Let $ABC$ be a triangle with semiperimeter $s$ and inradius $r$. The semicircles with diameters $BC$, $CA$, $AB$ are drawn on the outside of the triangle $ABC$. The circle tangent to all of these three semicircles has radius $t$. Prove that
\[\frac{s}{2}<t\le\frac{s}{2}+\left(1-\frac{\sqrt{3}}{2}\right)r. \]
Alternative formulation. In a triangle $ABC$, construct circles with diameters $BC$, $CA$, and $AB$, respectively. Construct a circle $w$ externally tangent to these three circles. Let the radius of this circle $w$ be $t$.
Prove: $\frac{s}{2}<t\le\frac{s}{2}+\frac12\left(2-\sqrt3\right)r$, where $r$ is the inradius and $s$ is the semiperimeter of triangle $ABC$.
Proposed by Dirk Laurie, South Africa
I just wanted to note that on the first German TST 2004 in Rostock, we've got a weaker version of this problem (in fact, slightly simpler because of the $\leq$ instead of the < sign in the left member, but yet nobody gave a complete solution):
Let ABC be a triangle with perimeter 2s and inradius r. Construct the semicircles with diameters BC, CA, AB outwardly (with respect to the triangle). The radius of the circle tangent to these three semicircles will be denoted by t. Prove
$\frac{s}{2}\leq t\leq \frac{s}{2}+\left( 1-\frac{\sqrt3}{2}\right) r$.
I got 2 of 7 points on this problem, I think nobody got more... it is a really sadistic problem.
Darij
darij grinberg wrote:
I got 2 of 7 points on this problem, I think nobody got more... it is a really sadistic problem.
It's not only "sadistic", it's useless too (not nice) (or at least in my opinion) . Besides the Soddy circles (which I never used at any problem ) I didn't learn anything out of it.
If either of you are interested, this problem also appeared in the 3rd MathLinks Contest (the current edition) in the 2nd Round. I also saw it come up on the South African Math Olympiad 2003.
Probably darij got 2 points by proving the left part of the inequality like this(or am I wrong??):
Notations:We denote by $\Omega_A,\Omega_B,\Omega_C$ the circles with diameters $BC,CA,AB$ respectively.Let $\Omega$ be the circle externally tangent to these three circles.Let $A'$ be the point of tangency of $\Omega_A$ with $\Omega$ and define $B',C'$ analogously.Let $A",B",C"$ be the midpoints of $BC,CA,AB$ and $O$ the center of $\Omega$.
Proof of the first part:It is clear that the points $(A'A",O),(B',B",O),(C',C",O)$ are collinear.Thus $OC"=t-\frac{c}{2},OA"=t-\frac{a}{2}$.We also have $A"C"=\frac{b}{2}$.So triangle inequality in $\triangle{A"C"O}$ gives
$A"O+C"O > A"C" \Rightarrow (t-\frac{a}{2})+(t-\frac{c}{2}) > \frac{b}{2} \Rightarrow t > \frac{a+b+c}{4}=\frac{s}{2}$
I shall prove that more difficult inequality (the RHS):
Let $ A', B', C' $ the midpoints of segments $ BC, CA, AB $ respectively. Let the incircle of $ \triangle{A'B'C'} $ touch segments $ B'C', C'A', A'B' $ at $ M, N, P $ respectively. Consider the circle internally tangent to the circles with centers $ A', B', C' $ and radii $ A'N, B'P, C'M $ respectively. This is called the inner Soddy circle of $ \triangle{A'B'C'} $ and has radius $ t - \frac{s}{2} $. Letting $ R $ be the circumradius of $ \triangle{ABC} $, it is well-known (and can be easily derived from Descartes' Theorem) that this inner Soddy circle also has radius $ \frac{rs}{2(4R + r + 2s)} $.
So it suffices to show that $ \frac{rs}{2(4R + r + 2s)} \leq \left(1 - \frac{\sqrt{3}}{2}\right)r \Longleftrightarrow \frac{s}{4R + r + 2s} \leq 2 - \sqrt{3} \Longleftrightarrow 4R + r \geq \sqrt{3}s $. Let the tangent lengths of $ \triangle{ABC} $ be $ x, y, z $ respectively.
Then this inequality is equivalent to $ \frac{(x + y)(y + z)(z + x)}{\sqrt{xyz(x + y + z)}} + \sqrt{\frac{xyz}{x + y + z}} \geq \sqrt{3}(x + y + z) $.
Upon expanding the denominators and applying some simple algebraic manipulations becomes $ (x^2y + x^2z + y^2z + y^2x + z^2x + z^2y + 3xyz)^2 \geq 3xyz(x + y + z)^3 $. By cancelling similar terms from both sides this becomes:
$ \sum_{sym}x^4y^2 + \sum_{sym}x^3y^3 \geq \sum_{cyc}x^4yz + \sum_{sym}x^3y^2z + 3x^2y^2z^2 $ which is trivial by some applications of Muirhead's inequality.
this is a stupid problem. It takes a lot of skill to write a beautiful geometry problem, even more for a beautiful solution, even more to carry through a stupid and ugly solution, and yet more to write such a stupid, ugly, and terrible and not worth solving problem such as this one. No deep beautiful geometrical results can be applied, and this problem can only be attacked by sufficient reading into algebraic literature and by enough computational fortitude. This problem is an insult to all of the beautiful geometry problems that have made their way onto the IMO.
I think the criticism is rather harsh. At least $LHS$ can be solved somewhat nicely.
Let the midpoints of $AB,BC,CA$ be $F,D,E$, and the centre of new circle be $X$.
From homothety we can determine that:
$XD+\frac{BC}{2}=XE+\frac{AC}{2}=XF+\frac{AB}{2}=t$. WLOG $X$ is not on $DE$, then we have $2t=XD+\frac{BC}{2}+XE+\frac{AC}{2}>DE+\frac{BC}{2}+\frac{AC}{2}=s$.
As for $RHS$, there is not really any synthetic involved, but not too messy computational wise too. Note that $XD,XE,XF$ are of the form $\frac{b+c-a}{4}+k,\frac{a+c-b}{4}+k,\frac{a+b-c}{4}+k$( rewrite as $u+k,v+k,w+k$).
Hence we need to show $XD+\frac{BC}{2}\leq \frac{s}{2}+(1-\frac{\sqrt3}{2})r\Leftrightarrow k\leq (1-\frac{\sqrt3}{2})r$. Draw the circles centered at $D$,$E$,$F$ with radius $u,v,w$, giving us the configuration needed to apply Descartes (the last circle is that centered at $X$ with radius $k$). We get that $k=\frac{uvw}{uv+vw+uw+2\sqrt{uvw(u+v+w)}}$. Further, the inradius of $DEF$ is half of that of $ABC$, so $r=\frac{2\sqrt{uvw(u+v+w)}}{u+v+w}$.
It remains to show that $\frac{uvw}{uv+vw+uw+2\sqrt{uvw(u+v+w)}}\leq (1-\frac{\sqrt3}{2})\frac{2\sqrt{uvw(u+v+w)}}{u+v+w}$
$\Leftrightarrow \sqrt{uvw(u+v+w)}\leq(2-\sqrt3)(uv+vw+uw+2\sqrt{uvw(u+v+w)})$
$\Leftrightarrow (2\sqrt3-3)\sqrt{uvw(u+v+w)}\leq (2-\sqrt3)(uv+vw+uw)$
$\Leftrightarrow (u+v+w)uvw\leq u^2v^2+v^2w^2+u^2w^2$ which is $AM-GM$
Let $A',B',C'$ denote the mid points of $BC,CA,AB$ respectively. Consider circles $\omega_A = \odot (A',\frac s2 - \frac a2)$. Similarly draw in $\omega_B$ and $\omega_C$.
Note that the centre of the requested circle, call it $O'$, is the center of the Inner Soddy circle $\omega$ of $\triangle A'B'C'$. Suppose $\omega$ has radius $x$. It follows that the circle in question has radius $x+\frac s2$ which readily proves the left part of the inequality because $x>0$.
We will now prove the right part of the inequality. Consider the incircle $\Omega$ of $\triangle A'B'C'$ , centered at $I$ , having a radius of $r'=\frac r2$. We wish to prove $x\leq (2-\sqrt 3)r'$. We will distinguish two cases.
Case 1 : $O\equiv I$.
Straightforward calculations reveal $x= (2-\sqrt 3)r'$ , so the bound holds in this case.
Case 2 : $O \neq I$.
We claim that there exists $X \in OI$ such that, under the inversion $\psi (X,1)$ , the circles $\psi(\omega)$ and $\psi (\Omega)$ become con-centric and $\psi(\omega)$ lies inside $\psi (\Omega)$.
Proof : Impose coordinates on the line $OI$ and let the points where $\omega$ and $\Omega$ hit $OI$ be denoted by $i,j,k,\ell$ with $i<j<k<\ell$.
Consider the function given by $$f(x)= \frac {\frac {1}{x-\ell}-\frac {1}{x-k}}{\frac {1}{x-j}-\frac {1}{x-i}}$$(for a point $X\in OI$.)
Note that we have $\lim_{x\rightarrow \ell^{+}}f(x) = \infty$ and $\lim_{x\rightarrow \infty}f(x) = \frac {\ell-k}{j-i}<1$. This means that by IVT , we can choose a point $X$ such that $f(x)=1$. Note that we have, in that case $:=$
$$ \frac {1}{x-\ell}-\frac {1}{x-k} = \frac {1}{x-j}-\frac {1}{x-i} \implies \frac {1}{x-\ell}+\frac {1}{x-i}=\frac {1}{x-j}+\frac {1}{x-k}$$
This means that the inversion $\psi(X,1)$ makes the circles $\omega$ and $\Omega$ concyclic. We also have $:=$
$$ \frac 1{x-i}<\frac 1{x-j}<\frac 1{x-k}<\frac 1{x-\ell} \implies \frac 1{(x-i)(x-\ell)}< \frac 1{(x-j)(x-k)}$$
So $\omega$ lies inside $\Omega$.
We now claim that the ratio of the radius of $\omega$ and $\Omega$ increases after this inversion.
We have $:=$
$$ \frac { \frac 1 {x-k} - \frac 1 {x-j} } {\frac 1 {x-\ell} - \frac 1 {x-i} }= \frac {k-j}{\ell-i} \cdot \frac {(x-i)(x-\ell)}{(x-k)(x-j)} > \frac {k-j}{\ell-i} = \frac x{r'}$$
So since $\omega_A$,$\omega_B$,$\omega_C$ invert to circles tangent to $\psi (\omega)$ and orthogonal to $\psi (\Omega)$ , hence this ratio is exactly $2-\sqrt 3$ as noted in the previous case.
We are done. $\blacksquare$
Let $D$, $E$, $F$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. Let $O$ be the center of the circle with radius $t$ and let $OD$, $OE$, $OF$ intersect the circle at $A'$, $B'$, $C'$, respectively. Let the incircle of $DEF$ intersect $DE$ at $Z$, $EF$ at $X$, and $FD$ at $Y$. Let $DY=DZ=a$, $EX=EZ=b$, and $FX=FY=c$. Note that
\[t=OC'=OF+FC'=OF+AF=OF+ED=OE+FD=OD+EF\]so $OF-OE=FX-XE$. Since $OF+OE>FX+FE$, $OF>FX$ and $OE>EX$. Construct the circles $\omega_D$ centered at $D$ passing through $Y$ and $Z$ and $\omega_E$ and $\omega_F$ analogously. Then clearly there exists a circle centered at $O$ externally tangent to $\omega_D$, $\omega_E$, and $\omega_F$, whose radius is $OF-FX=OC'-ED-FX=t-a-b-c$. Note that $s$ is the half the perimeter of $\triangle ABC$ so it is the perimeter of $\triangle DEF$. $a+b+c$ is the semiperimeter of $\triangle DEF$ so $\tfrac{s}{2}$ so the radius of the small circle is $t-\tfrac{s}{2}>0$ as desired.
$~$
Now, we need to show that $u$, the radius of the small circle centered at $O$ is at most $(2-\sqrt{3})v$ where $v$ is the inradius of $\triangle DEF$. By Descartes' Theorem,
\[\frac{1}{u}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-2\sqrt {\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\]and we have
\[\frac{1}{v}=\frac{s}{A}=\frac{a+b+c}{\sqrt{abc(a+b+c)}}=\sqrt{\frac{a+b+c}{abc}}\]We want to show that
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-2\sqrt {\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\ge \sqrt{\frac{a+b+c}{abc}}\]which is equivalent to
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3\sqrt {\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\]and this is a trivial inequality.
