SMO(J) '24 Q1

(Dylan here.) I realised the general content on this blog is pitched at a rather high level. Here's an attempt to balance it.

(SMOJ '24 Q1) Let $ABC$ be an isosceles right-angled triangle of area $1$. Find the length of the shortest segment that divides the triangle into $2$ parts of equal area.

First, let’s draw some pictures and find out what the side lengths are: if $|AB| = |BC| = x$, then
$$\text{Area}(ABC) = \frac{1}{2}x^2 = 1 \implies x = \sqrt{2}.$$
And for good measure, the length of the hypotenuse is determined by Pythagoras: $|AC| = \sqrt{x^2+x^2} = \sqrt{2} x = 2$.

Okay, now let’s try to cut the triangle. I know I can divide the triangle into $2$ parts symmetrically, as in case (0). This gives us an upper bound of $1$ for the shortest segment.

But more generally, the line segment can touch any two of the $3$ sides, giving us $3$ cases. And, note that cases (I) and (II) are symmetric via reflection, so it suffices to consider cases (I) and (III). And from my drawings, it looks like (I) has a chance of being shorter (while (III) looks far from optimal).

Perhaps we eliminate case (III) first. Let's draw in the edge-cases (i.e. extremes) of (III): the two medians.

Okay, so any dividing line segment must interpolate between (i.e. be somewhere in the middle of) the two. Concretely, we may bound
$$|DE|^2 = |BD|^2+|BE|^2 \geq |BM|^2+|BN|^2 = 1$$
so it is impossible for case (III) to have a shorter segment than (0).

Now, let's study case (I)! at this point I am tilting the picture to its side, and imagining not a triangle, but a whole $45^\circ$-sector: this cup holds some water of volume $\frac{1}{2}$ (actually, area), and I am tilting it to minimize the surface area (actually, length).

So, what angle of the cup/position of the line segment minimises its length? I’m guessing that it is minimised when it is symmetric (i.e. the cup is as vertical as possible). For now, my intuition is pretty much Occam's razor (there's no reason to believe asymmetry is better).

Now, I need to try to understand and formalise why symmetry is optimal. This involves giving an expression of the area $S = \frac{1}{2}$ of the water, in terms of lengths in the diagram. There are a few ways that I can conceive of doing this:

1. (trigonometry) $S = \frac{1}{2} |AD| \cdot |AE| \sin 45^\circ$.
2. (Heron’s) $S = \sqrt{\frac{|AD|+|AE|+|DE|}{2}\cdot \frac{|AD|+|AE|-|DE|}{2}\cdot \frac{|AD|-|AE|+|DE|}{2}\cdot \frac{-|AD|+|AE|+|DE|}{2}}$.
3. (altitude) $S = \frac{1}{2}|DE|\cdot h$, where $h$ is the height from $A$ to the line through $DE$.

Among the $3$ options, I go for the simplest one that directly involves the quantity that I’m interested in (the side length $|DE|$). Notice that this involves the height $h$ that is not drawn into the diagram; so I simply draw it in. (Philosophically, the height $h$ is not any more complicated than the side lengths $|AD|$ and $|DE|$!)

Now, minimising the base length $|DE|$ is equivalent to maximising the height $h$. So let's overlay the asymmetric picture with the symmetric picture:

Consider the line $l$ through $DE$. Tilting the slanted cup upright will bound a region with $l$ of smaller area (pink-ish in the above diagram), since "an infinitesimal angle further away from the vertical, spans a larger infinitesimal length along $l$". Therefore, in order to preserve the area $S=\frac{1}{2}$, the symmetric height $h_0$ must be larger than $h$. Nice!

Finally, let's compute the length $|D_0E_0|$ that is optimal for case (I). It's straightforward with trigonometry, but I don't remember knowing much trigonometry at 13 years old, so let's try to Pythagoras our way through. We have the first equation, $|D_0E_0|^2 + (h_0/2)^2 = |AE_0|^2$. Dropping the perpendicular from $D_0$ to $AE$ at point $F$, say, we now also use that the angle is $45^\circ$, to get $|AF| = |D_0F| = |AD_0|/\sqrt{2}$. There's another equation of Pythagoras for triangle $D_0FE_0$. And finally, the area condition (expressing base-times-height in two different ways) is $|D_0E_0|\cdot h_0 = |AE_0|\cdot |FD_0| = 1$.

This bunch of equations should now be solvable. In any case, if we let $|D_0F| = s$, then all lengths are expressible in terms of $s$:

$s$ is determined by the area condition $\sqrt2 s \cdot s = 1$ (so $s = 1/\sqrt[4] 2$). Now we have
$$|D_0E_0| = \frac{\sqrt{1^2+(\sqrt 2 - 1)^2}}{\sqrt[4] 2}.$$

Now, is this actually less than $1$? Let's check.
$$|D_0E_0|^4 =\frac{(1^2+(\sqrt 2 - 1)^2)^2}{2} = \frac{(4-2\sqrt 2)^2}{2} = \frac{8(\sqrt2-1)^2}{2} = (2\sqrt 2-2)^2 < 1$$
(because $\sqrt 2 < 1.5$), so it is indeed less than $1$, and is the shortest such line segment. Hooray~

Conclusion: the shortest such segment is length $\sqrt{2\sqrt 2 - 2}$.

Alternative finish

The above argument is perhaps not too easy to formalise without calculus. With trigonometry, the optimisation problem (in case (I)) is: minimise $|DE|^2 = x^2+y^2-2xy\cos 45^\circ$ subject to $\frac{1}{2} xy \sin 45^\circ = \frac{1}{2}$. 

Then instead of calculus, one may appeal to AM-GM/standard inequalities: 

$$|DE|^2 \geq 2xy-2xy\cos 45^\circ = 2(1-1/\sqrt 2) \sqrt 2 = 2\sqrt 2 - 2$$

with equality when $x = y=\dots$ (some explicit value, that one may check to be achievable).

[edit: it was pointed out that I didn't actually finish the geometric argument above, justifying that the (asymmetric) blue triangle has greater area than the (symmetric) pink triangle; I only handwaved some calculus intuition (formalisable with polar integrals, but 13-year-olds are not polar bears). 

So, here's a hint to a direct argument. While I cannot move the blue triangle to cover the pink triangle, I can certainly move the "blue-minus-pink" triangle to cover the "pink-minus-blue" triangle. Concretely, one may either rotate or reflect one triangle onto the other, then show certain side lengths are greater than others. This proves the desired inequality of area. This translates to an inequality $h<h_0$ of heights, by observing the symmetric blue triangle is a dilated (i.e. scaled up) version of the pink triangle to be of the correct area.]