I think the following is an interesting approach, but this is only my opinion, of course . Let $\mathcal C_a,_b,_c$ be the circles with radius $\frac r2$ tangent to $AB,AC$, $BC,BA$, and $CA,CB$ respectively. Move $D$ around on $BC$, and let $E,F$ be the points where the tangent to $\mathcal C_c$ through $D$, different from $BC$, and the tangent to $\mathcal C_b$ through $D$ different from $BC$ cut $AC,AB$ respectively. The map $E\mapsto F$ is projective, so $EF$ is always tangent to some conic which is tangent to $AB,AC$. Let $B',C'$ be the midpoints of $AB,AC$ respectively. Let $t_1,t_2$ be the two internal common tangents of $\mathcal C_b,_c$. One of them, say $t_1$, cuts the segment $BB'$ in a point $P_1$, and the other one cuts the segment $CC'$ in $P_2$ (this need justification, but it's easy enough). Also, let $P=t_1\cap t_2$. Our conic is tangent to the five lines $B'C',C'P_2=AC,PP_1,P_2,B'P_1$, so it must be included in one region determined by these five lines. In particular, it cannot lie in the same half-plane determined by $B'C'$ as $A$, so it's tangent to $\mathcal C_a$ in at most one point, the point of tangency between $\mathcal C_a$ and $B'C'$, meaning that there is at most one point $D\in BC$ for which $EF$ is tangent to $\mathcal C_a$, and this is exactly what we wanted to show.

Assume $D',E',F'$ are the midpoints and let $D$ be closer to $B$ than $C$, it follows imediately that $E$ is closer to $C$ than to $A$ and $F$ is closer to $A$ than to $B$. Now let's intersect $DE$ with $D'E'$ at point $X$. Now the circle inscribed in $ACB$ with radius $\frac r2$ is tangent to both $DE$ and $D'E'$ so it is tangent to $D'X$ in its interior. So let's write Menelaos:\[\frac {XE'}{XD'}\frac{DD'}{CD}\frac {EC}{E'E}=1\] and write $\frac {XE'}{XD'}<\frac {p-a}{p-b}$. Write all 3 inequalities derived from Menelaos by using such inequalities and take their product. It will turn up that $\prod \frac{CD}{DB}<1$ but $CD>BD$ and the analogues from which we derive the contradiction. What do you think about this solution? As far as I know even the official solution used computations... This was also the solution I gave during the contest.

Dear Ikap , please explain to me why $ \frac {XE'}{XD'}<\frac {p-a}{p-b} $ ?

ikap wrote: Now the circle inscribed in $ACB$ with radius $\frac r2$ is tangent to both $DE$ and $D'E'$ so it is tangent to $D'X$ in its interior. That's why $\frac {XE'}{XD'}<\frac {YE'}{YD'}$, where $Y$ is the tangency point between the circle and $D'E'$. But the triangle $CD'E'$ is similar to $CBA$ so $\frac {YE'}{YD'}=\frac {p-a}{p-b}$

Sorry for the next dumb question... How do you solve such a problem through computations? What do you compute?

I have very little experience in solving geometry problems. Can anyone show me what path should I follow?

I BEG YOUR PARDON GENTLEMEN CAN ANYONE HAVE THE SOLUTION THAT IS FULLY CORRECT? PLEASE BECAUSE I HAVE BEEN TRYING TO SOLVE IT AND TILL NOW I CAN'T AND I WANT TO SEE THE SOLUTION FROM CURIOSITY THANK YOU

I have tried to demonstrate the problem and achieve inradii the triangle DEF is equal to half the inradius the triangle ABC, it shows that the area of triangle DEF is the fourth of the area of triangle ABC. With this you can prove that DEF is the medium triangle of ABC?

ikap wrote: Assume $D',E',F'$ are the midpoints and let $D$ be closer to $B$ than $C$, it follows imediately that $E$ is closer to $C$ than to $A$ and $F$ is closer to $A$ than to $B$. Now let's intersect $DE$ with $D'E'$ at point $X$. Now the circle inscribed in $ACB$ with radius $\frac r2$ is tangent to both $DE$ and $D'E'$ so it is tangent to $D'X$ in its interior. So let's write Menelaos:\[\frac {XE'}{XD'}\frac{DD'}{CD}\frac {EC}{E'E}=1\] and write $\frac {XE'}{XD'}<\frac {p-a}{p-b}$. Write all 3 inequalities derived from Menelaos by using such inequalities and take their product. It will turn up that $\prod \frac{CD}{DB}<1$ but $CD>BD$ and the analogues from which we derive the contradiction. What do you think about this solution? As far as I know even the official solution used computations... This was also the solution I gave during the contest. ??? I don't understand Quote: Write all 3 inequalities derived from Menelaos by using such inequalities and take their product. It will turn up that $\prod \frac{CD}{DB}<1$

And $\frac {EC}{EE'}>\frac {p-c}{p-a}$

??????????????????????????

Supermath555 wrote: ikap wrote: Assume $D',E',F'$ are the midpoints and let $D$ be closer to $B$ than $C$, it follows imediately that $E$ is closer to $C$ than to $A$ and $F$ is closer to $A$ than to $B$. Now let's intersect $DE$ with $D'E'$ at point $X$. Now the circle inscribed in $ACB$ with radius $\frac r2$ is tangent to both $DE$ and $D'E'$ so it is tangent to $D'X$ in its interior. So let's write Menelaos:\[\frac {XE'}{XD'}\frac{DD'}{CD}\frac {EC}{E'E}=1\] and write $\frac {XE'}{XD'}<\frac {p-a}{p-b}$. Write all 3 inequalities derived from Menelaos by using such inequalities and take their product. It will turn up that $\prod \frac{CD}{DB}<1$ but $CD>BD$ and the analogues from which we derive the contradiction. What do you think about this solution? As far as I know even the official solution used computations... This was also the solution I gave during the contest. ??? I don't understand Quote: Write all 3 inequalities derived from Menelaos by using such inequalities and take their product. It will turn up that $\prod \frac{CD}{DB}<1$

To add some closure to a thread titled "classical computational," here is a computational solution to the problem. Let the incircles of $AEF$, $CED$, and $BDF$ be $\omega_A$, $\omega_C$, and $\omega_B$ with centers $I_A$, $I_C$, and $I_B$. For convenience, let $d(P,Q)$ denote the length of $PQ$, and $p(P,l)$ denote the projection of point $P$ onto line $l$. Let also $L(ABC)$ and $[ABC]$ denote the perimeter and area of triangle $ABC$. If $M$, $N$, $P$ are the midpoints of $BC$, $AC$, and $AB$, it's easy to see that the four triangles thus produced are all congruent, with inradius equal to $1/2$ the inradius of $ABC$. It's also clear, then, that \[ d(p(I_A,AB),p(I_B,AB)) + d(p(I_B,BC),p(I_C,BC)) + d(p(I_A,AC),p(I_C,AC)) = \frac{L(ABC)}{2}. \] On the other hand, we have $d(F,p(I_A,EF)) = d(F,p(I_A,AF))$ and $d(E,p(I_A,EF)) = d(E,p(I_A,AE))$, both by equal tangents. Four similar equations hold, two for each other vertex. This immediately implies that $L(DEF) = L(ABC)/2$. Now, $[ANP] + [NCM] + [PBM] = (r/2)(AC+BC+AB+DE+DF+EF) = (r/2)(L(ABC) + L(DEF)) = (3r/4) L{ABC} = (3/4) [ABC]$, so $[DEF] = [ABC] / 4$. Together with $L(DEF) = L(ABC)/2$, this implies $DEF$ has inradius $r/2$ as well. Let the incircle and incenter of $DEF$ be $\omega_1$ and $I_1$, respectively. We take $DEF$ to be our base triangle, and we consider $A$, $B$, and $C$ to vary depending on the locations of $I_A$, $I_C$, and $I_B$. It suffices to show the only valid configuration for $A$, $B$, and $C$ occurs when $AC||DF$, etc. This is equivalent to $d(F,p(I_A,EF)) = d(E,p(I_1,EF))$, etc. since the four incircles have equal inradii. To show this, define and notice the following: \begin{align*} a &:= \cot (\angle EDF/2) \quad \text{and two similar relations...}\\ a_b &:= \cot (\angle AEF / 2) \quad \text{and five similar relations...} \\ a_c+b_c &= a+b \quad \text{because } PN \text{ is a fixed length.} \\ a_b + c_b &= a+c \\ b_a + c_a &= b+c \\ a + b + c &= abc \quad \text{because angles in a triangle sum to } 180^\circ \text{.} \\ a_c + a_b + a &= a_c a_b a \quad \text{because } BC \text{ is a straight line.} \\ b_a + b_c + b &= b_a b_c b \\ c_a + c_b + c &= c_a c_b c \\ \end{align*} Motivated by the desired result, we additionally set \begin{align*} a_b - c = a - c_b &= \epsilon_b \\ b_c - a = b - a_c &= \epsilon_c \\ c_a - b = c - b_a &= \epsilon_a \\ \end{align*} Substituting these relations into $a_c + a_b + a = a_c a_b a$, we get \begin{align*} b - \epsilon_c + c + \epsilon_b + a &= (b - \epsilon_c )( c + \epsilon_b ) a \\ &= abc + ab\epsilon_b - ac\epsilon_c - a\epsilon_b \epsilon_c \\ -a \epsilon_b \epsilon_c + ab\epsilon_b - ac\epsilon_c - \epsilon_b + \epsilon_c &= 0\\ \end{align*} We get two similar equations for the other two vertices. The three equations are \begin{align*} -a \epsilon_b \epsilon_c + ab\epsilon_b - ac\epsilon_c - \epsilon_b + \epsilon_c &= 0\\ -b \epsilon_a \epsilon_c + bc\epsilon_c - ba\epsilon_a - \epsilon_c + \epsilon_a &= 0\\ -c \epsilon_c \epsilon_b + ca\epsilon_a - cb\epsilon_b - \epsilon_a + \epsilon_b &= 0.\\ \end{align*} Now it's not hard to see that if any $\epsilon$ is zero, then the location of one of (wlog) $I_A$ is fixed at the right location, so $AC||DF$ and $AB||DE$, and the other two incenters are fixed at the right positions, so $BC||EF$, and the desired result holds. Assume none of the $\epsilon$ are zero. Then we may divide the first equation by $\epsilon_b\epsilon_c$ and so on, and, setting $\delta = 1/\epsilon$, we find \begin{align*} (ab-1)\delta_b + (-ac+1) \delta_c & = a \\ (-ab+1)\delta_a + (bc-1) \delta_c & = b \\ (ac-1)\delta_a + (-bc+1) \delta_b & = c \\ \end{align*} Multiplying the first equation by $bc-1$, the second by $ac-1$, and the third by $ab-1$, and adding, we get $0 = a(bc-1) + b(ac-1) + c(ab-1) = 3abc - a - b - c = 2abc$, which is a contradiction as $DEF$ is not degenerate. This contradiction shows that, in fact, some $\epsilon$ is nonzero, which implies the desired result, as noted above. In conclusion, while solutions like these are sometimes nice in their own sort of way, I think they illustrate the value and beauty of non-computational solutions...

We can use the following: Lemma 1:Suppose that $ D,E,F $ are points on the sides $ BC,CA,AB $ of a triangle $ ABC $, respectively, such that the inradii of the triangles $ AEF,BFD,CDE $ are equal to $ r_1 $ and inradii of $ DEF $ is equal to $ r_0 $, then inradii of $ ABC $ is equal to $ r_1 +r_0 $. Lemma 2:Let $ I_1 , I_2,I_3 $ are incenters of $ AEF,BDF,CED $ (with equal inradii) , then inradii of $ I_1I_2I_3 $ and $ DEF $ are equal.(moreover, areas and perimeters are equal).