Let $\triangle{ABC}$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Incircle $(I)$ of the $\triangle{ABC}$ is tangent to the sides $BC,CA,AB$ at $M,N,P$ respectively. Denote $\Omega_A$ to be the circle passing through point $A$, external tangent to $(I)$ at $A'$ and cut again $AB,AC$ at $A_b,A_c$ respectively. The circles $\Omega_B,\Omega_C$ and points $B',B_a,B_c,C',C_a,C_b$ are defined similarly. $a)$ Prove $B_cC_b+C_aA_c+A_bB_a \ge NP+PM+MN$. $b)$ Suppose $A',B',C'$ lie on $AM,BN,CP$ respectively. Denote $K$ as the circumcenter of the triangle formed by lines $A_bA_c,B_cB_a,C_aC_b.$ Prove $OH//IK$.
Problem
Source: VMO 2023 day 2 P7
Tags: vmo, geometry, homothety, vmo 2023, Inequality, mixtilinear, circumcircle
25.02.2023 09:49
$a)$ Easy to see $(I)$ is the $A-$mixtilinear circle (outside) $\implies$ $\triangle{DA_bP},\triangle{A_cA_bD},\triangle{A_cDN}$ are similar ($D$ midpoint $PN$ so $D$ is the excenter) From here using ratios and $AM-GM$ for $2$ variable we receive $A_bP+A_cN \ge NP$. Simiarly ,sum them up we get the result. $Q.E.D$ $b)$ Denote the triangle formed by 3 lines is $\delta$. Simple angle chasing gives $BA_bA_cC,AB_aB_cC,AC_aC_bB$ are cyclic $\implies$ $\delta$ is homothetic to the orthic triangle $\psi$. If $A_bA_c \cap B_aB_c$ at $L$ $\implies$ $\triangle{LA_bB_a}$ isosceles. Our mission now is to prove $PA_b=PB_a$ (Important part in the problem) $Z$ midpoint $PM$ By similar triangles $A_bP=PD\frac{A_bD}{DA_c}$ $B_aP=PZ\frac{B_aZ}{B_cZ}$ Construct the extriangle $\triangle{UVT}$ Easy to see $\triangle{UCB} \sim \triangle{DA_bA_c}$ and similar to other triangles. We need to show $\frac{PD}{PZ}\frac{A_bD}{DA_c}=\frac{B_aZ}{B_cZ}<=>\frac{PN}{PM}\frac{CU}{BU}=\frac{VC}{VA}$ $\triangle{MNP} \sim \triangle{UBC} \sim \triangle{AVC} \sim \triangle{ABT}$. Plug in the ratios $\implies$ $P$ midpoint $A_bB_a$. Similarly so $I$ is also the incenter of $\delta$. Observe $H$ is the incenter of the orthic circle $\psi$. $K,O_a$ are circumcenters of $\delta,\psi$($O_a$ is the nine-point circle center and it is well known that $\overline{O,O_a,H}$). By homothety, we are done. $Q.E.D$
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25.02.2023 18:45
Comment: A bit better than P4, but still tricky to find the solution, especially in the examination room. Also, someone please add these problem to the problem collection soon.
25.02.2023 18:51
Extra problem on this configuration (Old) (Teacher Nguyen Van Linh) $J$ is the radical center of $\Omega_A,\Omega_B,\Omega_C$. $G_e$ is the Gergonne point $\implies$ $\overline{G_e,I,J}$.
25.02.2023 19:34
Paramizo_Dicrominique wrote: $a)$ Prove $B_cC_b+C_aA_c+A_bB_a \ge NP+PM+MN$. It appears that $B_cC_b=B_cM+C_bM={\left[r\cot\frac{B}2-{\left(r\csc\frac{B}2-r\right)}\right]}+{\left[r\cot\frac{C}2-{\left(r\csc\frac{C}2-r\right)}\right]}$ and $NP=2r\cos\frac{A}2$; isn't it \(B_cC_b+C_aA_c+A_bB_a \le NP+PM+MN\)?
25.02.2023 21:12
b) Redefine $A_b, A_c$ as $AA_b \cdot AB=AA_c \cdot AC=AA' \cdot AM$. Then by inversion with radius $AP=AN$ we get $AA_bA'A_c$ is tangent to $(I)$. Thus $BA_bA_cC$ is cyclic which implies orthic triangle $DEF$ and $\Delta$ are homothetic (parallel sides) where $\Delta$ is the triangle formed by the lines $A_bA_c,B_cB_a,C_aC_b$. $OH$ is the $"OI"$ line of $DEF$ and $IK$ is the $"OI"$ line of $\Delta$ which gives the result.