tag:blogger.com,1999:blog-6933544261975483399.post5578312839970064011..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Math, Geometry Problem 984. Triangle, Circumcircle, Altitudes, Midpoint, Collinear PointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-33763544629379238162014-11-24T14:14:11.255-08:002014-11-24T14:14:11.255-08:00Let A1C1 meet A2C2 at M'. Using Law of sine,
...Let A1C1 meet A2C2 at M'. Using Law of sine, <br />M'C2=sin(<CC1A1)*C1C2/sin(<C1M'C2)<br />and M'A2=sin(<A2A1M')*A1A2/sin(<A1M'A2)<br />Evaluate M'C2/M'A2=(sin(<CC1A1)*C1C2)/(sin(<A2A1M')*A1A2),<br />but sin(<CC1A1)/sin(<A2A1M')=A1C/AC1, which simplifies M'C2/M'A2 to <br />(A1C*C1C2)/(AC1*A1A2)=A1C/A1A2*C1C2/AC1=1 because triangles AC1C2 and CA1A2 are similar.<br />This proves M' to be midpoint of A2C2Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70500866049506715852014-02-25T16:45:05.875-08:002014-02-25T16:45:05.875-08:00See below for the sketch of problem 984
http://s25...See below for the sketch of problem 984<br />http://s25.postimg.org/b91r2quv3/pro_984_1.png<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5376577022041391102014-02-25T13:55:57.537-08:002014-02-25T13:55:57.537-08:00Let CC1 cut AA1 and circle O at H and E
Let C1A1 c...Let CC1 cut AA1 and circle O at H and E<br />Let C1A1 cut A2C2 at M<br />1. Note that C1E=C1H<br />Since ∠(A1HC)= ∠(CDC2)= ∠ (B) => quadrilateral HDC2A2 is cyclic<br />So ∠(HA2C)= ∠(CC2D)<br />Triangle HCA2 similar to triangle EAC2…. (case AA)<br />With corresponding altitudes CA1 and AC1<br /><br />2. So A1H/A1A2=C1E/C1C2<br />Replace C1E=C1H in above => C1H/CC2=A1H/A1A2 <br />or C1H/CC2 x A1A2/A1H=1 …(1)<br /><br />3. Apply Menelaus’s theorem in triangle HA2C2 with secant C1A1M<br />C1H/C1C2 x MC2/MA2 x A1A2/A1H =1<br />Replace value of ( 1) to above expression we have MC2/MA2=1<br />So M is the midpoint of A2C2<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com