Home > Shapes > Geometry
Concurrent Rays (Posted on 20160820) 

Let I be the incenter of ΔA_{0}A_{1}A_{2}.
For i = 0, 1, and 2:
Let M_{i} be the midpoint of the side opposite vertex A_{i}.
Let S_{i} be the reflection of M_{i} about ray A_{i}I.
Prove that rays A_{0}S_{0}, A_{1}S_{1}, and A_{2}S_{2}, are concurrent.
No Solution Yet

Submitted by Bractals

Rating: 5.0000 (1 votes)


Possible solution

Comment 1 of 1

Writing
A_{0}, A_{1}, A_{2}, M_{0} as A, B, C, M, pro
tem, and denoting
the intersections of A_{0}I and A_{0}S_{0} with BC by P
and Q, and the
lengths MP and PQ by d and e; then using usual notation:
Since AP bisects /QAM: AM/AQ =
d/e (1)
Since /MAC = /QAB, the sine rule used in triangles MAC and
QAB gives: AM/AQ = (MC sinC)/(QB
sinB) = (MC/QB)(c/b)
then, using (1): d = e(MC*c)/(b*QB)
= e(a/2)c/(b(a/2
– d – e))
from which: e = bd(a –
2d)/(ac + 2bd) (2)
Since AP bisects /BAC: PC/BQ = (a/2
+ d)/(a/2 – d) = b/c
from which: d = a(b – c)/(2(b
+ c)) (3)
Using (3) to substitute for d in (2), after much simplifying
gives
e = abc(b – c)/((b
+ c)(b^{2} + c^{2})
so that BQ = a/2 – d – e then simplifies to give
BQ = ac^{2}/(b^{2}
+ c^{2}) and CQ = ab^{2}/(b^{2} + c^{2})
Thus BQ/CQ = c^{2}/b^{2}
Now reverting to the given notation, with Q becoming Q_{0},
and having corresponding points Q_{1} and Q_{2} on other sides,
it follows that (using scalar lengths):
(A_{1}Q_{0}/A_{2}Q_{0})(A_{2}Q_{1}/A_{0}Q_{1})(A_{0}Q_{2}/A_{1}Q_{2}) = (c^{2}/b^{2})(a^{2}/c^{2})(b^{2}/a^{2})
= 1
and, by the converse of Ceva’s theorem, A_{i}S_{i}, (i = 1,2,3)
are
concurrent.
….there must be a simpler way! – should BQ/CQ
= c^{2}/b^{2}
be more readily obtainable??

Posted by Harry
on 20160824 17:03:58 


Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ 
About This Site
Site Statistics
New Comments (11)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
