u can find easly by pitagoras at the triangle ODG ..

  3. Let the common tangent meet AB at N.
    Let BN = c so that DN = EN = b+c,being tangents to the smaller circle from N.
    The radius of the semicircle = (a+b)/2

    In the semicircle, NE^2 = NB.NA from whence (b+c)2 = c(a+b+c)
    This yields, upon simplification,
    c = b2 / (a-b) ….(1)

    OEN & DHN are similar triangles, hence

    x/{(a+b)/2} = (b+c)/{(a+b)/2 + c}
    = {b +b2/(a-b)}/{(a+b)/2 + b2/(a-b)}
    upon substituting from (1) for c.

    So x = (a+b){b(a-b) + b2}/{(a+b)(a-b) + 2b2)
    which simplifies to, x = ab(a+b)/(a2 + b2).

    (Due to software limitations a squared is shown as a2 and b squared as b2).

    Sumith Peiris
    Sri Lanka