Problem 49

CaptainBlack · #1 $Old$ October 29th, 2008, 11:45 AM

(minor variant of a problem due to Roy Barbara)

Let $a,\ b,\ c$ be three positive real numbers.

Find necessary and sufficient conditions on $a,\ b,\ c$ for there to exist an interior point $P$ in the equilateral triangle $ABC$ with unit side, such that $|PA|=a,\ |PB|=b,\ |PC|=c$ .

I have a solution to this, but as I have not looked it up so I cannot tell if it is the originator's solution but as it is not as neat as I would like it is probably clumsy compared to the best solution, so lets see what we can do

CB

CaptainBlack · #2 $Old$ October 30th, 2008, 05:58 AM

Quote:

Originally Posted by CaptainBlack $View Post$

(minor variant of a problem due to Roy Barbara)

Let $a,\ b,\ c$ be three positive real numbers.

Find necessary and sufficient conditions on $a,\ b,\ c$ for there to exist an interior point $P$ in the equilateral triangle $ABC$ with unit side, such that $|PA|=a,\ |PB|=b,\ |PC|=c$ .

I have a solution to this, but as I have not looked it up so I cannot tell if it is the originator's solution but as it is not as neat as I would like it is probably clumsy compared to the best solution, so lets see what we can do

CB

Ahh.. maybe its my modification to the question that is the reason my solution is clumsier than I would like??

I shall have to look into this possibility

CB

Laurent · #3 $Old$ November 12th, 2008, 02:37 PM

Quote:

Originally Posted by CaptainBlack $View Post$

(minor variant of a problem due to Roy Barbara)

Let $a,\ b,\ c$ be three positive real numbers.

Find necessary and sufficient conditions on $a,\ b,\ c$ for there to exist an interior point $P$ in the equilateral triangle $ABC$ with unit side, such that $|PA|=a,\ |PB|=b,\ |PC|=c$ .

Here is my solution, I let you appreciate its neatness/clumsiness... It is pretty simple anyway.

The idea is to use barycentric coordinates: for any point $P$ in the plane, there is a unique triplet $(\lambda,\mu,\nu)$ such that $\lambda+\mu+\nu=1$ and $P=\lambda A+\mu B+\nu C$ .
Given these coordinates, $P$ lies in the triangle $ABC$ if, and only if the three numbers $\lambda$ , $\mu$ , $\nu$ are positive (or zero, corresponding to bounderies).

It is easy to express $a,b,c$ in terms of $\lambda,\mu,\nu$ . We have $\overrightarrow{AP}=\mu\overrightarrow{AB}+\nu\overrightarrow{AC}$ , hence $a^2=AP=\|\mu\overrightarrow{AB}+\nu\overrightarrow{AC}\|^2=\mu^2 AB^2+2\mu\nu\overrightarrow{AB}\cdot\overrightarrow{AC}+\nu^2 AC^2$ .
Because $ABC$ is equilateral with unit sides, we conclude $a^2=\mu^2+\mu\nu+\nu^2$ .
By circular permutation of letters, we get similar expressions for $b^2$ and $c^2$ . Thus, $b^2=\nu^2+\nu\lambda+\lambda^2$ .

What we need in fine is expressions for $\lambda,\mu,\nu$ in terms of $a,b,c$ . This can be laborious, but I found a soft way to write it. We have $a^2-b^2=\mu^2-\lambda^2+\mu\nu-\lambda\nu=(\mu-\lambda)(\mu+\lambda+\nu)$ , hence $a^2-b^2=\mu-\lambda$ .
Again by circular permutation of the letters, we have $c^2-a^2=\lambda-\nu$ . We deduce $(a^2-b^2)-(c^2-a^2)=\mu+\nu-2\lambda=1-3\lambda$ .

Finally, we have $\lambda=\frac{1}{3}(1-2a^2+b^2+c^2)$ . And, similarly, $\mu=\frac{1}{3}(1-2b^2+c^2+a^2)$ and $\nu=\frac{1}{3}(1-2c^2+a^2+b^2)$ .

Remembering what I said first about barycentric coordinates, the conclusion is then straightforward: $P$ lies inside the triangle if, and only if $2a^2-b^2-c^2\leq 1$ , $2b^2-c^2-a^2\leq 1$ and $2c^2-a^2-b^2\leq 1$ .

If the triangle had side $r$ , it would suffice to replace 1 by $r^2$ in the conditions, making them more "homogeneous".

David24 · #4 $Old$ November 22nd, 2008, 01:24 AM

Quote:

Originally Posted by CaptainBlack $View Post$

(minor variant of a problem due to Roy Barbara)

Let $a,\ b,\ c$ be three positive real numbers.

Find necessary and sufficient conditions on $a,\ b,\ c$ for there to exist an interior point $P$ in the equilateral triangle $ABC$ with unit side, such that $|PA|=a,\ |PB|=b,\ |PC|=c$ .

I have a solution to this, but as I have not looked it up so I cannot tell if it is the originator's solution but as it is not as neat as I would like it is probably clumsy compared to the best solution, so lets see what we can do

CB

Is the solution to this problem not simply a = b = c??

Cheers,

David

CaptainBlack · #5 $Old$ November 22nd, 2008, 01:39 AM

Quote:

Originally Posted by David24 $View Post$

Is the solution to this problem not simply a = b = c??

Cheers,

David

No, see Laurents answer

CB

DangerMaths · #6 $Old$ January 4th, 2009, 03:10 AM

Well, I would draw three circles with centres at A,B,C, where P is on each of the circles, and then continue from there.