bluecarneal wrote: Let $AH_A$, $BH_B$ and $CH_C$ be the altitudes of triangle $\triangle ABC$. Prove that the triangle whose vertices are the intersection points of the altitudes of triangles $\triangle AH_BH_C$, $\triangle BH_AH_C$ and $\triangle CH_AH_B$ is equal to triangle $\triangle H_AH_BH_C$ HI same problem i think http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=424969

Yes, but different contest.

bluecarneal wrote: Let $AH_A$, $BH_B$ and $CH_C$ be the altitudes of triangle $\triangle ABC$. Prove that the triangle whose vertices are the intersection points of the altitudes of triangles $\triangle AH_BH_C$, $\triangle BH_AH_C$ and $\triangle CH_AH_B$ is equal to triangle $\triangle H_AH_BH_C$ what does that mean? equal in area or equal in what?

I agree that the wording is a bit vague, but it appears to be referencing area.

They mean they both are congruent... We can prove it also.. let $H1,H2,H3$ be the ortho centres of $AHBHC$,$CHBHA$ and $BHAHC$ RESP. OBSERVE THAT $HAHCH1H2$ IS A PARALLELOGRAM...(we can prove using ditance from orthocentre formula$2RcosA$ for each small triangle keeping in mind that therir circumradii a agin the ditances from the ortho centre of the parent triangle) Similar cyclic observation proves it using SSS rule [The very enlightening question i had seen]