Saturday, November 24, 2018

Geometry Problem 1402: Triangle, Three Tangent Circles, Parallel Diameter, 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 1402: Triangle, Three Tangent Circles, Parallel Diameter, Collinear Points, Tutoring.

Posted by Antonio Gutierrez at 3:19 PM
Labels: , , , , , , , , ,

9 comments:

  1. AnonymousNovember 26, 2018 at 8:15 AM

    Triunghiurile GMH și CMA sunt similare ,
    de unde GH/AC = GM/MC = MH/MA
    și cum unghiurile GMA = CMH rezultă GA // HC
    adică ACHG paralelogram, de unde
    GH = AC, GM = MC, MH = MA.
    Fie BM intersectat cu AC în punctul P.
    Cum M este punctul de mijloc al diagonalelor rezultă
    BM = MP, BG =BH, AP=PC.
    Fie acum punctul N intersecția lui DE cu BP.
    Atunci din teorema transversalei în triunghiul BAC rezultă că
    NP/NB *AC = (DA/DB)*PC + (EC/EB )*AP = AP*(DA/DB + EC/EB) =
    = AP*(DA+EC)/DB = DA + EC = AF + FC =AC . De aici NP/NB =1,
    Adică NP = NB și de aici punctele M=N.
    Florin Popa, Comănești , România.

    ReplyDelete
    Replies
    1. c.t.e.oNovember 29, 2018 at 11:28 AM

      ACGH is not a paralelogram ?

      Delete
    2. Peter TranOctober 18, 2019 at 11:06 AM

      I try to translate the solution of Florin to English . see below

      The triangles GMH and CMA are similar,
      whence GH / AC = GM / MC = MH / MA
      and how the angles GMA = CMH result in GA // HC
      that is, ACHG parallelogram, from where
      GH = AC, GM = MC, MH = MA.
      Let BM intersect with AC at point P.
      As M is the midpoint of the diagonals it results
      BM = MP, BG = BH, AP = PC.
      Let now be the point N the intersection of DE with BP.
      Then from the cross-sectional theorem in the triangle BAC it follows that
      NP / NB * AC = (DA / DB) * PC + (EC / EB) * AP = AP * (DA / DB + EC / EB) =
      = AP * (DA + EC) / DB = DA + EC = AF + FC = AC. From here NP / NB = 1,
      That is, NP = NB and hence the points M = N.
      Florin Popa, Comănești, Romania.
       

      In my opinion , there are some errors in above solution.
      GA is not parallel to HC and ACHG is not a parallelogram

      Peter Tran

      Delete
  2. c.t.e.oNovember 27, 2018 at 11:25 AM

    It is the same thing to be proved M, F, intersections of internal tg are collinear

    ReplyDelete
  3. AnonymousNovember 27, 2018 at 12:08 PM

    Mark Δ(delta) the line that goes through the points D and E.
    Δ ∩ AC =Q; Δ ∩ GH = P; Δ ∩ AB=D₀
    1. ΔAMC ͠ ΔMGH; AM/MH=CM/MG=AC/GH=(R₁+R₃/2R₂)
    2. ΔCEQ ͠ ΔBPE; (CQ/BP=CE/BE X/Y+R₂= R₃/R₂)
    3. ΔCMQ ͠ ΔMPG; (CQ/PG=CM/MG X/Y= R₁+R₃/2R₂)
    4. ΔAQD₀ ͠ ΔPBD₀; (AQ/PB=AD₀/D₀B R₁+R₃+X/R₂+Y= Z/R₁+R₂-Z)
    By solving the system:

    X= (R₁+R₃)R₃/R₁-R₃
    y= 2R₂*R₃/R₁-R₃
    Z=R₁
    AD₀=R₁=AD D₀ Ξ D

    E,M,D are colinear

    Erina, NJ

    ReplyDelete
  4. Pradyumna AgasheDecember 3, 2018 at 11:03 AM

    Also F,D,G and F,E,H are collinear. Problem would have been more challenging if FD and FE extended meets circle "b" at G and H. Then we have to prove GH is diameter and also parallel to AC.

    ReplyDelete
  5. AnonymousDecember 9, 2018 at 11:45 AM

    ACHG is a paralelogram if radius of circle B is arithmetic average of radius of the circle A and the circle C. In all the other cases ACGH trapezus.

    Erina,NJ

    ReplyDelete
  6. Peter TranOctober 19, 2019 at 11:01 AM

    https://photos.app.goo.gl/cRCLTb68yREoFXGg7

    note that triangles DBG and DAF are isosceles => F, D, G are collinear
    similarly F, E, H are colinear
    Per Pappus theorem applied to colinear points G,B,H and A, F, C .
    Pappus line formed by D,M,E will be colinear

    ReplyDelete
  7. Ludwig MerckNovember 10, 2019 at 5:48 AM

    Problem solve with Hyperbola (Adriaan van Roomen).

    Big Circle Radius: R
    Small Circle Radius: r

    Let the big Circle Midpont A(0,0) => F1 (Hyperbola Foci)
    Let the small Circle Midpoint C(R+r,0) => F2 (Hyperbola Foci)

    1.)
    We know the Hyperbola have the linear eccentricity length:
    2*e=R+r

    The Hyperbola Center (e,0).

    2.)
    The Hyperola curve will be construction on the Intersections Point with the big and Small Circle, so:

    (x-e)²/a² - y²/b² = 1 (hyp.)
    And: e-a=r => a=(R+r)/2 , b=√(R*r) | Because: a²+b²=e²

    3.)
    On the Hyperbola curve can be construction a other Circle (only touch Circle to the two Circle (F1 , F2)) with the Radius w and Center B(k,l).

    -> k²+l² = (R+w)²
    (k-e)²/a² - l²/b² = 1 (Point (k,l) on hyp.)

    => Solve (k,l) <=> (R; r; w) or B(k,l) (Point B)

    => Solve Intersect Point's (D and E) from the Circles F1 and F2, to Point B(k,l)
    x²+y²=R²
    (x-k)²+(y-l)²=w²
    -> Solve Point D

    (x-(R+r))² + y²=r²
    (x-k)²+(y-l)²=w²
    -> Solve Point E

    4.)
    Construction with the Parallel Line to Ox,
    so: y=l

    And now it can win two other Points, with Intersect of Line and the 3'rd Circle (k,l).

    (x-k)²+(y-l)²=w²
    y=l

    Solve x1 and x2: x1=k+w , x2=k-w, who x1 > x2, and x1=Point H, x2=Point G

    5.)
    Construction two Lines on Points x1 and F1, x2 and F2.
    The Intersect Point give the Point M

    6.)
    Point D, M and E can construction three Lines, who will Prove the three Lines are the same.

    ReplyDelete

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