Saturday, May 21, 2016

Geometry Problem 1217: Triangle, Circle, Excenter, Incenter, Angle Bisector, Cyclic Quadrilateral, Circumcircle, Tangent Line

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1217.

Geometry Problem 1217: Triangle, Circle, Excenter, Incenter, Angle Bisector, Cyclic Quadrilateral, Circumcircle, Tangent Line
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2 comments:

  1. Peter TranMay 21, 2016 at 9:21 PM

    http://s32.postimg.org/679nin5x1/pro_1217.png
    Denote (XYZ) angle (XYZ)
    We have (IBL)=(ICL)= 90 => quạ BICL is cyclic
    We have (BIK)= 1/2A + 1/2B
    (BJK)=1/2(BEC)+1/4B= ½(A+B/2) + 1/4B= 1/2A+1/2B = (BIK) => qua (BIJK) is cyclic
    Since quạ BIJK cyclic => (IBJ)=(JBQ)=(IKJ) => quạ BPQK is cyclic
    (RAL)=(RBL)= 90 => qua. ABLR is cyclic

    Since quạ. BICL is cyclic= > (JLB)= (BCI)= C/2
    Let x= (IBP)=(BPQ)=(PKQ)=(MKL)
    We have (BKI)=(BJI)= x+C/2
    In triangle KML we have (BMK)= x+C/2
    Triangles BKl similar to trị KML => LK ^2=LM.LB
    So LK tangent to circumcircle of triangle BKM

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  2. AnonymousMay 30, 2016 at 5:32 AM

    ∠IBL=∠IBC+∠CBL=1/2∠ABC+1/2(180-∠ABC)=1/2×180=90
    Similarly，∠ICL=90
    ∠IBL+∠ICL=90+90=180
    BICL concyclic
    ∠IKJ=∠AKE=∠KEC-∠KAC=1/2∠BEC—1/2∠BAC
    =1/2∠ABE=1/2∠EBC=∠IBJ
    BIJK concyclic
    ∠PBQ=∠IBJ=∠IKJ=∠PKQ
    BPQK concyclic
    BICL concyclic,similarly AICK concyclic
    ∠BLC=∠RIC=∠RAC
    ∠BLR+∠BAR=∠BLC+∠BAR=∠RAC+∠BAR=∠RAC+∠BAC+∠CAR=180
    ABLRconcyclic
    ∠BMK=∠MLK+∠LKM=∠PKQ+∠BCI=∠PBQ+∠BCI=∠BJI=∠BKI
    ∴AL is tangent to circumcircle of triangle BKM

    ReplyDelete

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