Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to enlarge the problem 942.

## Tuesday, December 17, 2013

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

Labels:
circle,
circumcircle,
excircle,
exradius,
incircle,
inradius,
metric relations,
sagitta,
triangle

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

ReplyDeleteDraw 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

ArcAF=ArcAC so BF bisect angleABC and hence BIFE collinear

ReplyDeleteLet 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

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

ReplyDeleteObserve 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