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Consider a general triangle, $\triangle ABC$. The lengths of all sides are known. We are looking for the largest possible rectangle that can be inscribed into the triangle in such a way that one side of this rectangle lies on one side of the triangle (in the exercise in the book, it lies on $c$)

(For numerical results, let $\overline{BC} = a = 249, \quad \overline{AC} = b = 151.8 \quad$ and $\quad \overline{AB} = c = 238.9$)

Here's a little drawing:

photo of drawing

What I got so far:

I found that $x = \frac{1}{2}c$ and thus $y = \frac{1}{2} H$ (where $H$ is the altitude that belongs to side $c$). I couldn't prove this rigorously though: the way I found this out is that I considered three cases:

  1. $x > \frac{1}{2}c \rightarrow$ the rectangle is 'taller'
  2. $x < \frac{1}{2}c \rightarrow$ the rectangle is 'wider'
  3. $x = \frac{1}{2}c$

For case 1: If I imagine folding the three smaller triangles around the rectangle over the rectangle, but they will overlap $\rightarrow T_{\text{rectangle}} < \Sigma T_{\text{small triangles}}$

For case 2: If I imagine folding again, I get the same inequality as for case 1

For case 3: This is the only case when there's no overlap and $T_{\text{rectangle}} = \Sigma T_{\text{small triangles}}$

Now I suppose I can say that $T_{\text{rectangle}} + \Sigma T_{\text{small triangles}}$ is constant (it's the area of the big triangle), so $T_{\text{rectangle}}$ is maximal for case 3 (How should I prove this rigoruously?)

Then of course from $x = \frac{1}{2}c$ it follows that $y = \frac{1}{2}H$ and also that the inscribed rectangle's area is half of the triangle's area.


So there, I somewhat solved it... But I would like to see a proof that is more concise, where there are no ambiguous steps. I know that I could also use calculus (at first I was trying to differentiate the area of the rectangle which is $T(x,y) = xy$, but to do that I'd have to express $y$ in terms of $x$ (or vice versa) and it didn't really work out... I'd appreciate to see how that would be done though, I'm sure it's not hard). However, this problem is in a secondary school book, I don't think it involves calculus at all.

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  • $\begingroup$ You are not allowed to use calculus? $\endgroup$ – user Mar 25 '18 at 21:12
  • $\begingroup$ @gimusi I was trying to help a friend who doesn't know calculus yet (and I failed pretty badly as you can see) --- I think I now know how to solve this using calculus, but I'm also looking for a pre-calculus solution that I can explain to said friend $\endgroup$ – bertalanp99 Mar 25 '18 at 21:13
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The problem can be tackled through the AM-GM inequality. For simplicity, let us regard the $ABC$ triangle as the union of two right triangles having legs $c_1,h$ and $c_2,h$, with $c_1+c_2=c$.
The height $y$ of the inscribed rectangle fixes its area $$ A = y\left(c-\frac{y}{h}c_1-\frac{y}{h}c_2\right)=\frac{y}{h}\left(1-\frac{y}{h}\right)hc $$ and $\frac{y}{h}\left(1-\frac{y}{h}\right)$ attains its maximum value when $y=\frac{h}{2}$. In such a case $A=\frac{hc}{4}=\frac{[ABC]}{2}$.
The fact that the area of the rectangle cannot exceed half the area of the triangle also follows from a nice origami argument: $ABC\setminus\text{rectangle}$ is made by three triangles that can be folded along three sides of the rectangle, and when we actually fold them... enter image description here

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Someone has posted a calculus hint but for some reason immediately deleted it --- nevertheless, I'd like to add that calculus solution for future reference for visitors of this question:

The area of the rectangle in terms of $x$ and $y$ is $T(x,y) = xy$

Let $H$ denote the height of the large triangle and let $h$ be the height of the top small triangle. Then of course $0 \leq h \leq H$

Since $\triangle GFC$ and $\triangle ABC$ are similar, the following stands:

$$ \frac{x}{c} = \frac{h}{H} $$

therefore $x = \frac{h}{H} c$

We can also express $y$ in terms of $h$ since $y = (H-h)$

Substituting back $x$ and $y$ into $T(x,y)$:

$$ T(h) = \frac{h}{H} c (H - h) \implies \frac{\mathrm{d} T}{\mathrm{d} h} = c - 2 \frac{c}{H} h $$

To find the maximum of $T$, we set $T'(h) = 0$, thus

$$ c = 2 \frac{c}{H} h \\ h = \frac{1}{2} H $$

($T(h)$ is a parabola where the coefficient of $h^2$ is negative, so it opens downwards and can only have a maximum which we have just found)

We have shown that $T$ is maximal when $h = \frac{1}{2} H \implies y = \frac{1}{2} H$

From this, the length of $x$ and the area of the rectangle follows.

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  • $\begingroup$ To fit secondary school program, apply the fact that the first coordinate of the vertex of parabola is arithmetic mean of zero points (if any), here $h={{0+H}\over 2}.$ $\endgroup$ – user376343 Nov 3 '18 at 23:28

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