Let $\triangle D'E'F'$ be the triangle whose medial triangle is $\triangle DEF$. We want to show $\triangle D'E'F'$ and $\triangle ABC$ have the same inradius. We will instead show the following stronger claim, which finishes. Key Claim: In fact, $\triangle ABC$ and $\triangle D'E'F'$ share the same incircle. Proof: We know $AF+DE=AE+DF$ by Pitot, so $AF+D'F=AE+D'E$. By the ``excircle'' version of the converse of Pitot, this implies the quadrilateral bounded by rays $FB$, $D'F$, $D'E$, and $EC$ has an incircle. Call this $\omega_A$, and the corresponding $B-$ and $C-$ analogs $\omega_B$ and $\omega_C$. By Monge on $\omega_A$, $\omega_B$, and $\omega_C$, we find that $D,E,F$ are collinear, contradiction. Hence two of $\{\omega_A,\omega_B,\omega_C\}$ are equal, which implies the desired claim, and finishes the proof. $\blacksquare$

@above, wow!! What an awesome solution!! I'll still post mine since its very different.

Now, direct DEF, ABC positively. Now, let $r$ be inradius of $\Delta DEF$, now expand every circle with radius $-2r$. Let this map be $f$. Now, consider the following $D'=f(DE^+)\cap f(FD^+)$ and define $E', F'$ similarly. Also let $f(BC^+)\cap f(DE^+)=D_Y, f(BC^+)\cap f(DF^+)=D_Z$ and midpoint of $D'D_Y=L_Y$ and midpoint of $D'D_Z=L_Z$. Similarly define $E_X, E_Z, F_X, F_Y,M_X,M_Z,N_X,N_Y$. Now, we have that $D'D_YD_YZ$ has $D'-$ excircle of radius $2r$ and $D'E'F'$ has incircle of radius $r$. Thus, excircle of $D'L_YL_Z$ is the incircle of $D'E'F'$. Also, we have from problem statement that the quadrilateral formed by $f(DE^+), f(BC^+), f(AC^+), f(EF^+)$ has an incircle but we that the quadrilateral formed by $f(DE^+), f(EF^+), L_YL_Z, N_XN_Y$ has incircle the same as that as $D'E'F'$ and the lines of the quadrilaterals are pairwise parallel. Thus, from $E'$, we must have that $L_YN_Y\parallel D_YF_Y$. Thus, $L_YN_Y\parallel AC$. Now, since $L_YL_ZN_XN_YM_ZM_X$ has an incircle, we have from Brianchon that $L_YN_Y, L_ZM_Z, M_XN_X$ concur. Let this point be $T$. Now, $D'L_YTL_Z$ is a parallelogram, thus the midpoint of $D'T$ lies on $L_YL_Z$, so $T$ is on $D_YD_Z$. But then we get that $D_XD_Z, E_XE_Z, F_XF_Y$ concur!! But then under $f$, the incircle of $ABC$ went to the point circle $T$. Thus, it must have had radius $2r$. Hence proved!

Consider the anticomplementary triangle \(D'E'F'\) of \(\triangle DEF\). We will show that the incircles of \(\triangle ABC\) and \(\triangle D'E'F'\) coincide. [asy][asy] size(7cm); defaultpen(fontsize(10pt)); pen pri=blue; pen pri2=lightblue; pen sec=lightred; pen tri=purple+pink; pen fil=invisible; pen fil2=invisible; pen sfil=invisible; pen tfil=invisible; pair Dp = (11.23550,1.19023), Ep = (8.32,-4.02), Fp = (15.,-4.), D = (11.66,-4.01), EE = (13.11775,-1.40488), F = (9.77775,-1.41488), I = (11.43393,-2.18789), A = (11.80061,3.90725), B = (8.49173,-4.79838), C = (13.64143,-3.51694); filldraw(incircle(Dp,Ep,Fp),tfil,tri); filldraw(A--B--C--cycle,sfil,sec); filldraw(D--EE--F--cycle,fil2,pri2); filldraw(Dp--Ep--Fp--cycle,fil,pri); dot("\(D'\)", Dp, N); dot("\(E'\)", Ep, SW); dot("\(F'\)", Fp, SE); dot("\(D\)", D, S); dot("\(E\)", EE, dir(30)); dot("\(F\)", F, dir(150)); dot("\(I\)", I, N); dot("\(A\)", A, N); dot("\(B\)", B, SW); dot("\(C\)", C, E); [/asy][/asy] First note that \(AF-AE=DF-DE=D'E-D'F\), so there is a circle \(\omega_A\) tangent to rays \(AE\), \(AF\), \(D'E\), \(D'F\). Define \(\omega_B\) and \(\omega_C\) analogously. Assume for contradiction the circles \(\omega_A\), \(\omega_B\), \(\omega_C\) do not coincide. One can check that the pairwise exsimilicenters of \(\omega_A\), \(\omega_B\), \(\omega_C\) are \(\overline{BC}\cap\overline{E'F'}=D\), \(\overline{CA}\cap\overline{F'D'}=E\), \(\overline{AB}\cap\overline{D'E'}=F\). By Monge's theorem, points \(D\), \(E\), \(F\) are collinear, contradiction.

pad wrote: By the ``excircle'' version of the converse of Pitot, this implies the quadrilateral bounded by rays $FB$, $D'F$, $D'E$, and $EC$ has an incircle. Could someone explain what that version states?

Darn those are nice solutions. How much would this get?

Loved the half talking you were doing in the solution. Is it advisable to write solutions this way?

What is a half taking?

Solution (Forgot to submit oops, same as above ones though): Take the anticomplementary triangle $D'E'F'$ of $DEF$ (twice the inradius). We claim that $ABC$ and $D'E'F'$ share a incircle, in which case we'd be done. Note that $AF+DE=AE+DF$, and since $D'F = DE$ and $DF=D'E$, we have $AF+D'F = AE+D'E$. Thus, the quadrilateral formed by the rays $AE, AF, D'E, D'F$ has an incircle, call this $w_a$ (I don't know what else to call ). We can do similar things with Pitot, and define $w_b, w_c$ equivalently. By Monge on those three circles, we see that $D, E, F$ are collinear, which is false, so two of the circles coincide, as desired. $\square$ Remark: I added the anticomplementary of $DEF$ as that was the only way I could actually deal with the "twice inradius" statement.

This solution is kinda weird. Use directed lengths throughout. Let $\omega_1$ and $\omega_2$ be the incircles of $DEF$ and $ABC$, and let $\lambda$ be the ratio of their inradii. Let $P$ be the center of the positive homothety mapping the $\omega_1$ to $\omega_2$. Notice by Monge's theorem on $\omega_1$, $\omega_2$, and the incircle of $AEDF$, we have that $P$ lies on $AD$. Similarly, it lies on $BE$ and $CF$. Now, suppose the homothety centered at $P$ mapping $\omega_1$ to $\omega_2$ maps $D$, $E$, and $F$ to $D'$, $E'$, $F'$. We have that $D'E'F'$ and $ABC$ share an incircle. By Poncelet, this implies that $ABCD'E'F'$ all lie on a conic. Hence, it suffices to show the following problem: new problem wrote: Let $ABC$ be a triangle, let $P$ be point in its interior, let $AP$ meet $BC$ at $D$, and extend $PD$ to $D'$ with $PD'=\lambda PD$. Define $E,E',F,F'$ similarly. Show that if $A,B,C,D',E',F'$ are on a conic, then $\lambda=2$. We establish a claim (hidden b/c it's sorta known).

By Pascal, $A,B,C,D',E',F'$ lying on a conic is equivalent to $X=AD' \cap BF'$, $Y=BE' \cap CD'$, and $Z=CF' \cap AE'$ being collinear. Note that $X,Y,Z$ are on $PD$, $PE$, and $PF$, Now, suppose $X$, $Y$, $Z$ are collinear. We have that for some fixed constants $\alpha,\beta,\gamma$, by the hidden claim, \[DBC\text{ collinear } \implies \frac{\alpha}{PD}+\frac{\beta}{PB}+\frac{\gamma}{PC}=0\]\[AEC\text{ collinear } \implies \frac{\alpha}{PA}+\frac{\beta}{PE}+\frac{\gamma}{PC}=0\]\[ABF\text{ collinear } \implies \frac{\alpha}{PA}+\frac{\beta}{PB}+\frac{\gamma}{PF}=0\]\[XBF'\text{ collinear} \implies \frac{\alpha}{PX}+\frac{\beta}{PB}+\frac{\gamma}{\lambda PF}=0\]\[D'YC\text{ collinear} \implies \frac{\alpha}{\lambda PD}+\frac{\beta}{PY}+\frac{\gamma}{PC}=0\]\[AE'Z\text{ collinear} \implies \frac{\alpha}{PA}+\frac{\beta}{\lambda PE}+\frac{\gamma}{PZ}=0\]\[XYZ\text{ collinear} \implies \frac{\alpha}{PX}+\frac{\beta}{PY}+\frac{\gamma}{PZ}=0\]Summing these up with coefficients $1,1,1,-2,-2,-2,2$ gives \[\left(1-\frac{2}{\lambda}\right) \left(\frac{\alpha}{PD}+\frac{\beta}{PE}+\frac{\gamma}{PF}\right)=0.\]Since $DEF$ are not collinear, the second term is nonzero, so we must have $\lambda=2$, as desired.

franzliszt wrote: Darn those are nice solutions. How much would this get? Incomplete bashes are generally worth 0 points.

Can someone complete this

@above i think you forgot to use a sharing link, so we can't see it

mira74 wrote: @above i think you forgot to use a sharing link, so we can't see it Oh, my bad. It should work now ^^

Solution with metrics and trig bash, by @diegoca1. Part 1

Solution with metrics and trig bash, by @diegoca1. Part 2

Solution with metrics and trig bash, by @diegoca1. Part 3

Here is a convoluted solution using Menelaus' Theorem Let $\Omega, \omega, \omega_A, \omega_B, \omega_C$ denote the incircles of $\triangle ABC$, $\triangle DEF$, $AFDE$, $BDEF$, $CEFD$, respectively. Let $I,J,I_A, I_B, I_C$ be the centers of $\Omega$, $\omega$, $\omega_A$, $\omega_B$, $\omega_C$. Claim: $AD, BE, CF$ concur at the exsimilicenter $P$ of $\omega, \Omega$. Proof: Follows from Monge's theorem applied on $\Omega, \omega, \omega_A$. $\blacksquare$ Claim: $D, I_B, I_C, J$ are concyclic. (Similarly, $E,I_A,I_C,J$ and $F,I_A,I_B,J$ are concyclic.) Proof: Angle chasing: $$\angle I_BDI_C = 180^{\circ} - \frac{\angle BDE+\angle CDF}{2} = 90^{\circ} + \frac{\angle EDF}{2}. \blacksquare$$ Before the next step, we would like to recall the well-known fact that $\frac{PD}{AD} + \frac{PE}{BE} + \frac{PF}{CF}=1$. This follows from summing $\frac{PD}{AD} = \frac{[BPC]}{[BAC]}$, etc. Claim: We also have the relation $$\frac{JI_A}{DI_A} + \frac{JI_B}{EI_B} + \frac{JI_C}{FI_C} = 1$$ Proof: Follows from inversion around $J$ an applying the above fact on $\triangle I_A^*I_B^*I_C^*$. $\blacksquare$ Claim: $J$ is the midpoint of $PI$. Proof: Let $t = \tfrac{PI}{JI}$. By Menelaus' theorem on $\triangle DJP$ and $\overline{AI_AI}$, we obtain $$\frac{PI}{IJ}\cdot \frac{JI_A}{I_AD}\cdot\frac{DA}{AP} = 1 \implies t\cdot \frac{JI_A}{I_AD} = \frac{AP}{AD}.$$Similarly, we obtain \begin{align*} t\cdot \frac{JI_B}{I_BE} &= \frac{BP}{BE} \\ t\cdot \frac{JI_C}{I_CF} &= \frac{CP}{CF}, \\ \end{align*}so summing gives $t=2$ as desired. $\blacksquare$ The first and the last claim combined imply the problem.

There is another way to solve this problem here i just give an sketch

iman007 wrote: There is another way to solve this problem here i just give an sketch

Nice. But $P'IPG$ is a parallelogram, in fact.

MZLBE wrote: iman007 wrote: There is another way to solve this problem here i just give an sketch

Nice. But $P'IPG$ is a parallelogram, in fact. no It's not look

You have seen a lot of normal solutions. Now, let's see this weird one, and I'm also not sure if it is totally correct. By Monge's Theorem with three incircles, $BC\cap EF,CA\cap FD, AB\cap DE$ are collinear, and by Desargue's Theorem, $AD,BE,CF$ are concurrent. Let $\omega$ be an incircle of $\triangle DEF$. Construct $\triangle A'B'C'$ which its incircle is $\omega$ and homothetic with $\triangle ABC$ and a homothetic center lies outside $AA'$. If we homothety with center at $D$ and send $\omega$ to an incircle of $AFDE$, it will send $A'B'\to AB$ and $A'C'\to AC$. Therefore, $A'$ lies on $AD$, and similarly, $B'$ lies on $BE$, $C'$ lies on $CF$. Let $X=B'C'\cap AD$. We want to prove that $X$ is a midpoint of $PD$ or $(P,D;X,\infty_{AD})$. The fun begins here. Let $T$ be a projective transformation that send $P$ to the center of $\omega$. For convenience, a name of point after this will referred to a point after the transformation as we will not convert it back. Clearly, $P$ is incenter of $\triangle DEF,\triangle A'B'C'$. Claim: $\triangle DEF,\triangle A'B'C'$ are symmetric with respect to $P$. Let $\triangle D'E'F'$ be symmetric with $\triangle DEF$ wrt $P$. $B'$ and $C'$ are defined by intersections of lines tangent to $\omega$ and $PE,PF$. We see that as $A'$ move along $AP$, a segment $B'C'$ r $BC$ will not intersect $E'F'$ or $EF$ except when $A'\in\{D',D\}$. Thus, $B'C'$ is tangent to $\omega$ iff $A'\in\{D',D\}$, but $A'\neq D$ so $A'=D'$ as desired. $\square$ Since $(A,P;EF\cap AD,D)=-1$ before the transformation, $\triangle ABC$ is now a excentral triangle of $\triangle DEF$. Let $K=BC\cap B'C',L=CA\cap C'A',M=AB\cap A'B'$, these point lie on image of a line at infinity. We want to show that $(D,P;X,KL\cap AD)=-1$. Pencil at $K$, $K(D,P;X,KL\cap AD)=(C,P;C',KL\cap CP)$. Now another unexpected thing come. By Reverse Pascal's theorem on $LA'DKB'E$, these six points lie on a conic $\mathcal{C}$. Moreover, because $A'B'$ and $DE$ are symmetric wrt $P$, a reflection of $L$ across $P$ lso lie on this conic. Denote this reflection by $L'$. Now $L(C,P;C',KL\cap CP)=L(E,L';A',K)=B'(E,L;A',K)$. We know that $B'E,B'L'$ bisect $\angle KB'A'$. (since $B'L'//EL$.) Therefore, the last cross-ratio is equal to $-1$ as desired.

starchan wrote: pad wrote: By the ``excircle'' version of the converse of Pitot, this implies the quadrilateral bounded by rays $FB$, $D'F$, $D'E$, and $EC$ has an incircle. Could someone explain what that version states? Can anyone explain it to me too, please?

Consider the structure based on triangle DEF, it is easy to see(calculating the tangent length) that the condition is equivalent to tangent length from D,E,F to $\omega_{a,b,c}$ are 1/4 perimeter of DEF ($\omega_{a,b,c}$ are 1/2 homothety of excircles of DEF mapped with center D,E,F). Moreover, the tangent of incircle of ABC at each side are the symmetry of D,E,F wrt the midpoint of tangent of $\omega_{a,b,c}$ on each side. The incircle radius of ABC is then easily solved by the area relation.

mira74 wrote: new problem wrote: Let $ABC$ be a triangle, let $P$ be point in its interior, let $AP$ meet $BC$ at $D$, and extend $PD$ to $D'$ with $PD'=\lambda PD$. Define $E,E',F,F'$ similarly. Show that if $A,B,C,D',E',F'$ are on a conic, then $\lambda=2$.

Let $\triangle D'E'F'$ be the anticomplementary triangle of $\triangle DEF$. We finish by the following claim. Claim: The incircles of $\triangle ABC$ and $\triangle D'E'F'$ coincide. Proof. By Pitot, $AF+DE=AE+FD$, so using medial triangle lengths, we rewrite this as \[ AF - AE = D'E-D'F, \]so by Pitot again we have that there is an excircle $\omega_A$ tangent to $AC, AB, D'F', D'E'$. Analogously, we draw $\omega_B$, $\omega_C$. Suppose that $\omega_A$, $\omega_B$, $\omega_C$ are pairwise distinct. By Monge's theorem on $\omega_A$, $\omega_B$, $\omega_C$, we have that $D$, $E$, and $F$ are collinear, a contradiction. Thus, at least two of these circles coincide, implying the claim. Remark: The excircle construction by Pitot very similar to that in 2022 China TST/4/2 (which also used Monge, with one of the three circles being the constructed one), although in the China TST problem the finish was more interesting; the claim that Monge induced resulted in the existence of an ellipse, and DIT was applied on the ellipse.

I was thinking about this diagram recently and reremembered the following (hard) fact. Problem: Show that the radical center of the incircles of $AFDE$, $BDEF$, and $CEFD$ is the incenter of $DEF$. My own solution is fairly long, so I would love to see if there are any shorter solutions.