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Can anyone provide an easy to understand proof as to why the orthic triangle of an acute triangle has the smallest perimeter of all inscribed triangles?

Thanks.

dfeuer
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Teddy
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2 Answers2

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Hint: Reflect $B$ about $CA$ to $B'$.

Hint: Reflect $C$ about $B'A$ to $C'$.

Hint: Now trace what happens to your triangle.

Refer to diagram

enter image description here

Show that $DE+EF+FD \geq DD''$.

Given $D$ on $BC$, how is $D''$ determined? What can you say about $\angle DAD''$?

Hint: There is a rotation of $BDC$ about $A$ which brings us to $B'D''C'$.

How would you minimize $DD''$? Show that it is minimal when $AD \perp BC$.

Hint: We want to minimize $AD$. When does this occur, if $D$ is on $BC$?

Hence conclude that the orthic triangle has minimum perimeter of all inscribed triangles.

Note: Where in this proof did we use the fact that $ABC$ is an acute triangle?

Calvin Lin
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  • Is it right for me to say that as I shift the position of DD", ∠DAD′′ remains the same? Also, length DD" can be found using cosine rule. Since ∠DAD′′ is obtuse, in the cosine rule you will be adding the cosine angle. AD and AD" are the same length in the isosceles so the length of DD" is now dependent on length AD, which is shortest when perpendicular. So DD" = perimeter of orthic triangle will also be shortest? – Teddy Sep 19 '13 at 23:42
  • @Teddy Yes. Can you express $\angle DAD''$ in terms of angles in the original triangle? Think of a rotation. There is no need to involve cosine rule, as the angle is fixed, so all the triangles that you get are similar. Hence, we want to minimize $AD$, which occurs when D$ is the foot of the perpendicular. – Calvin Lin Sep 19 '13 at 23:57
  • @CalvinLin: you've shown that $DD''$ is minimized when $AD\perp BC$ and that the perimeter of the orthic triangle is greater than or equal to the length of $DD''$. However, why does that show that the perimeter of the orthic triangle is minimized? – robjohn Sep 20 '13 at 22:11
  • @robjohn No, I've shown that the perimeter of any internal triangle is at least $DD'$. Next, I found that $DD'$ is minimized when $AD \perp BC$. Finally, it remains to show that the orthic triangle has perimeter that is this $DD'$, but it follows from the construction. – Calvin Lin Sep 20 '13 at 22:31
  • @CalvinLin: I assume the minimizing triangle you are considering is the pre-image of the lines through the three triangles from $D$ to $D''$. How do we know that is the orthic triangle? – robjohn Sep 20 '13 at 23:35
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The standard proof relies on reflections, and is quite straightforward. Have you seen that? There is a rather clear exposition in "Introduction to Geometry" by Coxeter. Look for Fagnano's Theorem.

Bill Kleinhans
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