Thread: Problem 49
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Old November 12th, 2008, 02:37 PM
Laurent Laurent is offline
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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".
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