Areas in triangle

Question

In the above isosceles right triangle ABC, with its two sides $AB = AC = 1$ unit, we take a random point D on the hypotenuse and draw perpendicular lines to the sides AB and AC, which intersect them at points E and F respectively. Show that the maximum of the three areas AFDE, EBD and CDF is always $\geq \frac {2}{9}$.

If we set $EB = x$ then, area $EBD = \frac {x^2}{2}$,

$AFDE = (1-x)*x$,

$CDF = \frac {(1-x)^2}{2}$.

Also $AFDE + EBD + CDF = \frac {1}{2}$.

Clearly the 3 areas can't be equal. We can only have 2 of them equal, when $x=0.5$ or $x=0.33$ or $x=0.66$.

When $x=0.33 = \frac {1}{3}$ then

$EBD = \frac {x^2}{2} = \frac{1}{18}$ and $AFDE = CDF = \frac{1}{2}*(\frac{9}{18}-\frac{1}{18}) = \frac{2}{9}$.

But I don't know if this is a sufficient proof.

Answer 1

Yes you are right to compare them in this way.

But in fact, you should have an array divided into intervals

$$[0,1/3],[1/3,2/3],[2/3,1]$$

showing which one is "on the top of the podium" on each interval. This is made rather convincing in terms of analysis, i.e., when you see it with curves.

Fig. 1: Curves with equations $y=\dfrac12 x^2, \ y=x(1-x), \ y=\dfrac12 (1-x)^2$ in red, black, blue resp. Their max function, in green, is always above the ordinate of the intersection of the two first curves, or the two last ones as well, which is $\tfrac29$. Therefore, you were right to compare the cases of equality.

Edit: These three polynomials, up to a factor $\frac12$ are the same as those in the 3rd figure here. They are particular cases of a general family called "Bernstein polynomials" useful in different part of mathematics such as Bezier curves or probability (order statistics).

Answer 2

That is correct, but not quite sufficient. Let's denote the areas of the three shapes $\alpha,\beta,\gamma$ (where $\alpha$ represents the upper triangle, $\beta$ the rectangle and $\gamma$ the lower.

Then: $$\max_{x\in[0,1]}\{\alpha,\beta,\gamma\}=\begin{cases}\gamma & 0\leq x\leq \frac13 \\ \beta & \frac13\leq x\leq \frac23 \\ \alpha & \frac23\leq x\leq 1\end{cases}$$

This is a result of the equalities you found (indeed, they indicate when the function behind the maximum changes).

So, you simply need to show that each function is $\geq \frac29$ on the domain on which it is maximal.

Answer 3

Start with D as the mid point of BC. WLOG, imagine D moves up by a vertical distance of $\Delta x$ .

\begin{align}S_{AEDF}=\frac{1}{4}-\Delta x^2, S_{\triangle DFC}=\frac{1}{2}(\frac{1}{2}+\Delta x)^2\end{align}

\begin{align}S_{AEDF}\geq \frac{2}{9}(\Delta x \leq\frac{1}{6})\\ S_{\triangle DFC}\geq \frac{2}{9}(\Delta x\geq \frac{1}{6})\end{align}

Similar conclusion can be obtained if D moves down by $\Delta x$.