# Squares in triangles ?

• November 27th 2012, 01:17 PM
mathsforce
Squares in triangles ?
Hi everyone !

I've got some trouble...

IJKL is a square and [IJ] is confunded with [BC]
h is the height from A.
C1 is the red square
C2 is the green square
C3 is the blue square
a= [BC]
b= [CA]
c= [AB]

1. How to find IJKL using a and h
2. a< b< c. Classify C1, C2, C3 from the largest.
3. Find ABC triangles (their surface, S) when the square IJKL has the largest surface.

• November 30th 2012, 06:57 PM
Stephen347
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 $\angle$ALK = $\angle$ABC and $\angle$AKL = $\angle$ACB. Therefore, $\triangle$ALK is similar to $\triangle$ABC. Since these triangles are similar:

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

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

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

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 $\triangle$ADK is similar to $\triangle$AHC. Therefore,

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

And,

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

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

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

And then, solving for s, gives:

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

Could you explain exactly what you mean in part 3. of your question a little more? Thanks.
• December 2nd 2012, 09:13 AM
Stephen347
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 $\triangle$ABC. Let the the altitude to side a be ha, the altitude to side b be hb, and the altitude to side c be hc. Then after a few seconds looking at the many right triangles created, I think you will agree that:

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

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

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

From these equations, it follows that:

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

And we know that, from my previous post:

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

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

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

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

First, we know that $\sin(C)\ <\ 1$ 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:

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

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

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

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

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

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

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

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

i.e.

C1 > C2. A similar procedure shows that C2 > C3. Thus, we have shown that:

$C_{3}\ <\ C_{2}\ <\ C_{1}$
• December 2nd 2012, 10:12 AM
Stephen347
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 $ah_{a}$ 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 $a+h_{a}$ to maximize C1. Since the area is constant, we have that:

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

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

Thus, our goal is to minimize:

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

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

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

Solving for a gives,

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

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

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

Thus,

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

And so, in conclusion, the side length of the square is maximized when $a\ =\ h_{a}\ =\ \sqrt{2A}$. 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 $a\ =\ h_{a}\ =\ \sqrt{2A}$, and so the C1 square inside any one of these triangles is the same size.