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Problem:

In a triangle $T$ , all the angles are less than 90 degrees, and the longest side has length $s$. Show that for every point $p$ in $T$ we can pick a corner $h$ in $T$ such that the distance from $p$ to $h$ is less than or equal to $s/\sqrt{3}$.

Source: Problem 2 in http://abelkonkurransen.no/problems/abel_1213_f_prob_en.pdf

Here's my try:

If $p$ is on any of the edges of $T$, it can't be further away than $s/2$. Then I thought that the point furthest away from any $h$ would be a point equidistant to all the vertices. I also know that the equidistant point has to be inside the triangle because no angle is obtuse.

I am assuming an equilateral triangle (I'm not sure if I can). It has no longest side, therefore all must be of length $s$.

Let $h_1,h_2,h_3$ be the vertices, and $z=h_1e=h_2e=h_3e$. To find the equidistant point $e$, I could half all the angles. By trigonometry I will get that, $\cos 30^\circ=\frac{\sqrt{3}}{2}=\frac{s/2}{z} \implies z=\frac{s}{\sqrt{3}}$. Since this is the length from the closest corner to the equidistant point it cannot be further away.


I am not sure if I can rightly assume the triangle is equilateral without loss of generality, probably not. However this is the closest I've gotten to proving this. Could you explain a better approach?

PS. Calculators would not be allowed.

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3 Answers 3

up vote 4 down vote accepted

You are right in suspecting that it is not sufficient to consider equilateral triangles. On the other hand this special example shows that the stated inequality cannot be improved.

Since the triangle $T$ in question is acute the midpoint $M$ of its circumcircle lies in the interior of $T$, and looking at a figure leads to the conjecture that $M$ is farthest away from the three vertices $A_i$.

Let $r$ be the radius of the circumcircle. We have to prove that the three disks $C_i$ centered at the $A_i$ and having radius $r$ cover $T$. Let $M_i$ be the midpoints of the sides of $T$. Then for each $i$ the four points $A_i$, $M_{i+1}$, $M$, $M_{i-1}$ are the vertices of a cyclic quadrilateral $Q_i$. This quadrilateral is covered by a disk $C_i'$ of radius ${r\over2}$ centered at the midpoint of $A_iM$. It is then obvious that $Q_i\subset C_i'\subset C_i$, and as the three $Q_i$ make up $T$ the claim follows.

When the longest side of $T$ is $s$ then the opposite angle $\sigma$ is the largest among the three. It follows that ${\pi\over3}\leq \sigma<{\pi\over2}$. By a well known formula for $r$ we now have $$r={s\over 2\sin\sigma}\leq{s\over 2\cdot{\sqrt{3}\over2}}={s\over\sqrt{3}}\ .$$

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Thank you for a great answer, I'm just wondering what the formula for $r$ is called? –  Hans Groeffen Sep 8 '13 at 11:33
    
The sine theorem for triangles says that ${s_i\over\sin\sigma_i}$ is the same for all three $i$. The common value is $2r$. –  Christian Blatter Sep 8 '13 at 12:37
    
Now I understand, thanks! –  Hans Groeffen Sep 8 '13 at 12:43

It seems the following.

Assume the converse, that all segments from the point $p$ to the corners of the triangle $T$ have length greater than $s/\sqrt{3}$. There exist segments $x$, $y$ from the point $p$ to the corners of the triangle $T$ such that the angle $\varphi$ between $x$ and $y$ is not less than $2\pi/3$. Then the length of the triangle side which vertices are the ends of the segmtnts $x$ and $y$ is $$\sqrt{|x|^2+|y|^2-2|x||y|\cos\varphi}>\sqrt{s^2/3+s^2/3-2(s/3)(s/3)(-1/2)}\ge s,$$ a contradiction.

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If you assume that all segments from $p$ to the closest corner have a length greater than $s/\sqrt{3}$, wouldn't a point in one of the corners have a length of zero, therefore not longer than $s/\sqrt{3}$? –  Hans Groeffen Sep 8 '13 at 11:37
    
Yes, of course. So in this case we obtain a contradiction more quickly. :-) –  Alex Ravsky Sep 8 '13 at 17:36

Rephrasing Christian Blatter: for any point interior to an acute triangle, measure the distance to each of the four vertices, and find the smallest value. Now find the point where the this minimum value is largest. Almost obviously, this is the circumcenter. Connect the circumcenter to each of the vertices and drop perpendiculars from the circumcenter to each of the three sides. This divides the original triangle into six right triangles. Any point will be in one of these, and will be closer or equidistant to the vertex than the circumcenter.

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