Triangle Minimization

[Kushballo7]

In triangle ABC, D lies on AB, CD| AB, |AD|=|BD|=4cm, and |CD|=5cm. Where should a point P be chosen on CD so that the sum |PA|+|PB|+|PC| is a minimum.[/i]

 

[galactus]

Since DC is 5, we can express DP as x and PC as 5-x.

We have two hypoteneuses and 5-x to add up.

Hypoteneuse PB would be \(\displaystyle sqrt{(DP)^{2}+4^{2}}=sqrt{x^{2}+16}\)

Hypoteneuse PA would be the same thing.

Length PC is 5-x.

So we have \(\displaystyle 2sqrt{x^{2}+16}+(5-x)\).

Differentiating using the product rule:

\(\displaystyle (2)(\frac{1}{2}(x^{2}+16)^{\frac{-1}{2}})(2x)-1\)

\(\displaystyle =\frac{2x}{sqrt{x^{2}+16}}-1\)

Set to 0 and solve for x.

\(\displaystyle 2x=sqrt{x^{2}+16}\)

Square both sides:

\(\displaystyle 4x^{2}=x^{2}+16\)

\(\displaystyle 3x^{2}-16=0\)

we get \(\displaystyle x=\frac{4}{sqrt{3}}\) or \(\displaystyle \frac{-4}{sqrt{3}}\)

Subbing back in we get PB=PA=\(\displaystyle \frac{8}{sqrt{3}}\), since PB and PA are the same we have \(\displaystyle \frac{16}{sqrt{3}}=9.24\)

\(\displaystyle 5-(\frac{4}{sqrt{3}})=2.69\)

Adding them equals \(\displaystyle 4sqrt{3}+5=11.93\)

Hey, we jive. Something must be right.

 

[soroban]

Hello, Kushballo7!

Galactus beat me to it . . . but I'll show you my trigonometric solution.

In \(\displaystyle \Delta ABC,\;D\) lies on \(\displaystyle AB,\;CD\,\perp\,AB,\;|AD|\,=\,|BD|\,=\,4\)cm, and \(\displaystyle |CD|\,=\,5\)cm.

Where should a point \(\displaystyle P\) be chosen on \(\displaystyle CD\) so that \(\displaystyle |PA|\,+\,|PB|\,+\,|PC|\) is a minimum?

               C
    -          *
    :         /|\
    :        / | \
    :       /  |  \
    :      /   |   \
    5     /    |    \
    :    /     *P    \
    :   /   *  |  *   \
    :  / * θ   |     * \
    - *--------+--------*
      A    4   D   4    B

We have isosceles triangle \(\displaystyle ABC\) with altitude \(\displaystyle CD\).

Let: \(\displaystyle \theta\,=\,\angle PAD\)

. . Then: \(\displaystyle PA\,=\,PB\,=\,4\sec\theta\)

. . Also: \(\displaystyle \,PC\:=\:CD\,-\,PD\:=\:5\,-\,4\tan\theta\)

Hence: \(\displaystyle \,S\:=\A\,+\,PB\,+\,PC\:=\:8\cdot\sec\theta\,+\,5\,-\,4\cdot\tan\theta\)


Differentiate: \(\displaystyle \,\frac{dS}{d\theta}\:=\:8\cdot\sec\theta\cdot\tan\theta\,-\,4\cdot\sec^2\theta\)

And we have: \(\displaystyle \,4\cdot\sec\theta(2\cdot\tan\theta\,-\,\sec\theta)\:=\:0\)

Since \(\displaystyle \,\sec\theta\,\neq\,0\), we have: \(\displaystyle \,2\tan\theta\:=\:\sec\theta\;\;\Rightarrow\;\;2\cdot\frac{\sin\theta}{\cos\theta}\,=\,\frac{1}{\cos\theta}\)

Since \(\displaystyle \,\cos\theta\,\neq\,0\) *, we have: \(\displaystyle \,2\cdot\sin\theta\,=\,1\;\;\Rightarrow\;\;\sin\theta\,=\,\frac{1}{2}\)


Therefore: \(\displaystyle \:\theta\,=\,\frac{\pi}{6}\,=\,30^o\)

\(\displaystyle P\) must be placed so that: \(\displaystyle \,PD\,=\,4\cdot\tan30^o\,=\:\frac{4}{\sqrt{3}}\:\approx\:2.31\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*
If \(\displaystyle \cos\theta\,=\,0\), then \(\displaystyle \theta\,=\,90^o\) and \(\displaystyle PA\,\parallel\,CD\).