The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $ BC, CA, AB$ at $ A_1 , B_1 , C_1 $ respectively. The line $AI$ meets the circumcircle of $ABC$ at $A_2 $. The line $B_1 C_1 $ meets the line $BC$ at $A_3 $ and the line $A_2 A_3 $ meets the circumcircle of $ABC$ at $A_4 (\ne A_2 ) $. Define $B_4 , C_4 $ similarly. Prove that the lines $ AA_4 , BB_4 , CC_4 $ are concurrent.
Problem
Source: North Korea Team Selection Test 2013 #1
Tags: geometry, circumcircle, trigonometry, North Korea
17.05.2014 17:43
$(A_{4},A,B,C)=-1$ $\Rightarrow$ $ AA_4 , BB_4 , CC_4 $ are concurrent in lemone poin of $ABC$
17.05.2014 19:21
nima1376, you are wrong, the quadrilateral $ABA_4C$ is not harmonic in general. The concurrency point is not the Lemoine point of $\triangle ABC$ but its center $X_{57},$ i.e. the perspector of $\triangle ABC$ and the orthic triangle of $\triangle A_1B_1C_1.$ Let $A_0,B_0,C_0$ be the projections of $A_1,B_1,C_1$ on $B_1C_1,$ $C_1A_1,$ $A_1B_1.$ $D$ and $M$ denote the midpoints of $\overline{BC}$ and the arc $BAC$ of the circumcircle $(O).$ From $(B,C,A_1,A_3)=-1,$ we get $A_3D \cdot A_3A_1=A_3B \cdot A_3C=A_3A_4 \cdot A_3A_2$ $\Longrightarrow$ $DA_1A_4A_2$ is cyclic $\Longrightarrow$ $\angle A_2A_4A_1=\angle A_2DA_1=90^{\circ}$ $\Longrightarrow$ $M \in A_1A_4.$ Furthermore, $A_1A_0 \perp A_3A_0$ implies that $B_1C_1$ bisects $\angle BA_0C$ externally or $\angle BA_0C_1=\angle CA_0B_1.$ Since $\angle BC_1A_0=\angle CB_1A_0,$ then $\triangle BA_0C_1 \sim \triangle CA_0B_1$ $\Longrightarrow$ $C_1A_0:B_1A_0=BC_1:CB_1=BA_1:CA_1,$ therefore isosceles $\triangle AB_1C_1 \sim \triangle MCB$ are similar with corresponding cevians $AA_0$ and $MA_1$ $\Longrightarrow$ $\angle MA_1C=\angle AA_0B_1.$ Hence since $A_0A_1A_4A_3$ is cyclic on account of the right angles at $A_0,A_4,$ we deduce that $A,A_0,A_4$ are collinear and analogously $B_0 \in BB_4$ and $C_0 \in CC_4.$ By Cevian Nest Theorem the conclusion follows.
17.05.2014 19:49
Luis González wrote: nima1376, you are wrong, the quadrilateral $ABA_4C$ is not harmonic in general. The concurrency point is not the Lemoine point of $\triangle ABC$ but its center $X_{57},$ i.e. the perspector of $\triangle ABC$ and the orthic triangle of $\triangle A_1B_1C_1.$ sorry for my bad mistake another solution let $O$ is circumcircle of triangle $ABC$ $A_{2}A_{1}\cap O=K$ . $(B,C,A_{1},A_{4})=-1\Rightarrow (K,A_{4},B,C)=-1\Rightarrow \frac{BA_{4}}{A_{4}C}=\frac{KB}{KC}$ but $K$ is center of spiral similar goes $BC_{1}$ to $CB_{1}$ $\Rightarrow \frac{BK}{CK}=\frac{BC_{1}}{CB_{1}}$ so we are done with ceva...
20.05.2014 03:47
North Korea is NOT Korea! "Korea" means South Korea. Why is this in Korea category?
20.05.2014 09:13
First we claim that $\dfrac{A_{4}B}{A_{4}C}=\dfrac{A_{3}B}{A_{3}C}$. For the sake of argument, let us assume that $A_3$ is on the ray $BC$. Now $\angle BA_{4}A_{3}=180^{\circ}-\dfrac{\angle A}{2}$. So, $\angle A_{4}A_{3}C=\angle A_{4}A_{3}B=180^{\circ}-\angle BA_{4}A_{3}-\angle A_{4}BA_{3}$ $=\dfrac{\angle A}{2}-\angle A_{4}BA_{3}=\dfrac{\angle A}{2}-\angle A_{4}BC=\angle A_{2}AC-\angle A_{4}AC$ $=\angle A_{2}AA_{4}=\angle A_{2}BA_{4}=\angle A_{2}BA_{4}$. This means $A_{2}B$ is tangent to the circle $BA_{3}A_{4}$ and $A_{2}C$ is tangent to the circle $CA_{3}A_{4}$. From these information, we have that $A_{4}B=\dfrac{A_{2}B\times A_{3}B}{A_{2}A_{3}}$ and $A_{4}C=\dfrac{A_{2}C\times A_{3}C}{A_{2}A_{3}}$. Since $A_{2}B=A_{2}C$ we have $\dfrac{A_{4}B}{A_{4}C}=\dfrac{A_{3}B}{A_{3}C}$ as we claimed. By similar arguments, we can show that, $\dfrac{B_{4}C}{B_{4}A}=\dfrac{B_{3}C}{B_{3}A}$ and $\dfrac{C_{4}A}{C_{4}B}=\dfrac{C_{3}A}{C_{3}B}$. Now apply Menelaus's theorem for the lines $A_{3}B_{1}, B_{3}C_{1}, C_{3}A_{1}$. Then we have three equations: 1) $\dfrac{BA_{3}}{A_{3}C}\cdot \dfrac{CB_{1}}{B_{1}A}\cdot \dfrac{AC_{1}}{C_{1}B}=1$ 2) $\dfrac{CB_{3}}{B_{3}A}\cdot \dfrac{AC_{1}}{C_{1}B}\cdot \dfrac{BA_{1}}{A_{1}C}=1$ 3) $\dfrac{AC_{3}}{C_{3}B}\cdot \dfrac{BA_{1}}{A_{1}C}\cdot \dfrac{CB_{1}}{B_{1}A}=1$ Multiply (1),(2),(3) and use the fact $AC_1=AB_1, BC_1=BA_1, CA_1=CB_1$ to get that, $\dfrac{A_{3}B}{A_{3}C}\cdot \dfrac{B_{3}C}{B_{3}A}\cdot \dfrac{C_{3}A}{C_{3}B}=1$ Thus by our previous argument, 4) $\dfrac{A_{4}B}{A_{4}C}\cdot \dfrac{B_{4}C}{B_{4}A}\cdot \dfrac{C_{4}A}{C_{4}B}=1$ Now,by sine law, we have that $\dfrac{A_{4}B}{A_{4}C}=\dfrac{\sin \angle BAA_4}{\sin \angle CAA_4}$. Deriving similar expressions for $\dfrac{B_{4}C}{B_{4}A},\dfrac{C_{4}A}{C_{4}B},$ and using them in (4), we get that , $\dfrac{\sin \angle BAA_4}{\sin \angle CAA_4}\cdot \dfrac{\sin \angle CBB_4}{\sin \angle ABB_4}\cdot \dfrac{\sin \angle ACC_4}{\sin \angle BCC_4}=1$. This the trigonometric form of Ceva's theorem. So $AA_4, BB_4, CC_4$ are concurrent.
23.05.2014 13:34
Dear Mathlinkers, the problem have something to do with the A-mixtilinear incircle of ABC... see http://perso.orange.fr/jl.ayme , A new mixtilinear incircle adventure I, G.G.G. vol. 4, p. 20-21. Sincerely Jean-Louis
13.12.2014 15:22
is this too hard for a problem 1 on an olympiad?
13.12.2014 17:05
To Konigsberg; depends To jayme, yes indeed, my solution shows this directly. Let the perpendicular to $I$ through $AI$ meet $BC$ at $A_5$, then note $A(A_5, A_4; B, C) = -1$. We know that $A_5B_5C_5$ are collinear and perpendicular to $HI$, so $AA_4, BB_4, CC_4$ are concurrent.
13.12.2014 17:18
My solution (for the original problem) : Since $ AA_1, BB_1, CC_1 $ are concurrent at $ X_{7} $ of $ \triangle ABC $ , so from Desargue theorem we get $ A_3 , B_3 , C_3 $ are collinear . ... $ (\star ) $ Since $ A_4A_2, B_4B_2, C_4C_2 $ is the external bisector of $ \angle CA_4B, \angle AB_4C, \angle BC_4A $, respectively , so combine with $ ( \star ) $ we get $ \frac{CA_4}{A_4B} \cdot \frac{AB_4}{B_4C} \cdot \frac{BC_4}{C_4A}=\frac{CA_3}{A_3B} \cdot \frac{AB_3}{B_3C} \cdot \frac{BC_3}{C_3A}=1 $ . ie. $ AA_4, BB_4, CC_4 $ are concurrent Q.E.D __________________________________________________ My solution ( for $ AA_4 \cap BB_4 \cap CC_4 \equiv X_{57} $ ) : Let $ D $ be the projection of $ A_1 $ on $ B_1C_1 $ and $ X $ be the midpoint of arc $ BAC $ . Since $ (B, C; A_1, A_3)=-1 $ , so $ \frac{CA_4}{A_4B}=\frac{CA_3}{A_3B}=\frac{CA_1}{A_1B} $ , hence we get $ A_4A_1 $ is the bisector of $ \angle CA_4B $ and $ X \in A_4A_1 $ . Since $ D(B, C; A_1, A_3)=-1 $ , so $ DA_1 $ is the bisector of $ \angle BDC $ , hence we get $ \triangle DC_1B \sim \triangle CB_1D $ . Since $ \frac{B_1D}{DC_1}=\frac{CD}{DB}=\frac{CA_1}{A_1B} $ , so we get $ \triangle AC_1B_1 \cap D \sim \triangle XBC \cap A_1 $ , hence from $ \angle BAD=\angle BXA_1=\angle BAA_4 $ we get $ D \in AA_4 $ . ie. $ X_{57} $ of $ \triangle ABC $ lie on $ AA_4 $ Similarly, we can prove $ X_{57} \in BB_4 $ and $ X_{57} \in CC_4 $ , so we get $ AA_4, BB_4, CC_4 $ are concurrent at the $ X_{57} $ of $ \triangle ABC $ . Q.E.D
15.12.2014 12:06
To prove $AA_4, BB_4, CC_4$ are concurrent at $X_{57}$ from my post, we can do this: Let $\ell$ be the radical axis of $\odot ABC, \odot A_1B_1C_1$. Let the medial triangle of $A_1B_1C_1$ be $A_6B_6C_6$ and note that $\ell$ is the radical axis of $\odot A_6B_6C_6$ by harmonic conjugates (fact 1). So, note $AA_5 \cap B_1C_1 \in \ell$ for obvious reasons, and because of the fact 1 we conclude $AA_5$ meets $B_1C_1$ at the same point where the orthic axis of $A_1B_1C_1$ meets $B_1C_1$, so $AA_4$ contains the foot of the $A$ altitude of $A_1B_1C_1$.
18.12.2015 10:34
This problem is really weak in the sense that we don't need $A_1,B_1,C_1$ to be the tangency points of the incircle with the sides of $\triangle{ABC}$; it is enough to have $AA_1,BB_1,CC_1$ concurrent. Let $\{X_A\}=A_1A_2\cap (ABC)$. As $(A_3,A_1,B,C)=-1$, by perspectivity from $A_2$ we get that $X_ABA_4C$ is harmonic. This yields $\dfrac{A_4B}{A_4C}=\dfrac{X_AB}{X_AC}$, but $X_AA_2$ is the bisector of $\widehat{BX_AC}$, whence $\dfrac{A_4B}{A_4C}=\dfrac{A_1B}{A_1C}$. Writing the analogous relations and multiplying them we get that $$\dfrac{A_4B}{A_4C}\cdot \dfrac{B_4C}{B_4A}\cdot \dfrac{C_4A}{C_4B}= \dfrac{A_1B}{A_1C}\cdot \dfrac{B_1C}{B_1A}\cdot \dfrac{C_1A}{C_1B}=1$$which is equivalent to the fact that $AA_4,BB_4,CC_4$ are concurrent.
23.05.2016 08:10
Let $(J_a)$ be a circle tangent to $BC$ at $A_1$ and tangent to $(ABC)$ at some point, say $T_a$. Define $J_b,J_c;T_b,T_c$ similarly. We claim that $A_4\equiv T_a$ etc. Let $X_a$ denote the midpoint of arc $BAC$ of $(ABC)$. If $\mathcal{H}$ denotes the homothety centered at $T_a$ that takes $(J_a)$ to $(ABC)$, then it takes $BC$ to the line $\ell$ parallel to $BC$ and tangent to $(ABC)$, clearly at $X_a$. Then $\mathcal{H}$ takes $A_1$ to $X_a$; so $T_a,A_1,X_a$ are collinear. Since $X_aA_2$ is a diameter of $(ABC)$, we have $X_aT_a\perp T_aA_2\implies A_1T_a\perp T_aA_2 \;(\star )$. Next, observe that $(B,C,A_1,A_3)=-1\implies T_a(B,C,A_1,A_3)$ is a harmonic pencil. But $T_aX_a$ and hence $T_aA_1$ is the bisector of $\angle BT_aC$, so we have $A_1T_a\perp T_aA_3$. Combining this with $(\star )$ gives $A_3,T_a,A_2$ are collinear, so $T_a\equiv A_4$. But from Concurrency on OI we have $AT_a$ etc. are concurrent at the Isogonal Mittenpunkt, as desired. $\blacksquare$
19.08.2016 05:37
laFiesta wrote: North Korea is NOT Korea! "Korea" means South Korea. Why is this in Korea category? I guess they is not good geography
01.03.2017 14:53
$\textbf{Proof :}$ Points $A_3, B_3, C_3$ lie on same line $l$. Consider projective transformation which sends circle $(ABC)$ into circle and line $l$ to infinity line. Name $A', B',\ldots, A_2', \ldots , A_4'$ images of points $A, B,\ldots, A_2, \ldots , A_4$ wrt this projective transformation. So we have that $A_2'A_4'|| B'C'$, $B_2'B_4'||A'C'$, $C_2'C_4'||A'B'$. Let lines $A'A_2'$, $B'B_2'$, $C'C_2'$ concurrent at point $X$. Then lines $A'A_4'$, $B'B_4'$, $C'C_4'$ concurrent at point which is isogonal to $X$ wrt $A'B'C'$. $\Box$
04.03.2017 19:04
Let $A_2A_1\cap \odot ABC=\{A_5\}$ $(A_5,A_4;B,C)=-1$ $\frac{A_5B}{A_5C}=\frac{CA_1}{BA_1}=\frac{BA_4}{CA_4}$ and hence $BA_4\cdot CB_4\cdot AC_4=A_4C\cdot AB_4 \cdot BC_4$ so we're done.
03.10.2017 18:57
27.02.2020 20:21
Nice Problem!!! North Korean TST 2013 P1 wrote: The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $ BC, CA, AB$ at $ A_1 , B_1 , C_1 $ respectively. The line $AI$ meets the circumcircle of $ABC$ at $A_2 $. The line $B_1 C_1 $ meets the line $BC$ at $A_3 $ and the line $A_2 A_3 $ meets the circumcircle of $ABC$ at $A_4 (\ne A_2 ) $. Define $B_4 , C_4 $ similarly. Prove that the lines $ AA_4 , BB_4 , CC_4 $ are concurrent. Clearly $\{A_2,B_2,C_2\}$ are the midpoints of $\widehat{BC},\widehat{AC},\widehat{AB}$ respectively and let $\{X,Y,Z\}$ be the midpoints of $\widehat{BAC},\widehat{CBA},\widehat{ACB}$ respectively. Then $(A_3,A_1;B,C)\overset{A_4}{=}(A_2,A_1A_4\cap\odot(ABC);B.C)$. So, $A_4A_1\cap\odot(ABC)= X$. Now if $P$ is the foot of perpendicular from $A_1$ to $B_1C_1$ then $AP,XA_1$ concur at $A_4$ from this problem $\longrightarrow$Two Lines meet on a circle. Analogously we get that if $\triangle PQR$ is the orthic triangle of $\triangle A_1B_1C_1$, then $\{\overline{A-P-A_4}\},\{\overline{B-Q-B_4}\},\{\overline{C-R-C_4}\}$. Now as $\{AA_1,BB_1,CC_1\}$ and $\{A_1P.B_1Q,C_1R\}$. Hence by Cevian Nest Theorem we conclude that $AA_4,BB_4,CC_4$ are concurrent. $\blacksquare$
27.02.2020 20:52
Suppose that $A_2A_1$ intersects the circle again at $A'$, and $A_1A_4$ intersects at $A_5$. We have $$(A_3A_1;BC)\stackrel{A_2}{=}(A_4A';BC)\stackrel{A_1}{=}(A_5A_2;BC)=-1$$Thus, $A_4A_5$ is an angle bisector, and $\frac{\sin\angle BAA_4}{\sin\angle CAA_4}=\frac{BA_4}{CA_4}=\frac{BA_1}{A_1C}$. Multiplying similar expressions for $B_4,C_4$, we are done by Trig Ceva and Ceva on the Gergonne point.
15.02.2021 05:01
[asy][asy] import graph; size(15cm); real a=105; real b=205; real c=335; real r=10; path omega=circle(origin, r); pair A=r*dir(a); pair B=r*dir(b); pair C=r*dir(c); pair I=incenter(A,B,C); pair A1=foot(I,B,C); pair B1=foot(I,C,A); pair C1=foot(I,B,A); dot(A1); dot(B1); dot(C1); draw(A--B--C--cycle); draw(omega,blue); pair X=10*C1-9*B1; pair P=10*B-9*C; pair A3=intersectionpoints(B1--X,B--P)[0]; pair A2=r*dir((b+c)/2); pair B2=r*dir((a+c)/2+180); pair C2=r*dir((a+b)/2); dot(A2); dot(B2); dot(C2); pair A4=intersectionpoints(A2--A3,omega)[1]; draw(A2--A3); dot(A4); draw(B1--A3); draw(A3--A2, magenta); pair D=(B+C)/2; dot(D); draw(B--A3); draw(A1--A4--A2--D--cycle,red+ linewidth(1)); draw(A--A4,green); label("$A$",A,dir(90)); label("$B$",B,dir(-135)); label("$C$",C,dir(-45)); label("$D$",D,dir(-45)); label("$A_1$",A1,dir(90)); label("$A_2$",A2,dir(-90)); label("$A_3$",A3,dir(-90)); label("$A_4$",A4,dir(-90)); label("$B_1$",B1,dir(45)); label("$C_1$",C1,dir(-45)); draw(B--A4--C,dotted); label("$B_2$",B2,dir(45)); label("$C_2$",C2,dir(135)); [/asy][/asy] As mentioned earlier, $A_1,B_1,C_1$ may be any points on their respective sides, not necessarily the touch points, as long as $AA_1,BB_1,CC_1$ are concurrent. First, note that it is well-known that $(A_3A_1;BC)=-1$, and let $D$ be the midpoint of $\overline{BC}$. By EGMO lemma $9.17$(midpoints of harmonic bundles) combined with PoP, $A_3A_4\cdot A_3A_2=A_3B\cdot A_3C=A_3A_1\cdot A_3D$, thus $A_1A_4A_2D$ is cyclic. Because $\angle A_1DA_2=\pi/2$, this implies $\angle A_1A_4A_2=\pi/2$. By another well-known property of harmonic bundles, $\overline{A_4A_2}$ bisects $\angle BA_4C$. By angle bisector theorem and law of sines, we have $\prod_{\text{cyc}}^{}\frac{\sin\angle BAA_4}{\sin\angle A_4AC}=\prod_{\text{cyc}}^{}\frac{2R\sin\angle BAA_4}{2R\sin\angle A_4AC}=\prod_{\text{cyc}}^{}\frac{BA_4}{A_4C}=\prod_{\text{cyc}}^{}\frac{BA_1}{A_1C}$(angles and lengths directed, $R$ is the radius of $(ABC)$) Because $AA_1,BB_1,CC_1$ concur(at the Gergonne point for the specific case of the problem), the last cyclic product is $1$ by Ceva's theorem. Because the first cyclic product is 1, by trig Ceva, the lines $AA_4,BB_4,CC_4$ are concurrent, as desired.
25.05.2021 23:15
laFiesta wrote: North Korea is NOT Korea! "Korea" means South Korea. Why is this in Korea category? Why? North Korea is still Korea. It encompasses the entire peninsula, should it not?
25.05.2021 23:49
Let $D_A = AA_4 \cap BC$ , $S = (AC_1IB_1) \cap (ABC)$ and $E = AS\cap BC$. It is well-known that $S,A_1,A_2$ are collinear. Now we have $-1 =(A_3,A_1;B,C) \stackrel{A_2} = (A_4,S;B,C) \stackrel{A} = (D_A,E;B,C)$. By radical axis theorem on $(ABC) , (BIC) , (ASC_1IB_1)$ we get that $SI$ is tangent to both $(BIC)$ and $(ASC_1IB_1)$. Hence, we have $\triangle EIB \sim \triangle ECI$. So, $\frac{EB}{EI} = \frac{EI}{EC} = \frac{IB}{IC} \implies \frac{EB}{EC} = \frac{IB^2}{IC^2}$. So, $-1 = (D_A,E;B,C) \implies \frac{D_AB}{D_AC} = \frac{EB}{EC} = \frac{IB^2}{IC^2}$. Writing up similar expressions for the others, we get that $AA_4 , BB_4 , CC_4$ are concurrent by the converse of Ceva's Theorem. $\square$
18.09.2021 22:02
By Ceva-Menelaus, we have $-1 = (A_3, A_1; B, C)$. Now, let $A_5$ denote the midpoint of arc $BAC$. Because $$\angle A_3A_4A_5 = 180^{\circ} - \angle A_2A_4A_5 = 90^{\circ}$$and $A_4A_5$ bisects $\angle BA_4C$, the Right Angle and Bisectors Lemma implies $$- 1 = (A_3, A_4A_5 \cap BC ; B, C)$$so $A_1, A_4, A_5$ are collinear. Let $P, Q, R$ be the feet of the $A_1$-altitude, $B_1$-altitude, $C_1$-altitude respectively of $A_1B_1C_1$. A well-known lemma implies $P \in AA_4, Q \in BB_4, R \in CC_4$. Now, let $H$ denote the orthocenter of $A_1B_1C_1$. Applying the Cevian Nest Lemma on the Gergonne Point of $ABC$ and $H$ implies $AP, BQ, CR$ are concurrent, which finishes. $\blacksquare$ Remark: The well-known lemma I cited is proven in many solutions to TSTST 2020/2.
14.04.2023 18:45
Let $X=(AI)\cap (ABC)$. So by spiral similarity we have $\triangle XBC_1\sim\triangle XCB_1$. So we get $XB/XC=BA_1/CA_1$. Hence $A_1$ lies on the bisector of $\angle BXC$, and that gives $X$, $A_1$, $A_2$ are collinear. So \[ -1=(BC;A_3A_1)\stackrel{A_2}{=}(BC;A_4X)\implies \frac{BX}{CX}=\frac{BA_4}{CA_4} \]Now we have \[ \frac{\sin \angle BAA_4}{\sin \angle CAA_4}=\frac{\sin \angle BCA_4}{\sin \angle CBA_4}=\frac{BA_4}{CA_4}=\frac{BX}{CX}=\frac{BA_1}{CA_1} \]Now by sine ceva we are done.
10.06.2024 15:37
Excellent Problem! Kim Jong Un’s math problems never disappoints me. Step 1 Redefine $A_4,B_4,C_4$ Note that $AA_1,BB_1,CC_1$ are concurrent $\implies$ $(A_3,A_1,B,C)=-1$ Let $M$ be a midpoint of segment $BC$; $A_3A_1\cdot A_3M=A_3B\cdot A_3C=A_3A_4\cdot A_3A_2$. We conclude that $A_1,M,A_2,A_4$ are concyclic which means $\angle A_1A_4A_2= 90^\circ$. So, $A_4,A_1,A’$ are collinear where $A’$ is a midpoint of arc $BC$ containing $A$. Similarly, $B_4,B_1,B’$ and $C_1,C_4,C’$ are collinear where $B’$ is a midpoint of arc $CA$ containing $B$ and $C’$ is a midpoint of arc $AB$ containing $C$. Step 2 Finish by simple trig-ceva calculation Observe that $A_1A_4$ bisects $\angle BA_4C$ $\implies \frac{BA_4}{A_4C}=\frac{BA_1}{A_1C}$ Apply law of sines yields $\frac{sin BAA_4}{sin A_4AC}=\frac{BA_4}{A_4C}=\frac{BA_1}{A_1C}$ Apply trigonometric ceva’s theorem to $\triangle ABC$ yields $AA_4, BB_4$ and $CC_4$ are concurrent, as desired.
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18.06.2024 22:56
bored In accordance with the absolutely outrageous point naming, let $A_5$ be the $A$-sharkydevil point, and define $B_5$, $C_5$ similarly. Now $A_2$, $A_1$, $A_5$ collinear, so \[ (A_3, A_1; B, C) \overset{A_2}{=} (A_4, A_5; B, C). \]Thus, if $A_6$ is the foot of the altitude from $A_1$ to $\overline{B_1C_1}$, then $\overline{AA_6A_4}$ collinear (this is well-known), and Cevian Nest finishes.
26.10.2024 03:50
laFiesta wrote: North Korea is NOT Korea! "Korea" means South Korea. Why is this in Korea category? DPRK is the only Korea. What do you mean South Korea ????
19.12.2024 21:49
Let $A_5$ be the $A$-Sharkydevil point, or $(AB_1C_1) \cap (ABC)$. Inverting around the incircle yields that $A$ goes to the midpoint of $B_1C_1$ and cyclic variants, implying that $(ABC)$ goes to the nine point circle of $A_1B_1C_1$, since the image of $A_5$ lies on $B_1C_1$ and is not $A$, we know that it is the foot of the altitude from $A_1$ to $B_1C_1$ (calling it $A_6$), giving the result that $A_5, A_6,I$ are collinear. By spiral similarity, we know that $A_5I$ maps to $A_5A_2$, and $B_1C_1$ goes to $BC$, and the $A_6$ goes to $A_1$, since $A_6A_1 \parallel IA_2$ and $A_1$ lies on $B_1C_1$, so $A_5, A_1, A_2$ are collinear. Since $AA_1, BB_1, CC_1$ are concurrenct by Ceva, harmonic lemmas give $(A_3, A_1; B,C)$ harmonic, projecting over $A_2$ gives $(A_4, A_5; B,C)$ harmonic, projecting through $A$ gives $(A_5A \cap B_1C_1, A_4A \cap B_1C_1;B_1,C_1)$ harmonic, projecting through $A_5$ gives $(A, (A_4A \cap B_1C_1)A_5 \cap (AI); B_1, C_1)$ harmonic, so $(A_4A \cap B_1C_1), A_5$ are collinear with $I$, so $A_4A \cap B_1C_1 = A_6$. By Cevian Nest, it is obvious that $AA_6, BB_6, CC_6$, concur, so we are done.