Saturday, December 17, 2016

Geometry Problem 1295 Right Triangle, Incenter, Incircle, Excenter, Excircle, Congruence, Angle

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 1295 Elements: Right Triangle, Incenter, Incircle, Excenter, Excircle, Congruence, Angle.



    Draw points F, N, P, Q and M per attached sketch
    Observe that BN=CP=CQ=IF= inradius of triangle ABC
    Triangle CFI congruent to EQC ( case ASA)
    So CI=CE => ICE is isoceles right triangle => CM/IE= ½
    In right triangle ACM , CM/AC= CM/IE= ½
    So ACM is 30-60-90 triangle
    And angle A= 60 and angle C= 30

  2. Draw from I parallel to AB, from E parallel to BC they meet at K
    Tr IEK congr to Tr ABC => ang EIK = ang C, but ang EIK = 1/2 ang A
    => ang C = 30

  3. Problem 1295
    In-radius r = ½(c+a-b) and ex-radius ra = ½(b+c-a)

    (ra+r)2 + (ra-r)2 = b2
    2(ra2 + r2) = b2
    (c+a-b)2 + (b+a-c)2 = 2b2
    Which simplifies to a2+c2 =2bc
    So b2 = 2bc and so b =2c

    Hence ABC is a 30-60-90 triangle and <C = 30

    Sumith Peiris
    Sri Lanka

  4. Let M,N be the contact points of incircle, excircle respectively with AB; easily MN=BC, so BC is projection of AC onto BC, MN projection of IE onto AB. Since EI=AC, <A=2<C, thus <C=30.

    Best regards

  5. BICE are concyclic with IE being the diameter. Let F be the center of this circle
    Let G be the circumcenter of ABC and is the mid-point of AC
    Since AC=IE => the above two circles are congruent with BC being the common chord
    Connect FG and observe that FG perpendicular to BC and FG=FC=GC=Radius of the circles
    => m(FGC)=A=60
    => m(ACB)=30