Let $ABC$ be an acute, non-isosceles triangle which is inscribed in a circle $(O)$. A point $I$ belongs to the segment $BC$. Denote by $H$ and $K$ the projections of $I$ on $AB$ and $AC$, respectively. Suppose that the line $HK $ intersects $(O)$ at $M, N$ ($H$ is between $M, K$ and $K$ is between $H, N$). Let $X, Y$ be the centers of the circles $(ABK),(ACH)$ respectively. Prove the following assertions: a) If $I$ is the projection of $A$ on $BC$, then $A$ is the center of circle $(IMN)$. b) If $XY\parallel BC$, then the orthocenter of $XOY$ is the midpoint of $IO$.
Problem
Source: 2016 Saudi Arabia GMO TST level 4+, III p1
Tags: geometry, orthocenter, midpoint, parallel, Circumcenter
Inconsistent
02.08.2020 08:05
For part A, consider the inversion $\psi(A,AI^2)$, by similar triangles $\psi(C)=K, \psi(B)=H$ so $\psi((ABC))=HK$, $\psi(HK)=(ABC)$, thus since ${M,N}=(ABC)\cap HK$, $\psi(M)=M, \psi(N)=N$, so $AM=AN=AI$ and A is the center of $(IMN)$ As for part B, construct perpendicular bisectors of AH, AB, AK, AC, note that extending perpendicular lines to AC and AB at A to intersect BC at B' and C' gives ratio $\frac{AB'}{AC'}=\frac{\frac{b\sin(\angle C)}{\cos(\angle C)}}{\frac{c\sin(\angle B)}{\cos(\angle B)}}=\frac{\cos(\angle B)}{\cos(\angle C)}$ by the law of sines, but since the intersections of the aforementioned perpendicular bisectors for a parallelogram with vector sides with direction equal to the perpendiculars off AB and AC, and since the vector,XY has the same direction as vector BC, let the p.b. of AH intersect the p.b. of AK at Z, then by proportionality, we know $\frac{ZX}{ZY}=\frac{\cos(\angle C)}{\cos(\angle B)}$, but if $\theta = \angle (M(A,K))ZY$, then $\frac{\cos(\angle C)}{\cos(\angle B)}=\frac{ZX}{ZY} = \frac{1/2\sin(\theta)CK}{1/2\sin(\theta)BH}=\frac{CI}{BI} \frac{\cos(\angle C)}{\cos(\angle B)}$, thus BI=CI, so I must be the midpoint of BC. Now that we know I, the rest should be trivial. Consider Homothety $\mathcal{H}(O,2)$, since the midpoint of A'K' has distance $\frac{b+a\cos(\angle C)}{2}$ along side AC from C, and the projection of the midpoint of AB onto AC has distance $\frac{c\cos(\angle A)}{2}+a\cos(\angle C)$ from C, using the identity $a\cos(\angle C)+c\cos(\angle A)=b$ from dropping the altitude in a triangle, we know these values are the same, and thus the image of the perpendicular bisector of AK intersects AB at the midpoint of AB, but since the perpendicular bisector of AB passed through O, it is fixed in the homothety and also intersects line AB at the midpoint of AB, thus the image of X is Mid(AB), the image of Y is Mid(AC), and the image of the midpoint of OI goes to I = Mid(BC). Since O is the orthocenter of the medial triangle ($OI \perp BC$, etc.), I is the orthocenter of the triangle of the midpoints of AB, AC, and O ($OI \perp X'Y', X'I \perp OY', Y'I \perp OX'$), and thus applying the inverse homothety $\mathcal{H}(O,\frac{1}{2})$, this maps to the fact that the midpoint of OI is the orthocenter of XOY since homothety preserves angles, and we are done. OOOHHHH BOY, I did my first real oly geo!!!!!!!!!!!!!!! <3 <3 <3 <3 <3 (I don't care if it's P1, shush)(It's might be trivial but I don't care)