Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, December 17, 2016
Geometry Problem 1295 Right Triangle, Incenter, Incircle, Excenter, Excircle, Congruence, Angle
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https://goo.gl/photos/cbTJjt1P5rtiULgY9
ReplyDeleteDraw 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
Draw from I parallel to AB, from E parallel to BC they meet at K
ReplyDeleteTr IEK congr to Tr ABC => ang EIK = ang C, but ang EIK = 1/2 ang A
=> ang C = 30
Problem 1295
ReplyDeleteIn-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
Moratuwa
Sri Lanka
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.
ReplyDeleteBest regards
BICE are concyclic with IE being the diameter. Let F be the center of this circle
ReplyDeleteLet 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
It is not clear why FG= FC . please explain
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