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

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

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## Friday, April 24, 2015

### Geometry Problem 1112: Triangle, Excenters, Circumcircle, Circle, Hexagon, Area

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Let I is the incenter of triangle ABC

ReplyDeleteWe have IB⊥BE1, IC⊥CE1

Per the result of problem 482 A1, B1, C1 are centers of circles IBE1C , ICE2A and IAE3B

So IA1/IE1=IC1/IE3=IB1/IE2= ½= factor of dilation

So S(E1E2E3)= 4. S(A1B1C1)

We have IA1=CA1, IB1=CB1 and IC⊥ A1B1

So Triangle IA1B1 is the symmetry of CA1B1 .

Similarly for other 2 triangles

So triangle of hexagon AC1BA1CC1 is 2x Area of triangle A1B1C1

how can i compute the area of the triangle formed from E1E2 and E3?

ReplyDeletePer the result of problem 482 A1, B1, C1 are centers of circumcircles of quadrilaterals IBE1C , ICE2A and IAE3B

ReplyDeleteSo IA1=A1E1 , IB1=B1E2 and IC1=C1E3

So IA1/IE1=IC1/IE1=IB1/IE2 = ½ and A1C1//E1E3 , A1B1//E1E2 and B1C1//E2E3

And triangles A1B1C1 similar to triangle E1E2E3 with ratio of similarity= ½

And Area of triangle A1B1C1= ¼ area of triangle E1E2E3