# A question on elementary plane geometry

Given a triangle $ABC$, let $S$ be an inner point of this triangle. Let $P$, $Q$, $R$ be the orthogonal projection of $S$ respectively on the three sides of this triangle. Are there beautiful methods (those which involve geometric insight with the least calculation) to find out the maximum among all $S$ of the product of non oriented lenghts $SP$, $SQ$, and $SR$?

Thank you!

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Do you mean the maximum among all $S$? – Lord_Farin Oct 10 '12 at 20:50
Yes, thanks for asking! – HYL Oct 10 '12 at 20:55
If I'm not mistaken, the $S$ that maximizes the product ought to be the intersection of the medians. But my derivation of that involves a bit of algebra and rudimentary calculus -- as well as some diagrams that I cannot reproduce in an answer right now. – Henning Makholm Oct 10 '12 at 21:16
Yes, that's what I was thinking (well, actually I thought the answer would be the, equivalent, center of the inscribed circle). – Lord_Farin Oct 10 '12 at 21:19
@Lord_Farin: To see that the center if the inscribed circle doesn't work, consider a very narrow isosceles triangle (with base angles very close to right). The inscribed circle has a center very close to the short side, but if you start S there and move towards the apex, the distance to the short side can easily double with hardly any corresponding decrease in the two other distances -- so their product will also become larger at the beginning of that movement. – Henning Makholm Oct 10 '12 at 21:27

## 2 Answers

I don't know if you'd consider this as an algebraic or geometric solution.

$$SP \cdot AB = 2\mbox{Area} (SAB)$$ and similarly the other two.

Thus

$$SP \cdot SQ \cdot SR = 8 \frac{ \mbox{Area} (SAB) \mbox{Area} (SBC) \mbox{Area} (SAC)}{AB \cdot BC \cdot AC }$$

Hence, the product is maximal if the product $\mbox{Area} (SAB) \mbox{Area} (SBC) \mbox{Area} (SAC)$ is maximal. Maximizing thsi product is very easy with some algebra:

$$\sqrt[3]{\mbox{Area} (SAB) \mbox{Area} (SBC) \mbox{Area} (SAC)} \leq \frac{ \mbox{Area} (SAB) + \mbox{Area} (SBC) + \mbox{Area} (SAC)}{3}= \frac{ \mbox{Area} (ABC) }{3}$$

with equality if and only if $\mbox{Area} (SAB) = \mbox{Area} (SBC) = \mbox{Area} (SAC)$. Thus the maximum is obtained when

$$\mbox{Area} (SAB) =\mbox{Area} (SBC)= \mbox{Area} (SAC)$$

It is easy to conclude that $S=G$ .

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Thank you, this is the kind of solution that I was looking for. – HYL Oct 11 '12 at 5:21

Here's a partially geometric solution. For brevity let's write $x$, $y$, $z$ for the distances $SP$, $SQ$, $SR$, and $b$, $a$, $c$ for the corresponding side lengths, as in the following diagram. Furthermore, let $h$ be the length of altitude $CH$, parallel to the line $z$:

We want to find an $S$ that maximizes $xyz$.

From calculus we know that if $S$ is at a local extremum at we move it an infinitesimal distance in any direction, $xyz$ must not change (to first order). In particular, this must be true if we imagine moving $S$ a small distance $\varepsilon$ to the left (away from point $F$ in the diagram). When we do so, $x$ increases and $y$ decreases, but since we're moving $S$ horizontally, $z$ stays the same.

Because triangles $HAC$ and $PFS$ are similar we can compute the increase in $x$ as $\Delta x = \varepsilon h/b$. Mutatis mutandis we find $\Delta y = -\varepsilon h/a$ (with the opposite sign because an increase in $x$ corresponds to a decreas in $y$).

A bit of simple algebra now gives $$\Delta\, xyz = (x+\Delta x)(y+\Delta y)z - xyz = \frac{\varepsilon hyz}b - \frac{\varepsilon hxz}{a} + \text{terms involving }\varepsilon^2$$

Because $\varepsilon$ is small we can ignore the $\varepsilon^2$ term. Since $\Delta xyz$ must be $0$ we then have $$\frac{\varepsilon hyz}b = \frac{\varepsilon hxz}a$$ Canceling common factors (all nonzero, since $z>0$ inside the triangle) and cross-multiplying gives us $$\tag{1} ya = xb$$ which we recognize as the equation of a line in the $x,y$ coordinate system. (This coordinate system is not rectangular, but equations of lines look the same in skew coordinates). If we can find two different solutions to $(1)$, the line connecting them will be a locus that every local extremum of $xyz$ inside the triangle must lie on.

The corner $C$ is is clearly a solution to $(1)$ because there we have $x=y=0$. Where can we find another one? The midpoint between $A$ and $B$ must be one, because there both sides of $(1)$ happen to be the area of triangle $ABC$.

So every local extremum of $xyz$ must lie on the median from $C$, and repeating the argument with another orientation we can see that it must lie on the other medians too. So the intersection of the medians is the only place in the inside of a triangle that can be a local extremum.

Since there quite obviously must be a global maximum somewhere in the triangle, the intersection of the medians has to be it.

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