Need help simplifying an equation

[DrToddMatthews]

I'm enrolled in a calculus class at City Colleges of Chicago; it's a community college math course.

I have a solution to a review problem, but I can't seem to simplify it.

The problem is to "Find a function h(P) that inputs the perimeter of a right triangle with area 25 ft^2 and outputs the length of the hypotenuse".

I won't bother to type all of my steps, but if you need to see them please let me know. My function does work for all of the test values that I tried. Here is my function:

h(P)=3P/4-25/P-1/(4P) times the square root of (P^4-600P^2+10000) minus 200P/(P^2+100 plus the square root of (P^4-600P+10000))

All of my attempts to simplify this function have actually made things worse, not better.

 

[Denis]

What's up Doc? :roll:

The problem is to "Find a function h(P) that inputs the perimeter of a right triangle with area 25 ft^2 and outputs the length of the hypotenuse".

Let sides = a (shorter leg), b (other leg) and c (hypotenuse); p = perimeter(given).

Since area = 25, then ab / 2 = 25 ; ab = 50; b = 50/a;

then c = p - a - 50/a
also c = sqrt[a^2 + (50/a)^2]

Combining above equalities leads to:
ap - a^2 - 50 = sqrt(a^4 + 2500)

Square both sides, simplify:
2a^2p - a(p^2 + 100) + 100p = 0

Solve above quadratic:
a = [p^2 + 100 + sqrt(p^4 - 600p^2 + 10000)] / (4p)

So hypotenuse c = sqrt[a^2 + (50/a)^2] , where a = [p^2 + 100 + sqrt(p^4 - 600p^2 + 10000)] / (4p)

Pretty well the same as what you got...

 

[pka]

Note that: \(\displaystyle \L
P = a + b + c\) and \(\displaystyle \L
c^2 = a^2 + b^2 \;{\rm{and}}\;a + b = P - c\) .

Now \(\displaystyle \L
\begin{eqnarray*}
P^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc \\
= 2c^2 + 100 + 2c(P - c) \\
\end{eqnarray*}\)

Solve for c we get \(\displaystyle \L
f(P) = \frac{{P^2 - 100}}{{2P}}\)

 

[DrToddMatthews]

How Embarrassing

Thank you pka. It took me a few minutes to see what you did, but now I understand it. Your approach is much nicer; I could not even begin to simplify my version to match yours.

There is an excerpt in my text about George Polya (the author of "How To Solve It."). He is quoted as saying often to his students and colleagues: "Yes, I see that your proof is correct, but how did you find it?"

I would like to ask you the same question. How did you come up with your elegant solution?

 

[pka]

Todd Matthews wrote: “How did you come up with your solution?”
It is nice of you to ask! The fact is I have done mathematics all my life.
I’ve taught for many years. But I had never seen this problem before!
To be quite honest when I first read you problem I doubted it was true.

Here is how things worked for me. I live in small village in the American South. Each morning I go for coffee with the same group of friends. Yesterday was a slow day; I was just sitting by myself. It dawned on me that the solution would a function of the perimeter and the area. Moreover, I realized that I needed to eliminate the a & b.
That is what I did. I really wrote on a paper napkin.

 

[Denis]

Wow...that's really "elegant", pka;
particularly substituting a+b = P-c

And your result is really a general case formula given perimeter and area;
with area = A: hypotenuse = (P^2 - 4A) / 2P.
You need a good supply of those napkins :wink:

 

[galactus]

That's what makes a nice proof, pka. Simplistic, yet clever. kudos.

 

[tkhunny]

Some of my favorites:

"Thought is only a flash between two long nights, but this flash is everything."
Jules Henri Poincare

"The subliminal self is in no way inferior to the conscious self. It knows how to choose and to divine."
Jules Henri Poincare

"Often nothing good is accomplished at the first attack. One takes a rest; and then all of a sudden the decisive idea presents itself to the mind."
Jules Henri Poincare

 

[Denis]

Well, TK, haven't had a "flash", plus got nowhere after a few "attacks" on this
one, which is quite similar:

given: the area and perimeter of an isosceles triangle;
what is length of the longest side?

I thought it would be easy, cheating by using pka's general method,
since we really have two identical right triangles stuck together,
like using a,b,c as one of them, then p = 2a + 2c, so we have the isosceles
with sides c,c,2a and b being the height....

Can someone please put me to shame by coming up with something
simple...I keep getting stuff that ain't too pretty

 

[pka]

Dennis, there ain't any pretty solutions.
But here goes. Let c, c, & a be the lengths of the sides of the triangle.
The perimeter, P=2c+a & a=P−2c; and the area is A.
Suppose that Θ is the base angle then
\(\displaystyle A = \left( {1/2} \right)\left( {ac} \right)\sin (\theta )\quad \Rightarrow \quad \sin (\theta ) = \frac{{2A}}{{ac}}\).

Also \(\displaystyle \cos (\theta ) = \frac{a}{{2c}}\quad \Rightarrow \quad \left( {\frac{{4A^2 }}{{a^2 c^2 }}} \right) + \left( {\frac{{a^2 }}{{4c^2 }}} \right) = 1\quad \Rightarrow \quad 16A^2 + a^4 = 4a^2 c^2\).

Making some substitutions we get \(\displaystyle 16A^2 + \left( {P - 2c} \right)^4 = 4\left( {P - 2c} \right)^2 c^2\).

There are no pretty solutions period.

 

[soroban]

Hello, Denis!

I have a slightly neater equation . . .

Given the area and perimeter of an isosceles triangle,
what is length of the longest side?

          *
         /|\
        / | \
       /  |  \
      /   |h  \
     /    |    \ 
    * - - + - - *
          b

Let \(\displaystyle b\) be the base and \(\displaystyle h\) be the height.

Then: \(\displaystyle \,A\,=\,\frac{1}{2}bh\;\;\Rightarrow\;\;b\,=\,\frac{2A}{h}\;\) [1]

Using Pythagorus, the equal sides have length: \(\displaystyle \sqrt{h^2\,+\,\left(\frac{b}{2}\right)^2}\:=\:\frac{\sqrt{4h^2\,+\,b^2}}{2}\;\)

The perimeter is: \(\displaystyle \\:=\:b\,+\,2\cdot\frac{\sqrt{4h^2\,+\,b^2}}{2}\;\;\Rightarrow\;\;\sqrt{4h^2\,+\,b^2}\:=\\,-\,b\)

Square: \(\displaystyle \,4h^2\,+\,b^2\:=\P\,-\,b)^2\;\;\Rightarrow\;\;4h^2\:=\^2\,-\,2Pb\)

Substitute [1]: \(\displaystyle \:4h^2\;=\;P^2\,-\,2P\left(\frac{2A}{h}\right)\)

And we have the cubic: \(\displaystyle \:4h^3\,-\,P^2h\,+\,4AP\:=\:0\)


Still not pretty . . . but no longer coyote-ugly.

 

[Denis]

And we have the cubic: \(\displaystyle \:4h^3\,-\,P^2h\,+\,4AP\:=\:0\)
Still not pretty . . . but no longer coyote-ugly.

Ok; but what does that do?
equal sides = c: we're looking for c in terms of A and P