Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 984.
Sunday, February 16, 2014
Math, Geometry Problem 984. Triangle, Circumcircle, Altitudes, Midpoint, Collinear Points
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Let CC1 cut AA1 and circle O at H and E
ReplyDeleteLet C1A1 cut A2C2 at M
1. Note that C1E=C1H
Since ∠(A1HC)= ∠(CDC2)= ∠ (B) => quadrilateral HDC2A2 is cyclic
So ∠(HA2C)= ∠(CC2D)
Triangle HCA2 similar to triangle EAC2…. (case AA)
With corresponding altitudes CA1 and AC1
2. So A1H/A1A2=C1E/C1C2
Replace C1E=C1H in above => C1H/CC2=A1H/A1A2
or C1H/CC2 x A1A2/A1H=1 …(1)
3. Apply Menelaus’s theorem in triangle HA2C2 with secant C1A1M
C1H/C1C2 x MC2/MA2 x A1A2/A1H =1
Replace value of ( 1) to above expression we have MC2/MA2=1
So M is the midpoint of A2C2
See below for the sketch of problem 984
ReplyDeletehttp://s25.postimg.org/b91r2quv3/pro_984_1.png
Peter Tran
Let A1C1 meet A2C2 at M'. Using Law of sine,
ReplyDeleteM'C2=sin(<CC1A1)*C1C2/sin(<C1M'C2)
and M'A2=sin(<A2A1M')*A1A2/sin(<A1M'A2)
Evaluate M'C2/M'A2=(sin(<CC1A1)*C1C2)/(sin(<A2A1M')*A1A2),
but sin(<CC1A1)/sin(<A2A1M')=A1C/AC1, which simplifies M'C2/M'A2 to
(A1C*C1C2)/(AC1*A1A2)=A1C/A1A2*C1C2/AC1=1 because triangles AC1C2 and CA1A2 are similar.
This proves M' to be midpoint of A2C2