Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 668.
Sunday, September 11, 2011
Problem 668: Right triangle, Incircles, Parallel, Common Tangent, Mind Map
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http://img819.imageshack.us/img819/5083/problem668.png
ReplyDeleteConnect CO and AO
Note that CO and CO2 are angle bisector of angle (BCA) so
C, O, O2 are collinear
Triangle OEC congruence with triangle OHC ( right triangles with common hypotenuse and CO is angle bisector)
So OC is the angle bisector of angle (HOE) >>Incircle O2 tangent to OE also tangent to OH.
Similarly AO is angle bisector of angle HOF) and incircle O1 tangent to both OF and OH
Peter Tran
Let circle O₁ touch AB at J, AC at K & FG at L.
ReplyDeleteComplete the square O₁JHM
O₁M = KH = AH - AK = AF - AJ = KF = radius of O₁ (note FJ O₁K is a square)
follows OH touches circle O₁.
Similarly we can show OH touches circle O₂
Refer my solution to Problem 668:
ReplyDelete"FJO₁L is a square" (not FJO₁K as previously typed)
another point of vue
ReplyDeletein ABC r=s-b
triangles AFG and ABC are homothetic,center A ratio (s-a)/c
the incircle of AFG is the image of the incircle of ABC;r1=(s-a)r/c=(s-b)(s-a)/c
line OH becomes the parallel line O1H1
distance H1H=(s-a)-(s-a)²/c=(s-a)(s-b)c=r1
incircle O1 tangent to OH
The same way CDE and ABC are homothetic with center C and ratio (s-c)/a