Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view more details of problem 1276.
Sunday, October 23, 2016
Geometry Problem 1276 Equilateral Triangle, Circumcircle, Internally Tangent Circles, Tangent Line, Measurement
Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view more details of problem 1276.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view more details of problem 1276.
ABTC is cyclic quadrilateral, By Ptolemy's theorem AT*BC = BT*AC + CT*AB,
ReplyDeleteSince AB=BC=AC, we have AT = BT + CT,
Lets assume AT,BT and CT meets inner circle O1 at X,Y and Z. We can see that Angle XTY = Angle XTZ = 60 Deg, also Angle YTZ=120 Deg, hence Triangle XYZ is equilateral.
Also Angle ATO = Angle XTO1, hence angle subtended by AT and XT on circle O and O1 are equal. We have XT/AT=YT/BT=ZT/CT=r/R (r is radius of inner circle O1, and R is radius of outer circle O).
Subtracting each term from 1. We get AX/AT=BY/BT=CZ/CT=1-(r/R)=k
AA1^2=AX.AT=k.AT^2 or AA1=sqrt(k).AT,similarly BB1=sqrt(k).BT,CC1=sqrt(k).CT
Since AT=BT+CT we get sqrt(k).AT=sqrt(k).BT + sqrt(k).CT or AA1=BB1+CC1
Problem 1276
ReplyDeleteWe apply the Casey’s theorem for zero circles A, B, C, and the circle with center O_1.Then
AC.AA_1=AC.BB_1+AB.CC_1 or AA1=BB1+CC1.(AB=BC=AC).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
Great!!!
DeleteVery nice solution.
-Pradyumna
Pradyumna thank you very much and your own solution is very good.
Delete