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

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

Click the figure below to view the complete problem 1043.

## Wednesday, September 10, 2014

### Geometry Problem 1043: Right Triangle, Incenter, Excenter, Congruence, Metric Relations

Labels:
congruence,
excenter,
incenter,
metric relations,
right triangle

Subscribe to:
Post Comments (Atom)

C, I, E are collinear

ReplyDeleteConnect I with E and A with B

AE = AI = 3√2

in the triangle AIC we have IC = 3√3 - 3

triangle EBC similar to triangle AIC, EB/AI = EC/AC

x = 3(1 + √3)/2

x= 4.098

Erina, NJ

Radius of incircle=r. Drop I onto ED to obtain point P.

ReplyDeleteAB=x+r=PI, so triangle PIE is congruent to triangle BAC.

BC=PE=x-r. DC=2x-r=6+r. Therefore

1) x-r=3.

Substitute this into (x-r)^2+(x+r)^2=36 in triangle PIE to get

2) x+r=3*sqrt(3).

Adding 1) and 2) yields 2x=3(1+sqrt(3))==>x=3(1+sqrt(3))/2