Friday, July 5, 2019

Geometry Problem 1441 Intersecting Circles, Perpendicular Bisector, Parallel Lines

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 1441 Intersecting Circles, Perpendicular Bisector, Parallel Lines, iPad apps, Tutoring.



    Define points I, J, K, L, P, S,T, X and Y as per attached sketch
    Note that I, J are midpoints of OD and CD
    L and K are midpoints of OF and EF
    QXJI and QYKL are rectangles
    We have OJ= ½(OC+OD)= 1/2OC + OI
    And JI= OJ-OI= 1/2OC= QX
    Similarly, KL= ½ OE=1/2OC= QY
    Since QX=QY => Chord GH= Chord MN=> Arc GH= Arc MN
    And Arc GM=Arc NH => GN//MH
    Let OS ⊥GN , we have quadrilaterals OSGJ and OSNK are cyclic
    We have ∠ (COS)= ∠ (SGJ)= u=∠ (SNK)= ∠ (SOP)
    So ∠ (COS)=u=∠ (SOE) => OS is the angle bisector of ∠ (TOP) and ∠ (COE)=> CE//GN

  2. From previous problem, A,C,H co line, M,E,B co line.
    Connect ACH, connect BEM.
    Connect AB;
    so angle AHM = angle ABM;
    also angle CEM = angle ABE (ACEM on the same circle O),
    so CE//MH

    Let GN and OF cross at P;
    OA=OB, so curve OA=curve OB;
    curve AG= curve GD; curve BN=curve NF; so angle(OA)+angle(OB)+angel(AG)+angle(GD)+angle(DMHF)+angle(NF)+angle(BN)=180;
    so angle NPF = angle(NF)+angle(OA)+angle(AG) = 90-1/2 angle(DMHF) =90- 1/2angleFOD = angle CEO.
    So CE//GN;
    So CE//GN//MH

  3. If Point's
    Q(R,0) and O(0,0)


    On Point O, can define two lines
    y=a*x (Line on Point's OCD) and y=b*x (Line on Point's OEF)

    => Point D [R/(1+a²),(a*R/(1+a²))]

    => Point F [R/(1+b²),(b*R/(1+b²))]

    => Point C [r²/(1+a²),(a*r²/(1+a²))]

    => Point B [r²/(1+b²),(b*r²/(1+b²))]

    Solve for the Slope k (germ. Steigung) at the Line on the Point's CE
    k(1)=a-b (i)

    On Point's CD can construction a Normal line, who will Intersect the (Big) Circle (-R,0)

    n: y=-(1/a)*(x-(R+r²)/(2*(1+a²)))+a*(R+r²)/(2(1+a²))

    => The Intersect Point's:
    G [(p+q)/(2(1+a²)),(a*(r²-R)-p/a)/(2(1+a²))] and H [(q-p)/(2(1+a²)),(a*(r²-R)+p/a)/(2(1+a²))]

    The same procedure can be define the Point's M and N.
    => The Intersect Point's:
    M [(u+v)/(2(1+b²)),(b*(r²-R)-u/b)/(2(1+b²))] and N [(v-u)/(2(1+b²)),(b*(r²-R)+u/b)/(2(1+b²))]

    In wich:
    p=a*√(R²(4a²+3)+r²(2R-r²)) , q=2a²R+R+r²

    u=b*√(R²(4b²+3)+r²(2R-r²)) , v=2b²R+R+r²


    Now it can define a new Lines with Point's G and N, M and H.
    The Slope from the new Lines, a solve will be show:

    k(2)=a-b, line with Point's G and N (ii)
    k(3)=a-b, line with Point's M and H (iii)

    For the three Lines, the Slope is a same.
    A prove for the Parallelity lines.

  4. We use the results of problem n°1440 applied to both lines (O,C,D) and (O,E,F), and consider alignements (A,C,H) and (B,E,M) to which we apply Reim’s theorem (see problem n°72).
    Thus: MH//CE.
    Similarly we obtain alignements (A,E,N) and (B,C,G) resulting in GN//CE.