0

For a given $\triangle ABC$, construct a $\triangle A'B'C'$, so that the $\triangle ABC$ is the extouch triangle of $\triangle A'B'C'$.

The extouch triangle is the triangle formed by the points of tangency of the excircles with their corresponding sides.

Thanks!

Blue
  • 75,673
User271828
  • 423
  • Is there a connection with the property given in: https://math.stackexchange.com/q/1623168 ? – Jean Marie Nov 20 '19 at 00:16
  • There might be, but it is still not enough for me to solve the problem. I will look more into it anyway. Thanks. – User271828 Nov 20 '19 at 00:39
  • For a given triangle ABC, the ex-center (wrt $\angle ABC$) can always be found. By dropping perpendiculars from that excenter to the sides (extended if necessary), the vertices of that extouch triangle can be found. – Mick Nov 20 '19 at 03:25
  • You are right, but my question is different. I ask, assume you are only given the intersection points of those perpendiculars from the excenters to the sides, can you find the original triangle? – User271828 Nov 20 '19 at 10:18
  • Please show the own effort to solve this problem. Else the problem is missing context. (There is no picture, there is no reference, no idea to attack. Else the notation is slightly irritating, since i always start with a triangle $\Delta ABC$, and construct from this one some $\Delta A'B'C'$, but this is my problem.) – dan_fulea Jan 14 '20 at 12:14

0 Answers0