Tuesday, December 5, 2017

Geometry Problem 1350: Triangle with three Intersecting Circles, Incircle, Collinear Points

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

Details: Click on the figure below.

Geometry Problem 1350: Triangle with three Intersecting Circles, Incircle, Collinear Points.


  1. https://photos.app.goo.gl/Yubn9z3lzvSmV5w03

    Denote A’, B’, C’ as points of contact of incircle I to triangle ABC
    a, b,c and 2p as 3 sides and perimeter of triangle ABC
    Denote k= B’C/B’A , s= C’A/C’B and q=A’C/A’B
    Define point B5 on line AC such that B4B’=B4B5 ( see sketch)
    Since B4 is on the radical line of circles I and B1 so
    So B4A.B4C=B4B’^2=B4B5^2 => ( ACB’B5)= -1
    And k= (B’C/B’A)=(B5C/B5A) => B’C/k=B’A/1=b/(1+k) => B’A= b/(1+k)
    Similarly B5A=b/(1-k)
    We have AB4= ½(AB’+AB5)= b/(1-k^2) and CB4=AB4- b= b.k^2/(1-k^2)
    And B4C/B4A= k^2
    Similarly we also have C4A/C4B= s^2 and A4B/A4C= 1/q^2
    And (B4C/B4A) x(C4A/C4B)x(A4B/A4C)= k^2 . s^2/q^2
    Replace k=(p-c)/(p-a) and s=(p-a)/(p-b) , q=(p-c)/(p-b) in above expression
    We will have (B4C/B4A) x(C4A/C4B)x(A4B/A4C)= 1
    And A4, B4 ,C4 are collinear per Menelaus’s theorem

  2. Let T1,T2,T3 be the points of contact of the in-circle
    with the sides BC,CA,AB resp.
    Let A4T1=x,B4T2=y,C4T3=z
    By Tangent-Secant property:
    x(x+a)=A4C.A4B=A4A3.A4A2 referred to circle A1 while
    A4A3.A4A2=A4T1^2 referred to in-circle.
    Note A4T1=A4C+CT1=x+(s-c)
    So x(x+a)=[x+(s-c)]^2,
    xa=(s-c)^2 + 2x(s-c),
    x=(s-c)^2/(c-b) and
    x+a=(s-b)^2/(c-b)on further simplification.
    Hence A4C/A4B=(s-c)^2/(s-b)^2
    By cyclic symmetry B4C/B4A=(s-a)^2/(s-c)^2
    and C4A/C4B=(s-b)^2/(s-a)^2
    Product of the above 3 ratios being 1 (numerically),
    by Menealau C4,A4,B4 are collinear.

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  4. Problem 1350
    Denote the points of contact of the incircle with the sides BC, CA, AB as P, Q, R respectively. We first evaluate the ratio BA4:A4C in terms of a, b, c, and s (respectively the lengths of the sides BC, CA, AB and the semi perimeter of the triangle). Cyclically permuting we get the values of the similar ratios CB4:B4A and AC4:C4B. We next show that the product of these 3 ratios is -1, from which fact follows the collinearity of the points C4, A4, and B4 by the converse of Menelau’s Theorem.
    First consider the in-circle I for which A4P is a tangent and A4 A3 A2 is a secant. By the Tangent – Secant Property we get A4P^2 = A4A3. A4A2
    Next consider the circle A1 for which both A4 A3 A2 and A4CB are secants. By the Secants Property we get A4A3 . A4A2 = A4C . A4 B.
    Hence it follows A4C . A4B = A4B3^2. It is well known that CP = s - c
    At this stage we let A4C = x . So A4P = A4C + CP = x + s -c
    So x(x + a) = (x + s - c)2,
    Cancelling x^2 on either side we get xa = (s – c)^2 + 2(s-c)x,
    x(a + 2c – 2s) = (s – c)^2,
    x( c – b) = (s – c)^2,
    x = (s – c)^2 / (c – b) ......(i)
    Also 4(x + a)(c – b) = 4x(c – b) + 4a(c – b)
    = 4(s – c)^2 + 4a(c – b) = (2s – 2c)^2 + 4ac – 4ab
    = (a + b –c)^2 + 4ac – 4ab
    = a^2 + b^2 + c^2 + 2ab -2bc – 2ca + 4ac – 4ab
    = a^2 + b^2 + c^2 – 2ab – 2bc + 2ac = (c + a – b)^2
    = (2s – 2b)^2= 4(s – b)^2, So
    x + a = (s – b)^2 / (c – b) .......(ii)
    Hence from (i) and (ii)
    BA4 / A4C = (x + a)/x = (s – b)^2 /(s – c)^2 in magnitude.
    Similarly by Cyclic Symmetry:
    CB4 / B4A = (s - c)^2 / (s – a)^2 ,
    AC4 / C4B = (s – a)^2 / (s – b)^2
    With due regard to sign we obtain on multiplication:

    [BA4 / A4C] [CB4 / B4A] [AC4 / C4B] = -1
    Hence by Menelau’s Theorem
    C4, A4, and B4 are collinear.

  5. Pravin: The above post is only a detailed version of my earlier post. Thank you.

  6. Referring to Problem 1350 and its figure.
    Let J be the center of the circle through B, C, and I
    and R' be the radius.
    Do the the following results hold good ?
    (i) R' = R / 2 cos A
    (ii)<OJI = C - A
    (iii) OI = R sin [(C - A)/2]/cos A