# Square in a Right Triangle

Discussion in 'General Math' started by Methuselah Jones, Nov 27, 2007.

1. ### Methuselah JonesGuest

I have a right triangle for which the sides are known, and want to
calculate a square which will have one corner in the right angle and the
opposite corner touching the hypotenuse. Could someone help me out with a
formula?

Methuselah Jones, Nov 27, 2007

2. ### Rick DeckerGuest

One way would be to draw your figure and identify two similar triangles.

Regards,

Rick

Rick Decker, Nov 28, 2007

3. ### MikeGuest

Do you mean something like s = ab/(a+b), where s, a, and b are the side of the square and the two sides of the triangle
respectively?

Mike, Nov 28, 2007
4. ### Methuselah JonesGuest

Carved in mystic runes upon the very living rock, the last words of Mike of
alt.math make plain:
Yep, something exactly like that! Thanks!

Methuselah Jones, Nov 28, 2007
5. ### Philippe 92Guest

Methuselah Jones a écrit :
But your 'concise' description allows for :
(see with fixed font)

.. A
.. |'-.
.. | -.
.. | -.
.. | -._
.. | ._
.. | R ._
.. | ,_ ._
.. | / --._ ._
.. | / -.._ -._
.. | .' -._ ._
.. | | '-..:._
.. | / '-=._ P
.. | / ]..
.. | / / -.
.. | ,' / -.
.. | .' / -.
.. | | .' -.
.. | / | -.
.. | / / -.
.. |/ / '-.
.. +,<-----------------------------/-----------------------------=+ C
.. B '-._ .'
.. -.._ |
.. -._ /
.. -.._ /
.. -.. /
..  Q

designed with JavE

Yeap, one corner in the right angle, opposite corner touches the
hypothenuse, true ;-)

Regards.

Philippe 92, Nov 28, 2007
6. ### MikeGuest

Which is why I started with the question "Do you mean...".

Mike

Mike, Nov 29, 2007
7. ### Methuselah JonesGuest

Carved in mystic runes upon the very living rock, the last words of
Philippe 92 of alt.math make plain:
Really? The corner remains in contact with the hypotenuse as you rotate the
square about the point of the angle? I cut out a couple of pieces of paper
and tried it, and the corner pulls away from the hypotenuse. Am I missing
something?

Methuselah Jones, Nov 30, 2007
8. ### Jim LangstonGuest

Consider the size of the square is not stated. Try a larger/smaller square
as it rotates.

Jim Langston, Nov 30, 2007
9. ### Methuselah JonesGuest

Carved in mystic runes upon the very living rock, the last words of Jim
Langston of alt.math make plain:
Ah, I see!

It turns out I'm going to have to do some more chopping of that
triangle. I need to know, given a length on either side (starting at the
right angle), the distance to the hypotenuse; in other words, a
rectangle in the right triangle. I'm going to try to figure it out
myself first, but will post here if I get stuck.

Methuselah Jones, Nov 30, 2007
10. ### Philippe 92Guest

Methuselah Jones wrote :
Jim wrote :
Hum... nothing to do with rectangles nor chopping of the triangle.

What I meant with my rotated square is that you omitted to state that
the corner in the right angle should be with the sides of the
square /along/ the sides of the triangle, or which is equivallent :
the square should be /inside/ the triangle.
A kind of joke about the accurate wording of statement (which is not
the title BTW). Sorry if you didn't take as I meant...

Otherwise there are infinitely many squares (with different sizes then)
which satisfy your original statement /as it is exactly written/.

Not only the one with Mike's formula s = ab/(a+b), but all squares
with side s between
smin = ab/(c.sqrt(2)), where c is hypothenuse
and smax = max(a,b)/sqrt(2)

(finding these limit values is a good exercice)

Regards.

Philippe 92, Nov 30, 2007
11. ### Methuselah JonesGuest

Carved in mystic runes upon the very living rock, the last words of
Philippe 92 of alt.math make plain:
I took no offense with your post, and I appreciate the importance of
stating a problem concisely. I just overlooked the possibility of
different-size squares, which of course my paper demonstration wouldn't
show me.

While Mike's formula did give me what I'm looking for at first, upon
further examination of the problem I'm trying to solve, I realized that
a square isn't going to do what I need.

To give some context, what I'm working on is building a house in Second
Life. I have a wall piece which is a right triangle (it abuts the roof).
The problem is, another wall intersects it at a right angle, meaning the
right-triangle piece spans more than one room, and I want to have all
wall pieces confined to a single room. So, I need to divide it up like
this: (please excuse any incorrect notations)

a
|\
| \
| \
| \
| \
| \
| \
| \
| \
| \d
|--------------------\
| | \
| | \
| | \
| | \
| | \
---------------------------------
b e c

The second wall intersects at ec. ab = 4.4, bc = 6 and thus ac = 7.44.
Segment be is 3 (I know, it doesn't look like it , but I want to
figure out a formula for any length of be that will give me the length
of de.

It's an interesting challenge. I've tossed around a few numbers and
ratios and such and have not come up with a solution, but it's been
interesting. What I could use now are some suggestions or hints. I'm not
ready to ask for the formula yet, but what should I be looking at? Give
me a clue to what numbers I should be looking at.

Methuselah Jones, Dec 1, 2007
12. ### Jim LangstonGuest

Lets look at your diagram, and we'll go as far as to say that fd (I added
the f) is the same length as de so it's a square. Now, how many triangles
do you have? I count 3 now. abc, afd, dec. Is there any formula that
talks about the ratios of the sides? Consider: ac = ad + dc. I think
that's as much info as I can give without giving the formula I would use,
there may be a better one. I'm not actually any type of math expert at all,
just a 3d game programmer, but when I see any trype of triangle at all I try
to throw the Pythagorean theorem at it. It usually works

Jim Langston, Dec 1, 2007
13. ### KenGuest

What is the name of this formula?

Ken, Dec 1, 2007
14. ### Philippe 92Guest

Jim Langston wrote :
Methuselah Jones wrote "I realized that a square isn't going to do
what I need".
But Pythagora is not a magic spell you throw against triangles.... ;-)

The key here is ... tadaaa ... "similar triangles" !
(arranged the diagram a little)

There are three triangles, yes, and they are /similar/. That is exacly
of the same shape, just different dimensions. Thus giving ratio of
corresponding sides.

abc, afd and dec similar (noting the vertices in this exact order, so
that they correspond)

gives :
ab/bc = af/fd = de/ec and ... from which you can expresss de as
function of be = bc - ec

Philippe 92, Dec 1, 2007
15. ### Methuselah JonesGuest

Carved in mystic runes upon the very living rock, the last words of
Philippe 92 of alt.math make plain:
I almost thought I had it, but then it all comes down to the unknown
length of de (and fb), which is what I'm trying to find. Thus, I can't
figure af/fd or de/ec. I know ab/bc, which is .7333 (how do you express
repeating in ASCII?), but that's it. be and ec are both 3, but I don't
know how to work back from that. Similarly, in what Jim wrote, I don't
know the length of ad or dc, because those all depend on de.

I think if I can figure out *why* s = ab/(a+b) works for a square, I
might be on to something.

Methuselah Jones, Dec 1, 2007
16. ### Methuselah JonesGuest

Carved in mystic runes upon the very living rock, the last words of
Philippe 92 of alt.math make plain:
Oh, duh!!! I see it now.

a
|\
| \
| \
| \
| \
| \
| \
| \
| \
|f \d
+--------------------+
| | \
| | \
| | \
| | \
| | \
+--------------------+-----------+
b e c

The triangles are all the same *ratio*. Thus (ab/bc) * fd gives us 2.2,
which makes sense, since ab = 4.4, bc = 6 and be = 3.

Got it! Thanks!

You know, I studied all this stuff once upon a time in high school, but
I think the one of the great failures of education has always been that
it too often doesn't teach application. Sure, I studied algebra and
geometry, but what could it be *used* for? Measuring flagpoles is about
the only thing I remember. Had we, say, constructed a model house, as
I'm doing now, or done some similar thing that would demonstrate
*application*, it might have been different. As it is, unless you're one
of the few interested in math for the sake of math, or who manage to
make the application leap oneself, not much sticks, and the student ends
up thinking, "Well, I guess it's just something I'm not good at," which
effectively shuts the door on that subject forever.

But along the way since school, I've learned that data is data -- if
you're capable of learning one thing, you're capable of learning
something else. I'm not stupid -- I'm a competent programmer, and in
fact have been succesfully self-employed as a freelance for over 15
years. Yet, the teacher who taught me computers in high school was the
same one who (tried to) teach me geometry and algebra, and in fact he's
one of the teachers I regard as having had a profound impact on my life.
So why was he able to teach me one but not the other? Well, a) I was
interested in computers for the sake of computers (it was fun!) and b) I
was able to make the application leap myself -- I remember writing my
first practical application program back then, not as an assignment, but
as a self-inspired project on my own time.

This is the first time I've applied the data=data principle to math,
though, and I'm finding it quite intriguing.

Methuselah Jones, Dec 1, 2007
17. ### Philippe 92Guest

Methuselah Jones wrote :
Glad to have helped you in realizing that :
And also with math here, there is another way of looking at this.
And this is helpfull too : there are generally several different ways
of solving a problem !

As you see on the little modified diagram (added x and y) you can
consider x = be and y = bf = ed as the coordinates of point d.

Hence line ac is the graph of y = function of x (that is de = function
of be). Being a straight line, this just means that the function is a
"linear relation" of the form :
y = A*x + B

We have just to find values of A and B...

For x = 0, y = B, that is B = ab = 4.4 (when d comes in a)
For x = bc = 6, we have y = 0 (when d comes in c)
So 0 = A*6 + 4.4 and solving this equation for A gives A = - 4.4/6
and finally the wanted relation y = de as function of x = be is

y = 4.4 - (4.4/6)*x

You can use algebra too to find for which position of d the area of
rectangle bfde is maximum (which may be usefull as we may want the
piece to be "as large as possible". But this is an other matter.
Not too hard but a little more.

Let's try :
Area is x*y that is Area = S = 4.4*x - (4.4/6)*x^2
for generality and ease of writing, I'll keep symbolic values, that is
S = B*x - A*x^2
The general way of finding the maximum of this is to use calculus...
Let's discard this high level method and find a simpler one.
That is just algebra.
First of all I'll rearange this equation so as to get easier
calculations, with x^2 coefficient = 1 :
-S/A = x^2 - (B/A)*x
In dealing with quadratic equations, there is a worthfull method named
"complete the square". That is consider x^2 - (B/A)*x as the beginning
of the square of (x - p) : (x - p)^2 = x^2 - 2*p*x + p^2
so we identify B/A to 2*p giving p = B/(2A) and
-S/A = (x - B/(2A))^2 - (B/(2A))^2
because the p^2 = (B/(2A))^2 term of the square is not in the original
S, and we have to substract it to cancel.

And returning to S itself :

S = B^2/(4A) - A*(x - B/(2A))^2

But now becomes clear what is the maximum of S :
the (x - B/(2A))^2 being >= 0 (a square) S will be maximum when we
substract from the constant value B^2/(4A) an A*(x- B/2A)^2 as little
as possible, that is 0.
Then the maximum area is Smax = B^2/(4A) when x - B/2A = 0, that is
when x = B/(2A).
Now just replace A = 4.4/6 and B = 4.4

There is also some geometric method to find the max area, but it is
not so easy to see : you have first to stretch the triangle so that
bc = ab. Then the perimeter of rectangle becomes constant, and the
maximum area for a constant perimeter rectangle is when it is a square.
Then inverse stretch this square to get the maximum area rectangle in
the original triangle.

Regards.

Philippe 92, Dec 2, 2007
18. ### Methuselah JonesGuest

Carved in mystic runes upon the very living rock, the last words of
Philippe 92 of alt.math make plain:
Actually, my formula was incomplete, which I realized when trying to
generalize it. It only works when be is exactly 1/2 of bc. With the
formula I came up with, be and de are in direct ratio, when they need to
be in inverse ratio, which subtracting from ab provides, as you show
above.

Oof! I barely made it this far!

As a programmer, my method of thinking is to find a problem and come up
with a solution to it. I can't follow what you wrote below, because I
don't have a problem I can apply it to right now, and my math knowledge
is insufficient to make the "application leap" without that. I do want
to expand my knowledge further, so I'll be working with building pieces
to see what other problems I can come up with, but this is sufficient
for now.

Methuselah Jones, Dec 2, 2007
19. ### Philippe 92Guest

Methuselah Jones wrote :
Take care that "inverse ratio" exactly means de*be = some constant.
Here it is not properly "inverse ratio", they just vary in opposite
directions : just one increases while the other decreases, but without
stating in exactly how much.

The exact formula is the one I gave above, or what can be deduced
directly from triangle ratios :

ab/bc = de/ec = de/(bc-be) that is de = ab - (ab/bc)*be = the same.
I don't know for Jim, but remember the first formula by Mike :
s = a*b/(a+b) for a square. Or with present notations s = de = be
and a = ab, b = bc :
de = be = ab*bc/(ab+bc)

A square is when be = de,
that is from the general formula de = ab - (ab/bc)*de
multiply by bc and get de*bc = ab*bc - ab*de or
de*(ab+bc) = ab*bc, that is de = ab*bc/(ab+bc), yeah !

Philippe 92, Dec 3, 2007