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: $\displaystyle \,PD\,=\,4\cdot\tan30^o\,=\:\frac{4}{\sqrt{3}}\:\approx\:2.31$

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