Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.
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Tags: USA(J)MO, USAMO, geometry, geometric transformation, homothety, parallelogram, geometry solved
30.09.2005 23:10
When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.
08.05.2007 21:30
MithsApprentice wrote: Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_{1}$ and $E_{1}$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_{2}$ and $E_{2}$ the points on sides $BC$ and $AC$, respectively, such that $CD_{2}=BD_{1}$ and $CE_{2}=AE_{1}$, and denote by $P$ the point of intersection of segments $AD_{2}$ and $BE_{2}$. Circle $\omega$ intersects segment $AD_{2}$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_{2}P$. SOLUTION Because of the homothety and the equality of segments (isotomic points and tangents), we get: $\frac{AQ}{QD_{2}}=\frac{AE_{1}}{E_{1}Z}=\frac{CE_{2}}{E_{1}C+CD_{2}}=\frac{CE_{2}}{CD_{1}+BD_{1}}=\frac{CE_{2}}{BC}$ $(1)$ Now we apply Menelaus's Theorem to the triangle $AD_{2}C$: $\frac{PD_{2}}{PA}=\frac{BD_{2}\cdot E_{2}C}{BC \cdot E_{2}A}=\frac{CE_{2}}{BC}\cdot \frac{BD_{2}}{E_{2}A}=\frac{CE_{2}}{BC}$ $(2)$ From $(1)$ and $(2) \Longrightarrow q.e.d.$ Nick Rapanos
06.08.2010 06:31
I'm sorry for reviving an old thread, but I have an alternate solution (though not as short as the above one) to this problem. It suffices to prove that $\frac{AQ}{AD_2}=\frac{PD_2}{AD_2}$. We will find the values of these fractions separately, and prove that they are equal. Before we begin, for the sake of convenience, let us have $AE_1=a$, $BD_1=b$, and $D_1D_2=c$. First, we find the value of $\frac{AQ}{AD_2}.$ Since $BD_1=CD_2$, the $A$-excircle meets $BC$ at $D_2$, and so $Q$ is diametrically opposite to $D_1$. Hence, the line through $Q$ parallel to $BC$ is tangent to the incircle. Furthermore, let the homothety centered at $A$ that takes $Q$ to $D_2$ take $B$ to $T$. We have $\frac{AQ}{AD_2}=\frac{AB}{AT}=\frac{a(a+b)}{a^2+3ab+ac+2b^2+bc}$; the length of $AT$ can be found easily using the relation $\frac{a}{a+b}=\frac{a+2b+c}{AT}$. Now, we find the value of $\frac{PD_2}{AD_2}$. Assign a mass of $a, b, b+c$ to points $A, B, C$, respectively. Then $P$ has a mass of $a+2b+c$. This implies $\frac{AP}{PD_2}=\frac{2b+c}{a}$. Adding one to each sides, $\frac{AD_2}{PD_2}=\frac{a+2b+c}{a}$. Reciprocating both sides, we have $\frac{PD_2}{AD_2}=\frac{a}{a+2b+c}$. Simple algebra reveals that $\frac{a(a+b)}{a^2+3ab+ac+2b^2+bc}=\frac{a}{a+2b+c}$, and we have $AQ=D_2P$, as desired. $\blacksquare$
06.08.2010 07:25
Quote: Let $ABC$ be a triangle and let $w(I)$ be its incircle. Denote by $D_{1}$ and $E_{1}$ the points where $\omega$ is tangent to sides $BC$ , $AC$ respectively. Denote by $D_{2}$ , $E_{2}$ the points on sides $BC$ and $AC$ respectively, such that $CD_{2}=BD_{1}$ , $CE_{2}=AE_{1}$ and denote by $P$ the point of intersection of segments $AD_{2}$ and $BE_{2}$ . Circle $w$ intersects segment $AD_{2}$ at two points, the closer of which to the vertex $A$ is denoted by $Q$ . Prove that $AQ=PD_{2}$ . Proof. Is well-known property $Q\in ID_1\ \ (*)$ . Therefore, $\frac {AQ}{AD_2}=\frac {h_a-2r}{h_a}=\frac {ah_a-2ar}{ah_a}=\frac {2pr-2ar}{2pr}=\frac {p-a}{a}\ \ (1)$ . Apply Menelaus' theorem to $\overline {BPE_2}/AD_2C\ \ :\ \ \frac {BD_2}{BC}\cdot\frac {E_2C}{E_2A}\cdot\frac {PA}{PD_2}=1$ $\implies$ $\frac {p-b}{a}\cdot\frac {p-a}{p-c}\cdot\frac {PA}{PD_2}=1$ $\implies$ $\frac {PA}{p}=\frac {PD_2}{p-a}=\frac {AD_2}{a}$ $\implies$ $\frac {PD_2}{AD_2}=\frac {p-a}{a}\ \ (2)$ . From the relations $(1)$ , $(2)$ obtain $AQ=PD_2$ . ================================== $(*)\ \blacktriangleright$ Denote $\{D_1,Q_1\}=w\cap D_1I$ .Prove easily that $\frac {AD}{Q_1D_1}=\frac {h_a}{2r}=\frac {ah_a}{2ar}=\frac {2pr}{2ar}=\frac pa$ and $\frac {DD_2}{D_1D_2}=\frac {p|b-c|}{a}\cdot \frac {1}{|b-c|}=\frac pa$ . In conclusion, $\frac {AD}{Q_1D_1}=\frac {DD_2}{D_1D_2}$ , i.e. $Q_1\in w\cap AD_2$ $\implies$ $Q_1\equiv Q$ .
07.08.2010 01:01
In my solution, I used also the fact that $Q\in D_1I$ which is new for me, I just remarqued it from the diagram and proved it, and I think if we knew this result, the problem becomes easy.
07.08.2010 08:54
Dear mathlinkers, this problem come from Lalesco a geometer of the beginning of the XX centuries... Why P which is the Nagel point is not used in order to have a synthetic proof? Sincerely Jean-Louis
07.08.2010 09:22
jayme wrote: Dear mathlinkers, this problem come from Lalesco a geometer of the beginning of the XX centuries... Why P which is the Nagel point is not used in order to have a synthetic proof? Sincerely Jean-Louis I am trying to get a proof using the fact that $P$ is the nagel point but I am not able to complete it. May be someone can. From an homothety with centre A and ratio $\frac{s-a}{s},$ the point $D_2$ maps to $Q$ and hence $\frac{AD_2}{AQ}=\frac{s}{s-a}$ and hence $\frac{AP}{AQ}+\frac{D_2P}{AQ}=1+\frac{a}{s-a}$ It suffices to prove that $\frac{AP}{AQ}=\frac{a}{s-a}\Longleftrightarrow \frac{AP}{AD_2}=\frac{a}{s}$ and from here, I am not able to complete the proof. If someone knows the length of the nagel point from the vertex...
07.08.2010 10:24
Nagel point $P$ of $\triangle ABC$ is the incenter of its antimedial triangle $\triangle A_0B_0C_0$ and orthogonal projections $M,M_0$ of $A,A_0$ on $BC$ are symmetric about the midpoint of $BC.$ Let the incircle $(P,2r)$ of $\triangle A_0B_0C_0$ touch its side $B_0C_0$ at $D_0$ and let $R \equiv AP \cap A_0M_0.$ Since $A_0D_1$ is the $A_0$- Nagel ray of $\triangle A_0B_0C_0,$ it follows that $A_0D_1 \parallel AR$ $\Longrightarrow$ quadrilateral $QRA_0D_1$ is a parallelogram, hence $A_0R=QD_1=PD_0=2r.$ $T$ is the orthogonal projection of $P$ onto $BC,$ since $D_0T=A_0M_0.$ Then $PT=RM_0$ $\Longrightarrow$ $\overline{D_2T}=\overline{D_2M_0}=\overline{D_1M}.$ If the projections of $\overline{PD_2},\overline{AQ}$ on $BC$ are equal, then $PD_2=AQ,$ as desired.
25.07.2011 06:34
Here is a rather nice solution using barycentric coordinates: Let $A$ be $(1,0,0)$, $B$ be $(0,1,0)$, and $C$ be $(0,0,1)$. Let the side lengths of the triangle be $a,b,c$ and the semi-perimeter $s$. Now, $CD_1=s-c$, $BD_1=s-b$, $AE_1=s-a$, $CE_1=s-c$. Thus, $CD_2=s-b$, $BD_2=s-c$, $AE_2=s-c$, $CE_2=s-a$. Therefore, the non-normalized coordinates of $D_2$ are $(0,s-b,s-c)$ and of $E_2$, $(s-a,0,s-c)$. Clearly then, the non-normalized coordinates of $P$ are $(s-a,s-b,s-c)$. Normalizing, we have that $D_2$ is at $\left(0,\frac{s-b}{a},\frac{s-c}{a}\right)$, $E_2$ at $\left(\frac{s-a}{b},0,\frac{s-c}{b}\right)$, and $P$ is at $\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right)$ Now, we find the point $Q'$ inside the triangle on the line $AD_2$ such that $AQ'=D_2P$. It is then sufficient to show that this point lies on the incircle. $P$ is the fraction $\frac{s-a}{s}$ of the way "up" the line segment from $D_2$ to $A$. Thus, we are looking for the point that is $\frac{s-a}{s}$ of the way "down" the line segment from $A$ to $D_2$, or, the fraction $1-\frac{s-a}{s}$ of the way "up". Thus, $Q'$ has normalized $x$-coordinate $1-\frac{s-a}{s}=\frac{a}{s}$. As the line $AD_2$ has equation $(s-c)y=(s-b)z$, it can easily be found that $Q'$ lies at $\left(\frac{a^2}{as}, \frac{(s-a)(s-b)}{as}, \frac{(s-a)(s-c}{as}\right)$, or in non-normalized coordinates with sum $as$, at $(a^2,(s-a)(s-b),(s-a)(s-c)$. Recalling that the equation of the incircle is $a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0$, we must show that this equation is true for $Q'$'s values of $x,y,z$. Plugging in our values, this means showing that $a^2(s-a)^2(s-b)(s-c)+a^2(s-a)[b^2(s-c)+c^2(s-b)]+as[a^2(s-a)^2+(s-a)(s-b)^3+(s-a)(s-c)^3]=0$. Dividing by $a(s-a)$, this is just $a(s-a)(s-b)(s-c)+a[b^2(s-c)+c^2(s-b)]+a^2s(s-a)+s(s-b)^3+s(s-c)^3=0$. Plugging in the value of $s$: $\frac{a(-a+b+c)(a-b+c)(a+b-c)}{8}+\frac{ab^2(a+b-c)}{2}+\frac{ac^2(a-b+c)}{2}+\frac{a^2(a+b+c)(-a+b+c)}{4}+\frac{(a+b+c)(a-b+c)^3}{16}+\frac{(a+b+c)(a+b-c)^3}{16}=0$. $2a(-a+b+c)(a-b+c)(a+b-c)+8ab^2(a+b-c)+8ac^2(a-b+c)+4a^2(a+b+c)(-a+b+c)+(a+b+c)(a-b+c)^3+(a+b+c)(a+b-c)^3=0$ $2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0$ The first bracket is just: $-a^3-b^3-c^3+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b-2abc+4ab^2+4b^3-4b^2c+4ac^2-4bc^2+4c^3=-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc$ and the second bracket is $-2a^3+4a^2b+4a^2c+6ab^2+6ac^2-12abc$. Dividing everything by $2a$ gives $-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc+(a+b+c)(-a^2+2ab+2ac+3b^2+3c^2-6bc)$, which is $0$, as desired. As $Q'$ lies on the incircle and $AD_2$, $Q'=Q$, and our proof is complete.
02.02.2013 18:39
Some days ago i saw another problem solved by using this Lemma but I couldn't solve.This one can be solved by following Lemma too: Lemma: $I$(incenter) , $G$(centroid),$N$(Nagel point) lie in that order on a line (Known as the Nagel line) & $GN=2IG$ Proof:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=50&t=518404 Draw a line through $D_{1}$ & $G$ so it intersects$PQ$ at$ F$.Name midpoint of $BC$ : $D"$ As $PI$ bisects $QD_{1}$ & $G$ is on $PI$ such that $GP=2IG$(according to theLemma) so $G$ is the centroid of $\vartriangle PQD_{1}$ so$F$ is the midpoint of $ QP$ (1) In $AD_{1}D_{2}$: $AD"$ bisects $D_{1}D_{2}$ and $G$ is on it such that $AG=AGD"$ so $G$ is the centroid of$\vartriangle AD_{1}D_{2}$ and $F$ is the midpoint of $AD_{2}$. (2) By using (1) & (2) we can conclude .
31.03.2013 04:25
The above lemma is awesome. See here and here. The key point is that the Nagel point of a triangle (in this case, $P$) is the incenter of its anticomplementary triangle.
24.04.2013 04:51
17.02.2015 17:28
Let $\omega_a$ be the excircle of $\triangle{ABC}$ opposite to vertex $A.$ $\omega_a$ touches $BC$ at $D_2$ Let $\omega_a$ touches $AC$ at $H.$ Then $\angle{D_2HE}=\frac{\angle{C}}{2}$ It is well known lemma that $Q,I,D_1$ are collinear. $\angle{E_1D_1C}=\angle{D_1E_1C}=\pi-\frac{\angle{C}}{2}$ So, $\angle{AE_1Q}=\frac{\angle{C}}{2}=\angle{D_2HE}$ Hence $D_2H || QE_1$ So $\frac{AQ}{AD_2}=\frac{AE_1}{AH}=\frac{s-a}{s}$ Applying menelus theorem ($\triangle{AD_2C},\overline{E_2PB}$) it is easy to obtain $\frac{PD_2}{AD_2}=\frac{s-a}{s}$ Hence $AQ=D_2P$
13.05.2015 12:20
Let $ M$ is the midpoint of $ BC \implies D_1M=D_2M $ , $D_1Q $ is the diameter and $ I $ is the center of $\omega$, implies $IM \parallel AD_2$ by using midpoint theorem in $ \triangle QD_1D_2 $ and $ 2IM=QD_2$ $ AD_2 \cap BE_2 =P$ (Nagel point ) $ AM \cap IP =G $(Centroid) and by using the fact that $ 2IG=GP$ we get that $AP=2IM=QD_2 \implies AQ=D_2P $
13.05.2015 12:23
This statement $ AQ = D_2P$ can be used to prove the existence of nagel line and $ 2IG=GN$ see here:http://www.artofproblemsolving.com/community/c3141h1088525_nagel_line__aq__d_2p
21.05.2015 02:43
Solution found with K6160 . For convenience, let us rephrase the problem: Let $\triangle ABC$ be a triangle, and let $D, A_1$ be the points of tangency of the incircle, $A$-excircle, respectively, with side $\overline{BC}.$ Let $D'$ be the intersection of $AA_1$ and the incircle that is closer to $A$, and let $N$ be the Nagel point of $\triangle ABC.$ Then $AD' = A_1N.$ Solution: Let $I, G$ be the incenter, centroid, respectively, of $\triangle ABC.$ We proceed with complex numbers: WLOG let $I$ be the origin, and assign complex numbers $a, b, c, d, d', a_1, n$ to $A, B, C, D, D', A_1, N$, respectively. It is well-known that $\overline{DD'}$ is a diameter of $\omega$, so $d' = -d.$ Now, in order to show that $AD' = A_1N$, it suffices to prove that $\overline{AA_1}$ and $\overline{D'N}$ share the same midpoint, i.e. $d' + n = a + a_1.$ This is equivalent to $n = a + a_1 + d$, but since $\overline{A_1D}$ and $\overline{BC}$ share the same midpoint, we have $a_1 + d = b + c.$ Therefore, it suffices to prove that $n = a + b + c = 3g.$ Because $I$ is the origin, this is equivalent to $IN = 3IG$, but this is just a well-known property of the Nagel line. $\square$
28.09.2015 21:37
From the lemma we know that $D_1Q$ is a diameter of the incircle. Let the incenter of ABC be I, its centroid be G, and the midpoint of BC be M. Note that P is the Nagel point of ABC. From the previous problem, we know that the dilation centered at G with ratio −2 sends M to A and I to P, and hence sends segment IM to PA, thus PA = 2IM. On the other hand, a dilation centered at $D_1$ with ratio 2 sends IM to $QD_2$, so $QD_2$ = $2IM$ = $PA$. Therefore, AQ = PA−QP = $QD_2$ − $QP$ =$D_2P$. (Lemma used is the diameter of incircle)
04.10.2015 04:56
Clearly, $P$ is the Nagel point of triangle $ABC$. Reflect $A$ about the incenter to $A'$. If we can show that $A'D_1PD_2$ is a parallelogram, we will be done, for reflecting back about $Q$ yields $A'D_1=AQ=PD_2$, as it is well known that $Q$ is the antipode of $D_1$ with respect to the incircle. We claim that triangles $AA'P$ and $ABC$ have concurrent centroids $G$. Indeed, by the Nagel line, $I$, $G$, and $P$ are collinear, but $I$ is the midpoint of $AA'$, so $G$ is indeed the centroid of triangle $AA'P$. Denoting $M$ as the midpoint of $BC$, it is now clear that $A', M$ and $P$ are collinear, whence $D_1A'D_2P$ is a parallelogram, so we are done.
31.12.2015 00:34
11.12.2021 03:22
WLOG let $b>c$. Let $D$ be the foot of the altitude from $A$ to $BC$ and let $D_3$ be the foot of the altitude from $P$ to $BC$. It suffices to show that $D_2D_1=D_3D=b-c$. By mass points we have $D_3D=D_2D\left(\frac{2a}{a+b+c}\right)$ and since $BD=\frac{c^2-b^2+a^2}{2a}$ we know that $D_2B=\frac{a+b-c}{2}$ we have that $D_2D=\frac{ab-ac-c^2+b^2}{2a}$ and $D_3D=b-c$, done.
15.12.2021 22:08
rcorreaa wrote: We use Barycentric Coordinates with $ABC$ being the reference triangle. Since $P$ is the Nagel Point, $$P= (s-a : s-b : s-c)= (\frac{b+c-a}{a+b+c},\frac{c+a-b}{a+b+c},\frac{a+b-c}{a+b+c}) \qquad (\star)$$ Let $I$ the incenter of $ABC$. It's well known that $I$ is the midpoint of $D_1Q$ and $I=(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c})$. Hence, $$Q=(\frac{2a}{a+b+c},\frac{2b}{a+b+c}-\frac{s-c}{2a},\frac{2c}{a+b+c}-\frac{s-b}{2a})$$so $\frac{P+Q}{2}= (1,1-\frac{s-c}{2a},1-\frac{s-b}{2a})= \frac{A+D_2}{2}$, since $D_2=(0 : s-c : s-b)$. Thus, the midpoint of $PQ$ coincides with the midpoint of$AD_2$, so $AQ=D_2P$, as desired. $\blacksquare$ Very ingenious solution.
05.01.2022 07:20
Note that $P$ is the Nagel point and let $I$ and $G$ be the incenter and centroid, respectively. Then, $Q$ is the antipode of $D_1$ with respect to $\omega.$ Also, $\triangle GAP\sim\triangle GMI$ since $AG/GM=PG/GI=2.$ Hence, $QD_2=2IM=AP.$ $\square$
16.03.2022 19:10
This screams bary lmao. Note that $Q$ is the antipode of $D_1$ on the incircle. We consider bary with $\triangle ABC$ as the reference triangle. Then, $I = (a:b:c), D_1 = (0:s-c: s-b)$, and thus, $Q= 2I - D = (\frac{2a}{a+b+c}, \frac{2b}{a+b+c} - \frac{s-c}{a}, \frac{2c}{a+b+c} - \frac{s-b}{a})$. Then, note that $P = (s-a: s-b: s-c)$ is the Nagel point. Then, it suffices to show that $P+Q = A+D_2$, and since they're collinear we check the first coordinate, so \[\frac{-\frac12 a + \frac12 b + \frac12c}{\frac{a+b+c}{2}} + \frac{2a}{a+b+c} = 1 + 0\]which is true, so $AQ = D_2P$ and we're done. $\blacksquare$.
18.07.2022 19:59
02.09.2022 21:28
$AP=2IM=QD_2$, where $M$ is the midpoint of $BC$.
25.12.2022 04:44
Note that $Q$ is the antipode of $D$ in the incircle. $I$ is the midpoint of $QD$ and $M$ the midpoint of $D_1D_2.$ Thus, $IM=\frac12 QD_2.$ Also, $I,G,P$ are collinear with $G$ one-third of the way from $I$ to $P$. Since $MI\parallel AP$, $\triangle MIG\sim \triangle APG\implies AP=2MI$ which implies result.
26.01.2023 22:29
20.05.2023 02:03
actually human solution: Claim: $Q$ is the reflection of $D_1$ over $I$ Consider the $A$-excircle, and let the tangents from $A$ to the $A$-excircle have length $x$, then considering the tangents from $B,C$ gives $x - b + x - c = a$, so $x = s$, then $D_2$ is the tangency of the $A$-excircle and $BC$ by length equivalence. Then by Fact 5, $A$, and the centers of the incircle and the $A$-excenter are collinear, so we consider the homothety mapping the incircle to the $A$-excircle, this also maps $Q$ to $D_2$, so we are done. Now, we prove the result that $\frac{AQ}{QD_2} = \frac{D_2P}{PA}$, which finishes. Notice that $\frac{D_2P}{PA} = \frac{CE_2}{E_2A} \frac{BD_2}{BC} = \frac{s - a}{a}$ by ratio lemma. Then we use ratio lemma again, get $\frac{AQ}{QD_2} = \frac{AB}{BD_2} \frac{\sin ABQ}{\sin QBD_2}$, then drop perpendiculars $D_3, Q_2$ from $D_1,Q$ to $AB$, then we have $\frac{AB}{BD_2} \frac{\sin ABQ}{\sin QBD_2} = \frac{c}{s - c} \frac{QQ_2}{QD_1} $, now notice (where undefined points are to be defined analgously) that $2F_1F_2 = D_1D_3 + QQ_2$, so we can evaluate $QQ_2 = 2r - (s-b)\sin B$, so our desired equality is $\frac{s-a}{a} = \frac{c}{s-c} \frac{2r - (s-b) \sin B}{2r}$, which we can reduce to $\frac{b^2 - (a-c)^2}{4ac} = 1 - \frac{\sin B}{2 \tan \frac B2}$, which reduces further to $\frac{b^2-a^2-c^2}{4ac} = \frac{1}{2} - \cos^2 \frac B2$, which we can reduce further to $\frac{b^2 - a^2 - c^2}{4ac} = \frac 12 - \frac{1 + \cos B}{2}$, which rearranges to the Law of Cosines, so because all steps are reversible, we are done.
31.07.2023 18:40
Very easy especially for U2. By Menelaus on triangle $AD_2C$ with respect to $\overline{BPE_2}$, we obtain $\frac{AP}{PD_2} = \frac a{s-a}$. Also, $$\frac{D_2Q}{D_2A} = \frac{2r}{h_A} = \frac as,$$so the result follows.
17.09.2023 21:38
$Q$ is the antipode of $D_1$ with respect to the incircle, so the tangent to the incircle at $Q$ is parallel to $\overline{BC}$. If the triangle $\triangle$ formed by it and $\overline{AB},\overline{AC}$ has similarity ratio $k$ from $[ABC]$, then we have $$k=\frac{AQ}{AD_2}=\sqrt{\frac{[\triangle]}{[ABC]}}=\sqrt{\frac{k(s-a)}{s}} \implies \frac{AQ}{AD_2}=\frac{s-a}{s}.$$On the other hand, in barycentric coordinates is clear that $D_2=(0:s-b:s-c)=(0,\tfrac{s-b}{a},\tfrac{s-c}{a})$ and $P=(s-a:s-b:s-c)=(\tfrac{s-a}{s},\tfrac{s-b}{s},\tfrac{s-c}{a})$, hence $\frac{PD_2}{AD_2}=\frac{s-a}{s}$ as well, so we're done. $\blacksquare$
11.04.2024 03:54
Note that $D_2$, $E_2$ are the respective extouch points. By Menelaus on $\triangle AD_2C$, we find that (in non-directed lengths) $$\frac{D_2P}{PA} = \frac{E_2C}{AE_2} \cdot \frac{BD_2}{CB} = \frac{s - a}{s - c} \cdot \frac{s - c}{a} = \frac{s - a}{a}.$$Also, by a homothety at $A$, we find that $Q$ is the antipode of $D_1$ wrt the incircle of $\triangle ABC$. Therefore if $E_3$ is the projection of the $A$-excenter of $\triangle ABC$ onto $\overline{AC}$, then $$\frac{AQ}{QD_2} = \frac{AE_1}{E_1E_3} = \frac{s - a}{a} = \frac{D_2P}{PA},$$which implies the conclusion.
11.04.2024 13:59
Oh wow did I overkill this. Denote $BC, CA, AB, $ the semiperimeter, inradius, and $A-$exradius of $\Delta ABC$ by $a, b, c, s, r, R$ respectively. Observe that $D_2, E_2$ are the $A, B$ extouch points respectively (i.e. where the $A-, B-$ excentres are tangent to $BC, CA$). Introduce the $F-$intouch and extouch points respectively to be $F_1, F_2$. Now observe that $AD_2, BE_2, CF_2$ concur due to Ceva's theorem and the existence of the incentre. Therefore, from Van Aubel's theorem and some well-known lengths, we get $$\frac{AP}{PD_2} = \frac{AE_2}{E_2C} + \frac{AF_2}{F_2B} = \frac{CE_1}{E_1A} + \frac{BF_1}{F_1A} = \frac{(s-b)+(s-c)}{(s-a)}= \frac{a}{s-a},$$so $$\frac{PD_2}{AD_2} = \frac{s-a}{s}.$$Also, it is well known that the homothety at $A$ taking the incircle to the $A-$excircle also takes $Q$ to $D_2$, so we get $$\frac{AQ}{AD_2} = \frac{r}{R} = \frac{s-a}{s},$$where the final part follows by the exradius length formula given in EGMO Chapter 2. Therefore, we get $$\frac{PD_2}{AD_2} = \frac{s-a}{s} = \frac{AQ}{AD_2},$$and the result follows. $\square$
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18.04.2024 19:17
We set $A=(1,0,0)$ and etc. Now, we know that $D_1=(0:a+b-c:a-b+c)$ and $E_1=(a+b-c:0:-a+b+c)$. Thus, we also immediately obtain that $D_2=(0:a-b+c:a+b-c)$ and $E_2=(-a+b+c:0:a+b-c)$. Now, it is clear that $P=(-a+b+c:a-b+c:a+b-c)$ since it is the intersection point of the cevians $AD_2$ and $BE_2$. We now do some synthetic work. Claim : The incenter $I$ of $\triangle ABC$ is in fact the midpoint of $QD_1$. Proof : Consider the homothety centered at $A$ mapping the incircle to the $A-$excircle. This clearly maps $D_2 \to Q$. Since $BC$ is tangent to the $A-$excircle at $D_2$, it also follows that the line parallel to , $BC$ through $Q$ is tangent to the incircle. Thus, $Q$ is the diametrically opposite point of $D_1$ from which the claim clearly follows. We know that $I=(a:b:c)$. Further, since $Q$ lies on $AD_2$, $Q=(t:a-b+c:a+b-c)$. Thus, $[BQC]=2[BIC]$ which means, \begin{align*} \frac{t}{t+(a-b+c)+(a+b-c)} &= 2\frac{a}{a+b+c}\\ \frac{t}{t+2a} &= 2\frac{a}{a+b+c}\\ ta + t(b+c) &= 2at +4a^2\\ t &= \frac{4a^2}{b+c-a} \end{align*}Thus, we conclude that $Q=(4a^2: c^2 - (a-b)^2 : b^2 - (c-a)^2)$. Now, to show that $AQ = PD_2$, it clearly suffices to show that $[ABQ] = [PBD_2]$. Now, note that \[\begin{vmatrix} 0 &1 & 0 \\ 1 & 0 & 0\\ \frac{2a^2}{a^2+ab+ac} & \frac{c^2 - (a-b)^2}{2(a^2+ab+ac)} & \frac{b^2 - (c-a)^2}{2(a^2+ab+ac)} \end{vmatrix} = \frac{b^2 - (c-a)^2}{2a^2+ab+ac} = \frac{b^2 - (c-a)^2}{2a(a+b+c)}\]and \[\begin{vmatrix} 0 &1 & 0\\ 0 & \frac{a-b+c}{2a} & \frac{a+b-c}{2a}\\ \frac{-a+b+c}{a+b+c} & \frac{a-b+c}{a+b+c} & \frac{a+b-c}{a+b+c} \end{vmatrix} = \frac{(a+b-c)(-a+b+c)}{2a(a+b+c)} = \frac{b^2- (c-a)^2}{2a(a+b+c)}\]so we are done.
27.10.2024 20:13
Has no one lenght bashed this? Denote by $I$ the incenter of $\triangle ABC$ and $M=QE_1 \cap BC$. Also let $p$ denote the semiperimeter of the triangle. Menelaus in $\triangle ACD_2$ gives: $$ \frac{AE_2}{CE_2} \cdot \frac{BC}{BD_2} \cdot \frac{D_2P}{AP}=1$$$$ \Rightarrow \frac{p-c}{p-a} \cdot \frac{a}{p-a} \cdot \frac{D_2P}{AP} =1 (1) $$ It is well-known that $Q$ is the antipode of $D_1$ in the incircle. In $\triangle QD_1M$, $D_1M=2r\cot(\frac{\angle C}{2})$, therefore $CM=2r\cot(\frac{\angle C}{2}) - (p-c)$ and $D_2M= 2r\cot(\frac{\angle C}{2}) - (p-c)+(p-b)$. Applying Menelaus in $\triangle ACD_2$, we get: $$\frac{AQ}{QD_2} \cdot \frac{D_2M}{CM} \cdot \frac{CE_1}{AE_1} = 1 (2)$$From $(1)$ and $(2)$ it suffices to prove: $$1+\frac{p-b}{2r\cot(\frac{\angle C}{2})-(p-c)}=\frac{a}{p-c}$$$$\Rightarrow \frac{2r\cot(\frac{\angle C}{2})}{p-b}=\frac{p-c}{p-b} + \frac{p-c}{a+c-p} (3)$$ Using Heron’s formula and the metric formulas for $\sin\frac{\angle C}{2}$ and $\cos\frac{\angle C}{2}$ we can compute: $2r\cot(\frac{\angle C}{2})=2(p-c)$. Plugging this in $(3)$ and doing the calculations we get that the problem is equivalent to: $$\frac{1}{p-b} = \frac{1}{a+c-p}$$which is true, as desired. $\blacksquare$