The maximal value of the inradius of a an inscribed triangle

vess

Let $ABC$ be a triangle whose vertices lie in the interior or on the boundary of the unit square. Determine the maximal possible value of the inradius of $ABC$ .

**Posted:** Thu May 27, 2004 4:50 am

iura

I think the maximum is reached by the isosceles triangle whose vertices are A,B and midpoint of CD.
Please tell me is my claim true?

**Posted:** Thu May 27, 2004 9:08 am

harazi

Yes, iura, your claim is true. It seems that this is a classical super-hard problem. I wonder who solved it during the contest without knowing it. I find it simply impossible. Some hints:
Obviously, we must search for traingles with vertices on the sides of the square. Then it's not very difficult (though I don't find it trivial) to show that if we have such a triangle, then we can find another triangle with one vertex a vertex of the square and the other two vertices on the opposite sides of the square (relative to the vertex of the square which was already chosen) and having larger inradius. And so the problem "reduces" to proving the very hard inequality:
$\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}\geq(1+\sqrt{5})(1-xy)$ for any x and y between 0 and 1. This is a hell of an inequality! We fix y and treat this as a function in x, let it be f(x). Then it's easy to see f is convex. We have three cases:
First case:
y is at least $\frac{1}{4}$ . In this case it's easy to show that $f'(0)>0$ and since $f'$ is increasing we find that $f'(x)>0$ and so f is increasing. From now on it's easy.
Second case: first, a notation: take $r^2=1-\frac{6\sqrt{3}}{(1+\sqrt{5})^2}$ . The second case is when x is smaller than $\frac{1}{4}$ and y is smaller than r. It is not hard to find that $f'(\frac{1}{4})<0$ and so f is decreasing on $[0,\frac{1}{4}]$ . Using the first case we find that $f(\frac{1}{4})\geq 0$ . Due to the symetry we have proved that this is true for all points (x,y) outside the set A of those points with both coordinates between r and $\frac{1}{4}$ .
Third case : (x,y) is in A. In this case we consider the triangle O(0,0), M(1,y) and N(x,1). Then the sum of radicals is exactly its perimeter. And its area is $\frac{1-xy}{2}$ . Using the fact that in any triangle $12\cdot\sqrt{3}\cdot S\leq (perimeter)^2$ we find that the sum of radicals is at least $\sqrt{6\sqrt{3}}\cdot\sqrt{1-xy}$ which is at least $(1+\sqrt{5})(1-xy)$ since we have (x,y) in A.
This is not my solution. I'm not so smart. But it shows how extremeley difficult this problem is. And the inequality proved here is one of the most difficult I have ever seen.

**Posted:** Fri May 28, 2004 11:35 am

harazi

And to show how hard this problem is, I jus say it was a conjecture of Louis Funar (1984!!!!!!!!!!) one of our best olympics ever. This really says a lot. I wonder if anyone could solve this problem.

**Posted:** Fri May 28, 2004 11:51 am

iura

thank you. I will try to understand the solution.
I've spent more then 4.5 hours on it for sure!
i have some ideas, but I'm still not sure they are correct.
We get that one vertex is a vertex on a square, the other lying on two adjacent sides.
We use the following: From triangles with given area, the biggest inradius has the isosceles triangle(1), and from isosceles triangles with given area, the smaller the base the bigger the inradius( i'm not sure of the last)(2)
If the triangle is AMN we show MN>AM,AN or we could use (1) to increase inradius.
Now dilate AMN to become triangle with area 1/2, then make it isosceler, then decrease the base to 1. I wonder if all these operations increase the inradius.
If it's true, then we reach our desired triangle and we are done.

**Posted:** Sat May 29, 2004 3:43 am

ZERO-00 · New Member Offline Joined: 10 Aug 2008 Posts: 1

for harazi
i found a simple proof for the inequality

**Posted:** Sun Aug 10, 2008 1:59 pm

Marian Dinca · New Member Offline Joined: 04 Sep 2008 Posts: 16

Lema 1:
For triangle ABC to exist two triangle isosceles same perimeter and same area of trangle ABC.
Proof:
Let a, b, c sides of triangle ABC and $s = \frac {a + b + c}{2}$ . Area of triangle ABC: $F = \sqrt {s.(s - a)(s - b)(s - c)}$ , and aply AM-GM rezults:
$F\le \sqrt {s.(\frac {\sum{(s - a)}}{3})^3} = \frac {s^2}{3.\sqrt {3}}\Rightarrow 0 < F\le\frac {s^2}{3.\sqrt {3}}$ .
Let triangle $A_1B_1C_1$ of sides $x, x, 2s - 2x; x + x > 2s - 2x \rightarrow 4x > 2s \rightarrow x > \frac {s}{2}$ and $x > 2s - 2x \Rightarrow x > \frac {2s}{3} > \frac {s}{2}$ and $2s - 2x > 0 \Rightarrow x < S$ because $F_1 = \sqrt {s.(s - x)(s - x).(s - 2s + 2x)}\le \sqrt {s.(\frac {s - x + s - x + 2x - s}{3})^3} = \frac {s^2}{3.\sqrt {3}}$ .
Let $f: (\frac {2s}{3};s)\rightarrow(0;\frac {s^2}{3.\sqrt {3}}); \ f(x) = \sqrt {s.(s - x)^2.(2x - s)}$ , because f continue, f Darboux proprerty $\Rightarrow (\exists)x_0\in(\frac {2s}{3};s)$ and $f(x_0) = F$ .
Using same method, for: $g: (\frac {s}{2};\frac {2s}{3})\rightarrow (0;\frac {s^2}{3.\sqrt {3}}); g(x) = \sqrt {s.(s - x)^2.(2x - s)}$ , to exists $y_0\in(\frac {s}{2};\frac {2s}{3})$ and $g(x_0) = F$ . Lema 1 proved!

Lema 2:
For any triangle ABC, and a, b, c sides, exists triangle $A_1B_1C_1$ for sides: $a_1 = \sqrt {a.(b + c - a)}; b_1 = \sqrt {b.(a + c - b)}; c_1 = \sqrt {c.(a + b - c)}.$
The triangle $A_1B_1C_1$ are same area of triangle ABC and perimeter: $2.s_1 = a _1 + b_1 + c_1 < a + b + c = 2.s.$
Proof: let $a = y + z, b = x + z, c = x + y\Rightarrow$
$\Rightarrow a_1 = \sqrt {2x.(y + z)};b_1 = \sqrt {2y.(x + z)};c_1 = \sqrt {2z.(x + y)}$ ;
$a_1 + b_1 > c_1; a_1 + c_1 > b_1; b_1 + c_1 > a_1.$
$16.F^2 = (\sum a).\prod (b + c - a)} = 16.xyz.(x + y + z)$
and
$16.F_1^2 = 2.\sum {a_1^2b_1^2} - \sum{a_1^4} =$
$= 2.\sum{2x.(y + z).2y.(x + z)} - \sum {(2x.(y + z))^2} =$
$= 8.\sum{xy.(z + y).(z + x)} - 4.\sum{x^2.(y + z)^2} =$
$= 8.\sum{xy.(z^2 + xy + yz + xz)} - 4.\sum{x^2.(y^2 + 2.yz + z^2)} =$
$= 8.\sum{xy.(z^2 + xy + yz + xz)} - 4.\sum{x^2.(y^2 + z^2)} - 8.xyz.(x + y + z) =$
$= 8.xyz.(x + y + z) + 8.(\sum{xy})^2 - 8.\sum{x^2y^2} - 8.xyz.(x + y + z) =$
$= 8.(\sum{xy})^2 - 8.\sum{x^2y^2} = 16.\sum{(xy)(xz)} = 16.xyz.(x + y + z) = 16.F^2.$
Rezults: $16.F^2 = 16.F_1^2 \Leftrightarrow F = F_1$ and using AM-GM
$\Rightarrow a_1 + b_1 + c_1 = \sum{\sqrt {a.(b + c - a)}} < \sum{\frac {a + b + c - a}{2}} = \sum{\frac {b + c}{2}} = \sum...$
Acording to Lema 1, exists two triangle isosceles same perimeter and same area of triangle $A_1B_1C_1$ , conclude exists two triangle isosceles same area of triangle ABC and perimeter inferior of perimeter triangle ABC, and:
$r = \frac {F}{s} < \frac {F}{s_1} = \frac {F_1}{s_1} = r_1$ , the is inradius of triangle isosceles.
Let $\triangle {MNP} =$ isosceles, M=midpoint of [DC] and |DN|=CP|=x $\Rightarrow |MN| = |NP| = \sqrt {x^2 + \frac {1}{4}} \Rightarrow$
$r_{MNP} = \frac {\frac {x}{2}}{\frac {2.\sqrt {x^2 + \frac {1}{4}} + 1}{2}} = \frac {x}{2.\sqrt {x^2 + \frac {1}{4}} + 1} = \...$ .
Let: $f(x) = \frac {1}{2.\sqrt {1 + \frac {1}{4.x^2}} + \frac {1}{x}}; x\in(0;1]$ .
If $x_1 < x_2 \Rightarrow f(x_1) < f(x_2)$ , then: f-increasing $\Rightarrow max{f(x)} = f(1) \Rightarrow r_{max} = \frac {1}{2.\sqrt {1 + \frac {1}{4}} + 1} = \frac {1}{\sqrt {5} + 1} = \fr...$

**Posted:** Sat Oct 18, 2008 9:26 am