Re: Squares in triangles ?

Working on parts 2 & 3, but I figured I would post part 1. In your drawing, LK and BC are parallel, making $\displaystyle \angle$ALK =$\displaystyle \angle$ABC and $\displaystyle \angle$AKL =$\displaystyle \angle$ACB. Therefore, $\displaystyle \triangle$ALK is similar to $\displaystyle \triangle$ABC. Since these triangles are similar:

(1) $\displaystyle \frac{LK}{BC}\ =\ \frac{AK}{AC}$

If the side length of the inscribed square above side a is s, then the equation above says:

(2) $\displaystyle \frac{s}{a}\ =\ \frac{AK}{AC}$

Now we draw the height of the triangle down from vertex A to side a, and call it h, meeting at the point H on a. Let the intersection of AH and LK be D. Line LK splits AH into two segments, the upper (AD) which has length h-s and the lower (DH) which has length s. Then, we now know that $\displaystyle \triangle$ADK is similar to $\displaystyle \triangle$AHC. Therefore,

(3) $\displaystyle \frac{AK}{AC}\ =\ \frac{AD}{AH}$

And,

(4) $\displaystyle \frac{AD}{AH}\ =\ \frac{h-s}{h}$

Thus, by equations (2), (3), and (4),

(5) $\displaystyle \frac{s}{a}\ =\ \frac{h-s}{h}$

And then, solving for s, gives:

(6) $\displaystyle s\ =\ \frac{ah}{a+h}$

Could you explain exactly what you mean in part 3. of your question a little more? Thanks.

Re: Squares in triangles ?

Alright, so I think I've figured out part two. To start, draw the altitudes to each of the sides in the acute $\displaystyle \triangle$ABC. Let the the altitude to side a be h_{a}, the altitude to side b be h_{b}, and the altitude to side c be h_{c}. Then after a few seconds looking at the many right triangles created, I think you will agree that:

(1) $\displaystyle \sin(A)\ =\ \frac{h_{b}}{c}\ =\ \frac{h_{c}}{b}$

(2) $\displaystyle \sin(B)\ =\ \frac{h_{c}}{a}\ =\ \frac{h_{a}}{c}$

(3) $\displaystyle \sin(C)\ =\ \frac{h_{a}}{b}\ =\ \frac{h_{b}}{a}$

From these equations, it follows that:

(4) $\displaystyle ah_{a}\ =\ bh_{b}\ =\ ch_{c}$

And we know that, from my previous post:

$\displaystyle C_{1}\ =\ \frac{ah_{a}}{a+h_{a}}$

$\displaystyle C_{2}\ =\ \frac{bh_{b}}{b+h_{b}}$

$\displaystyle C_{3}\ =\ \frac{ch_{c}}{c+h_{c}}$

We will show that $\displaystyle C_{3}\ <\ C_{2}\ <\ C_{1}$.

First, we know that $\displaystyle \sin(C)\ <\ 1$ because C is a positive angle in an acute triangle. Then we multiply both sides of this inequality by the positive quantity (b-a) to get:

$\displaystyle (b-a)\sin(C)\ <\ (b-a)$

$\displaystyle b\sin(C)-a\sin(C)\ <\ b-a$

By (3), $\displaystyle b\sin(C) = h_{a}$ and $\displaystyle a\sin(C) = h_{b}$. So:

$\displaystyle h_{a}-h_{b}\ <\ b-a$

$\displaystyle a+h_{a}\ <\ b+h_{b}$

$\displaystyle \frac{1}{a+h_{a}}\ >\ \frac{1}{b+h_{b}}$

And since, by (4), $\displaystyle ah_{a}\ =\ bh_{b}$,

$\displaystyle \frac{ah_{a}}{a+h_{a}}\ >\ \frac{bh_{b}}{b+h_{b}}$

i.e.

C_{1} > C_{2}. A similar procedure shows that C_{2} > C_{3}. Thus, we have shown that:

$\displaystyle C_{3}\ <\ C_{2}\ <\ C_{1}$

Re: Squares in triangles ?

For part three, this problem is similar to minimizing the perimeter of a square while keeping the area constant. We know that $\displaystyle ah_{a}$ is constant because it is twice the area of the triangle (i.e. 2 times a constant A). Thus, we just have to minimize the quantity $\displaystyle a+h_{a}$ to maximize C_{1}. Since the area is constant, we have that:

$\displaystyle \frac{ah_{a}}{2}\ =\ A$

$\displaystyle h_{a}\ =\ \frac{2A}{a}$

Thus, our goal is to minimize:

$\displaystyle f(a)\ =\ a+\frac{2A}{a}$

One way to do this is to treat it as a function of a on the domain [0,$\displaystyle \infty$). This is a differentiable function on (0,$\displaystyle \infty$), so if it has a minimum it will be at a place where the derivative is zero. i.e.

$\displaystyle f'(a)\ =\ 1-\frac{2A}{a^2}\ =\ 0$

Solving for a gives,

$\displaystyle a\ =\ \sqrt{2A}$

This is a global min for f on [0,$\displaystyle \infty$). And so:

$\displaystyle h_{a}\ =\ \frac{2A}{a}\ =\ \frac{2A}{\sqrt{2A}}\ =\ \sqrt{2A}$

Thus,

$\displaystyle C_{1}\ =\ \frac{\sqrt{2A}^2}{\sqrt{2A}+\sqrt{2A}}\ =\ \frac{\sqrt{2A}}{2}$

And so, in conclusion, the side length of the square is maximized when $\displaystyle a\ =\ h_{a}\ =\ \sqrt{2A}$. When this happens, the area of the inscribed square is half the area of the triangle. As far as the type of triangle in which this can happen, there is a little choice. It can be an isosceles right, equilateral, or scalene triangle. To see this, draw a square of length a. The base of the square is the side BC and A can be any point on the top line of the square. Drawing the triangle between these three points gives a triangle with $\displaystyle a\ =\ h_{a}\ =\ \sqrt{2A}$, and so the C_{1} square inside any one of these triangles is the same size.