## Tuesday, December 17, 2013

### Geometry Problem 942: Triangle, Circles, Circumcircle, Sagitta, Incircle, Excircle, Inradius, Exradius, Metric Relations

Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 942.

1. http://img22.imageshack.us/img22/3034/goav.png

Draw points L, M, F’, I’ per sketch.
We have AL=CM= p-a
Since F is the midpoint of arc AC so B, I, F, E are collinear
And D and F are midpoints of LM and IE
DF=MF’= FI’- I’M= ½( 20+7)- 7 = 6.5

2. ArcAF=ArcAC so BF bisect angleABC and hence BIFE collinear
Let BF cut AC at P ; incircle and ex circle cut AC at I' and E'
Note Triangle II'P ~ PDF ~ PEE'
Further D is the midpoint of AC hence the midpoint of I'E'
By solving similar triangle ratios it is arrived that L = 13/2

3. Let IJ be the in-radius shown in the graph, and EK be that ex-radius shown.

Observe that JDK are collinear with mid-point D.
Also, IFE are collinear, which is the angle bisector.

Thus, DF = 1/2 (20 - 7) = 13/2