In $\triangle ABC$, points $D$, $E$, and $F$ lie on sides $BC$, $CA$, and $AB$, respectively, such that each of the quadrilaterals $AFDE$, $BDEF$, and $CEFD$ has an incircle. Prove that the inradius of $\triangle ABC$ is twice the inradius of $\triangle DEF$.
Problem
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Tags: geometry
24.06.2021 10:04
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. $\blacksquare$
24.06.2021 10:55
@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!
24.06.2021 12:39
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.
24.06.2021 15:57
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?
24.06.2021 18:04
Darn those are nice solutions. How much would this get?
Attachments:
2021 ELMO P6.pdf (231kb)
24.06.2021 18:18
Loved the half talking you were doing in the solution. Is it advisable to write solutions this way?
24.06.2021 19:03
What is a half taking?
24.06.2021 20:15
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.
24.06.2021 23:58
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.
25.06.2021 01:42
franzliszt wrote: Darn those are nice solutions. How much would this get? Incomplete bashes are generally worth 0 points.
25.06.2021 02:03
Can someone complete this
25.06.2021 05:52
@above i think you forgot to use a sharing link, so we can't see it
25.06.2021 08:19
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 ^^
25.06.2021 08:38
Solution with metrics and trig bash, by @diegoca1. Part 1
Attachments:
problem6 1-2.pdf (335kb)
problem6 3-4.pdf (453kb)
problem6 5-6.pdf (391kb)
25.06.2021 08:39
Solution with metrics and trig bash, by @diegoca1. Part 2
Attachments:
problem6 11-12.pdf (335kb)
problem6 7-8.pdf (344kb)
problem6 9-10.pdf (289kb)
25.06.2021 08:39
Solution with metrics and trig bash, by @diegoca1. Part 3
Attachments:
problem6 13.pdf (150kb)
25.06.2021 11:28
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.
25.06.2021 12:19
There is another way to solve this problem here i just give an sketch
08.07.2021 06:02
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.
08.07.2021 10:17
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
08.07.2021 16:45
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.
31.07.2021 12:48
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?
04.12.2022 16:49
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.
05.06.2023 16:20
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 $\lambda=2$.
06.09.2023 03:21
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.
29.06.2024 05:55
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.