Perimeter of triangle formed by connecting intersection points of altitudes.

Question

Given acute triangle $ABC$ with altitudes $AA_1, BB_1,CC_1$. How do we show that the perimeter of triangle $A_1B_1C_1$ is less than twice the length of any of the altitudes?

Answer 1

I think your question is incomplete. I am getting a condition when I solve. I am getting that the triangle must be acute(you must mention it). Now for the solution- let me denote the altitudes by AD,BE and CF respectively.(just for easiness). You can easily prove that $AE=ccosA$(in my complete solution, symbols have their usual meanings). In $\Delta AFE,EF/sinA=AE/sinC$. Manipulating you get $EF=Rsin2A$. Similarly you get the other two sides. Now you get perimeter$=R(sin2A+sin2B+sin2C)$. Using the formula in trigonometry of converting from sum to product, use it twice and you shall have the converted expression equal to $4RsinAsinBsinC$. Now prove it less than $4\Delta/a$(this is twice the length of altitude), you will lead to something like sinA>1/2. This implies that the triangle must be acute angled.