From $D$ on any side of $\triangle ABC$, drop perpendiculars $DP$, $DQ$ to other sides. Find minimum value of $|PQ|$.

Question

Israel Olympiad:

Lengths of the sides of $\triangle ABC$ are $4$, $5$, and $6$. At any point $D$ on any side, drop perpendiculars $DP$ and $DQ$ to the other sides. Determine the minimum value of $|PQ|$.

:( Maybe I misunderstood the problem, because it looks trivial.

Let us consider the case where $D$ lies on $AB$. Then consider the 4-gon $CPDQ$. It is inscribed into a circle. $PQ$ equals $2R\sin(C)$. This means that you need to minimize $R$, as the sine is constant. $2R$ equals $CD$, thus $CD$ must be minimal, thus $D$ is the foot of the altitude. Now perform the computations and consider two other cases.

Answer 1

Yes, it's indeed trivial.

Let $AB=6$, $AC=5$, $AC=5$ and $D\in AB$.

Thus, $QDPC$ is cyclic and $CD$ is a diameter of the circle.

Thus, by the law of sines we obtain: $$PQ=CD\sin\gamma\geq h_c\sin\gamma.$$

Id est, since our triangle is an acute-angled triangle, the needed value is $$\min\{h_a\sin\alpha,h_b\sin\beta,h_c\sin\gamma\}.$$ Maybe you missed that it's very important that our triangle is acute-angled?

Answer 2

Your understanding is correct, except that you may not realize that:

No matter which side of the triangle you work with, you always end up with the same minimum value for PQ.

So, if you are interested, you could try to prove the general result that the minimum value of PQ is given by,

$$ PQ_{min} = K/R$$

where $K$ is the area of the triangle and $R$ is the radius of the circle that circumscribes the triangle. This general result is invariant of the side on which you place the point D.