# A triangle in a square ?

• Nov 11th 2012, 06:27 AM
mathsforce
A triangle in a square ?
Hi !

I've got some trouble to resolve this problem :

What is the most little surface area for a triangle containing a square of 1mē ?

It is admitted that a square' side is confunded with a triangle' side.

Thank you !
• Nov 11th 2012, 01:19 PM
HallsofIvy
Re: A triangle in a square ?
Okay, set up a coordinate system so that your square has vertices at (0, 0), (1, 0), (0, 1), and (1, 1). The smallest area triangle that will fit around that is the right triangle having (0, 0) as the vertex of the right angle and hypotenuse passing through (1, 1). If we call the point at which the hypotenuse touches the x-axis (X, 0) and the point at which it touches the y axis (0, Y), then the area of the triangle is $\displaystyle \frac{1}{2}XY$ and, using "similar triangles", $\displaystyle \frac{Y}{X}= \frac{1}{X- 1}$ or $\displaystyle Y(X- 1)= XY- Y= X$.

So the problem becomes "minimize $\displaystyle \frac{1}{2}XY$ subject to the constraint $\displaystyle XY- Y= X$.
• Nov 11th 2012, 10:15 PM
earboth
Re: A triangle in a square ?
Quote:

Originally Posted by mathsforce
Hi !

I've got some trouble to resolve this problem :

What is the most little surface area for a triangle containing a square of 1mē ?

It is admitted that a square' side is confunded with a triangle' side.

Thank you !

Here comes a slightly different approach:

1. Draw a sketch (see attachment)

2. The grey right triangles and the blue right triangles are similar:
$\displaystyle \Delta(ADH) \simeq \Delta(HKC)$ ......... and ........... $\displaystyle \Delta(EBG) \simeq \Delta(KGC)$

3. Use proportions:

$\displaystyle \frac hs = \frac{k+x}{x}~\implies~x=\frac{k s}{h-s}$

$\displaystyle \frac hs = \frac{s-k+y}{y}~\implies~y=\frac{s(s-k)}{h-s}$

4. The area of the triangle $\displaystyle \Delta(ABC)$ is calculated by:

$\displaystyle a = \frac12 \cdot (x+s+y) \cdot h$

Replace the terms x and y by the terms of #2:

$\displaystyle a = \frac12 \cdot \left(\frac{k s}{h-s}+s+\frac{s(s-k)}{h-s}\right) \cdot h$

Simplify!

5. You have found out (I hope):

$\displaystyle a(h)=\frac s2 \cdot \frac{h^2}{h-s}$

Solve for h: $\displaystyle \frac{da}{dh}=0$

You should come out with $\displaystyle h = 2s$

6. Replace h by 2s in the equation at #5:

$\displaystyle a(2s)=\frac s2 \cdot \frac{4s^2}{2s-s}=\boxed{2s^2}$
• Nov 12th 2012, 09:08 AM
mathsforce
Re: A triangle in a square ?
Thank you for your help but I don't know how to use similar triangles...
How can I do without using this ?