Euclidea 9.7

Euclidea 9.7 Minimum Perimeter 2


Given triangle $$ \triangle{\rm ABC} $$, find the inscribed triangle $$ \triangle {\rm DEF} $$, whose perimeter is the minimum possible.

This problem heavily relies on a result that in fact the desired triangle has the vertices $$ {\rm D, E, F} $$ as the foot of the altitudes dropped from the vertices $$ {\rm A, B, C} $$, respectively. If you didn't know this result like me, then it takes a long time to do trial and error, until you find it out. The proof of this fact, which I did in retrospect, is actually quite cute. So I write it down here.



To see how this works, we start by picking an arbitrary point $$ {\rm D} $$ on $$ \overline{\rm BC} $$. We then find the mirror images of $$ {\rm D} $$ over $$ \overline{\rm AB} $$ and  $$ \overline{\rm AC} $$ as points $$ {\rm G} $$ and $$ {\rm H} $$.

Now for whatever choices of $$ {\rm E} $$ and $$ {\rm F} $$, it is clear that $$ \overline{\rm GF}= \overline{\rm DF} $$ and $$ \overline{\rm HE}= \overline{\rm DE} $$. The perimeter of $$ \triangle {\rm DEF} $$ is thus equal to the sum of three line segments, $$ \overline{\rm GF} $$, $$ \overline{\rm FE} $$, and $$ \overline{\rm EH} $$. It is thus clear that we should choose $$ {\rm E} $$ and $$ {\rm F} $$ to be on the straightline $$ \overline{\rm GH} $$.

Our next observation is that $$ \triangle {\rm AGH} $$ is an isosceles triangle, with sides $$ \overline{\rm AG}= \overline{\rm AH}= \overline{\rm AD} $$. On top of that, the angle $$ \angle {\rm GAH} = 2 \angle {\rm BAC} $$ is fixed, regardless of the choice of $$ {\rm D} $$. Now in order to minimize $$ \overline{\rm GH} $$, it is clear that we should choose $$ \overline{\rm AD} $$ to be as short as possible, which implies that $$ {\rm D} $$ has to be the foot of the altitude dropped from $$ {\rm A} $$.

With this fact, the 6L solution is trivial. We drop three perpendicular lines from the three vertices and then connect them.



The 8E solution takes a little bit more work:

We start by using bisector (3E) to find the mid-point of $$ \overline{\rm AB} $$ as $${\rm G} $$. Draw circle $$ \odot {\rm G}$$ with radius $$ \overline{\rm GA} $$, intersect the other two edges at $${\rm D} $$ and $${\rm E} $$. Note that since $$ \overline{\rm AB} $$ is the diameter of $$ \odot {\rm G}$$, we right away have $$ \angle {\rm AEB} = \angle {\rm ADB} = 90^\circ $$. Thus we have found two out of the three desired points.

Now instead of trying to repeat this success, we should realize that the problem is turned into finding a point $${\rm F} $$ on $$ \overline{\rm AB} $$ such that the sum $$ \overline{\rm DF} + \overline{\rm EF} $$ is minimized. That is perhaps the most classic minimum distance problem. We simply make an mirror image of $${\rm E} $$ over $$ \overline{\rm AB} $$ as $${\rm H} $$, and connect $$ \overline{\rm DH} $$ to find the last point.