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
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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
If IE=EC & ABC is a right triangle, then it can be proved that it is a 30-60-90 triangle (refer problem 1295)
ReplyDeleteSo BC=3 and AB=3√3, the result follows with 3+x being the semi-perimeter of ABC