$P$ and that one triangle

Geometry Level 5

Let $P$ be a point in or on a scalene triangle $ABC$ with integer side lengths, and let $f(P) = PA + PB + PC$ . For each triangle $ABC$ , let $g_{ABC} = \max_P f(P)$ and $h_{ABC} = \min_P f(P)$ .

Consider all triangles where $g_{ABC}$ and $h_{ABC}$ are both integers. Over this set, what is the minimum value of $g_{ABC} - h_{ABC}$ ?

The answer is 4.

1 solution

Efren Medallo
Apr 24, 2017

Note that $f(P)$ is at its minimum when $P$ coincides with the Fermat point of the triangle.

Now, for $f(P)$ to be an integer, we can set it such that $P$ lies along the sides of the triangle $ABC$ itself. This will only be possible when one of the interior angles is $\geq 120^{\circ}$ , with $P$ on its vertex. Now, the smallest integer triangle to have that is the $3-5-7$ triangle.

The maximum value of $f(P)$ can also be established in a similar way; since the distance cannot be maximized when $P$ is inside the triangle, it has to be located on the triangle. Using the same triangle, it can be discovered that $f(P)$ is maximum when $P$ is on the vertex opposite the smallest side.

So, for the $3-5-7$ triangle, being the smallest integer triangle to have a $120^{\circ}$ angle, $f(P)$ is minimum when $P$ is at the vertex opposite the longest side. In this case, $f(P) = 3 + 5 = 8$ .

On the other hand, $f(P)$ is maximum when $P$ is at the vertex opposite the shortest side, making $f(P) = 5 + 7 = 12$ .

And that gives us the $\max (f(P)) - \min (f(P)) = 12 - 8 = \boxed {4}$ .

Do you want "minimum value of g that is an integer", or do you want "minimum value of g when both max f and min f are integers"?

The question and the assumptions are different from each other.

Calvin Lin Staff - 4 years, 1 month ago

As was stated in my further details, it asks of the latter. I was also doubtful that my phrasing was correct, so I will edit it accordingly.

Efren Medallo - 4 years, 1 month ago

K. I've edited the problem.

Looking at your solution, I find it really hard to understand what you're doing, and it seems like there are several assumptions / mistakes that you make

Why must "for $f(P)$ to be an integer, we can set it such that it lies along the side of the triangle
Why must the minimum of $g$ occur for the 3-5-7 triangle? It seems like you're committing the mistake of trying to optimize one term, and then using that to condition the other. IE to find min $g = f - h$ , we cannot simply minimize $f$ and calculate the corresponding value of $h$ .
I think that the 2-3-4 triangle has a max of 7, and a min of 4 or 5, which gives a $g$ value of 2 or 3. (Haven't done the calculations yet).

Calvin Lin Staff - 4 years, 1 month ago

@Calvin Lin –

If $P$ is in the triangle, $f(P)$ will be irrational.
The minimum of $g$ will occur at the smallest integer triangle bearing $120^{\circ}$ , because for any obtuse triangle having an angle greater than or equal to $120^{\circ}$ , $P$ will lie on the vertex opposite the longest side. We can test this for other obtuse integer triangles with the largest angle exceeding $120^{\circ}$ , and see what happens with $g$ .
The 2-3-4 is an acute triangle. Thus $f(P)$ in this case is not an integer.

Efren Medallo - 4 years, 1 month ago

@Efren Medallo –

If $P$ is in the triangle, $f(P)$ will be irrational -> Are you sure? Note that $f(P)$ is a continuous function, so its image would be an interval, as opposed to "just irrational values".
That makes no sense, esp since 1 is not true.
The 2-3-4 triangle is obtuse. In particular, $2^2 + 3 ^2 < 4 ^ 2$ . (However, it is true that the obtuse angle is not larger than 120, so the minimium of f(P) occurs in the interior)

Reading this again, you seem to want "The minimium of f(P) is an integer", as opposed to "The minimium integer value of f(P)". do you see the difference?

Calvin Lin Staff - 4 years, 1 month ago

@Calvin Lin – Reading through your assumption again, that seems to be the case. I have edited that into the problem.

I can now believe that your answer is correct, but your solution still does not feel rigorous to me. I feel that a big assumption is " $h_{ABC}$ is integer if and only if $ABC$ has an angle $\geq 120 ^ \circ$ ". I agree with the if part, but the only if part isn't immediately obvious to me.

Calvin Lin Staff - 4 years, 1 month ago

@Calvin Lin – Thank you for the insights! I really had trouble phrasing the problem, and I was happy that somehow, what I was trying to say (albeit incorrect, perhaps due to my poor command of the language and/or the concept) got through you. I just got the inspiration for my next problem!

Efren Medallo - 4 years, 1 month ago

@Calvin Lin –

Uhm, yes. I may have assumed that. I have found the smallest case for a triangle to have $P$ inside of it and $g$ is an integrr. That is a triangle whose sides are $455$ , $399$ , and $511$ . This will give $g= 966 - 784 =182$ . Alternatively, if we allow one of the angles to be greater than $120^{\deg}$ , we can alllow $P$ to be along the vertices of the triangle whix h will ascertain the rationality of $g$ .
1. Forgive me, it is indeed obtuse. But the obtuse angle is less than $120^{\circ}$

Efren Medallo - 4 years, 1 month ago

P P P and that one triangle

The answer is 4.

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$P$ and that one triangle