Triangle Geometry
By Anthony Cardell Tony on Tuesday, January 15, 2002 - 02:32 am:

In triangle ABC, the perpendicular bisector of AC intersects AC at M and AB at T. If the area of triangle AMT is 1/4 the area of triangle ABC, and angle A+ angle C=128 degrees, compute the number of degrees in angle A.

I am not sure how you go about this... I think there may be some tricky substitution involved since one attempt got me down to: (cosA)(sinC)=sin52, though I couldn't find A from there when I "simplified" to an expression involving only trig functions of A.

Thanks for your help!

By Kerwin Hui on Tuesday, January 15, 2002 - 04:17 pm:
Hint: Obviously if C=90 degrees then the given condition holds. Now suppose C ¹ 90 degrees and we know M must be the mid-point of A B. A contradiction (recall mid-point theorem).

Kerwin

By Arun Iyer on Tuesday, January 15, 2002 - 07:19 pm:

if you have got cosAsinC=sin52 ,then you can find A....

since cosAsinC=sin52,
sin(A+C)-sin(A-C)=2sin52
sin(A-C)=sin(A+C)-2sin52
i.e
sin(A-C)=sin128-2sin52
A-C=sin-1 (sin128-2sin52)=-52
we know,A+C=128

adding the above two equations,we get,
2A=76
A=38...

love arun
P.S-> I haven't checked your result....so i am not sure about the answer either..

By Anthony Cardell Tony on Tuesday, January 15, 2002 - 09:01 pm:

I agree with both of you, I was working on it at school some, and getting down to cosAsinC=sin 52, I noted that the 52 was 180-(A+C), so cosAsinC=sin(180-(A+C))=sin(A+C)=sinAcosC+cosAsinC

so, cosAsinC=sinAcosC+cosAsinC, and canceling, sinAcosC=0, so the only thing that makes sense for an angle is clearly C=90, from which of course A=38. Of course, it would definitely be much easier to just use your method Kerwin!

Thanks for the help Arun and Kerwin!