Given that $ABC$ is a triangle, points $A_i, B_i, C_i \hspace{0.15cm} (i \in \{1,2,3\})$ and $O_A, O_B, O_C$ satisfy the following criteria: a) $ABB_1A_2, BCC_1B_2, CAA_1C_2$ are rectangles not containing any interior points of the triangle $ABC$, b) $\displaystyle \frac{AB}{BB_1} = \frac{BC}{CC_1} = \frac{CA}{AA_1}$, c) $AA_1A_3A_2, BB_1B_3B_2, CC_1C_3C_2$ are parallelograms, and d) $O_A$ is the centroid of rectangle $BCC_1B_2$, $O_B$ is the centroid of rectangle $CAA_1C_2$, and $O_C$ is the centroid of rectangle $ABB_1A_2$. Prove that $A_3O_A, B_3O_B,$ and $C_3O_C$ concur at a point. Proposed by Farras Mohammad Hibban Faddila
Problem
Source: Indonesian Stage 1 TST for IMO 2022, Test 2 (Geometry)
Tags: Triangle, geometry, midpoint, concurrence
cadaeibf
25.12.2021 14:45
Let $M_A,M_B,M_C$ be the midpoints of $BC,CA,AB$ respectively, let $G$ be the centroid of $ABC$, let $K$ be the symmedian point of $ABC$, and $K_A,K_B,K_C$ its projections on $BC,CA,AB$. Since the barycentric coordinates of $K$ are $(a^2:b^2:c^2)$ it follows that $|\vec{KK_A}|:BC=|\vec{KK_B}|:CA=|\vec{KK_C}|:AB$ and so there exists a (positive) real number $\lambda$ such that $\vec{M_AO_A}=\lambda\vec{KK_A}$ and similarly for the other sides. Now consider the point $X$ which partitions $A_3O_A$ in a ratio of $2:1$. This point must be (using vectors) $$\vec{X}=\frac{2\vec{O_A}+\vec{A_3}}{3}$$$$=\frac{2\vec{M_A}+\vec{A}}{3}+\frac{\vec{AA_1}+\vec{AA_3}+2\vec{M_AO_A}}{3}$$$$=\vec{G}+\frac{2\lambda}{3}(\vec{KK_A}+\vec{KK_B}+\vec{KK_C})$$Since this expression is symmetric, it belongs to all three lines.
Seicchi28
27.12.2021 20:23
The number of participants that got this one correct during the test is lower than expected... I think this is because of the format of the test, which consists of all the four topics ACGN, and is not ordered by difficulty. Anyway here are some solutions.
[asy][asy]
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(130), B = dir(210), C = dir(-30);
pair A2 = rotate(-90, A) * ((B + 2 * A)/3);
pair B1 = rotate(90, B) * ((A + 2 * B)/3);
pair B2 = rotate(-90, B) * ((C + 2 * B)/3);
pair C1 = rotate(90, C) * ((B + 2 * C)/3);
pair C2 = rotate(-90, C) * ((A + 2 * C)/3);
pair A1 = rotate(90, A) * ((C + 2 * A)/3);
pair A3 = A1 + A2 - A, B3 = B1 + B2 - B, C3 = C1 + C2 - C;
pair Oa = (B + C1)/2, Ob = (C + A1)/2, Oc = (A + B1)/2;
pair H = orthocenter(A, B, C);
pair O = circumcenter(A, B, C);
pair G = (A+B+C)/3;
pair Ma = (B + C)/2;
draw(A--A2--B1--B--cycle^^A2--A3^^B1--B3, blue+linewidth(1));
draw(B--B2--C1--C--cycle^^B2--B3^^C1--C3, deepcyan+linewidth(1));
draw(C--C2--A1--A--cycle^^C2--C3^^A1--A3, heavymagenta+linewidth(1));
draw(H--O, orange+linewidth(1));
draw(A3--Oa, black+dashed);
draw(O--Oa^^A3--H, orange+linewidth(1));
string[] names = {"$A$", "$B$", "$C$", "$A_1$", "$A_2$", "$B_1$", "$B_2$", "$C_1$", "$C_2$",
"$A_3$", "$B_3$", "$C_3$", "$O_A$", "$O_B$", "$O_C$", "$H$", "$O$", "$G$", "$M_A$"
};
pair[] points = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc, H, O, G, Ma};
pair[] ll = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc, H, O, G, Ma};
for (int i=0; i<names.length; ++i)
dot(names[i], points[i], dir(ll[i]));
[/asy][/asy]
We claim the desired concurrency point is point $G$, the centroid of $\bigtriangleup ABC$. It suffices to show that $A_3O_A$ passes through $G$, and the rest will be similar.
Let $M_A$ be the midpoint of $BC$. It is well-known that $AH=2OM_A$.
Since $\angle AA_2A_3 = 180^{\circ} - \angle A_2AA_1 = 180^{\circ} - (360^{\circ} - 90^{\circ} - 90^{\circ} - \angle A) = \angle A$, and $\frac{AA_2}{A_2A_3} = \frac{AA_2}{AA_1} = \frac{AB}{AC}$, we have $\bigtriangleup AA_2A_3 \sim \bigtriangleup BAC$ (and denote $k$ as their similarity ratio, i.e. $k = \frac{AA_2}{AB}$). Therefore, $$\angle A_3AA_2 + \angle A_2AB + \angle BAH = \angle B + 90^{\circ} + (90^{\circ} - \angle B) = 180^{\circ},$$$\rightarrow A_3, A, H$ is collinear, and $AA_3 = k \cdot BC = BB_2 = 2 * M_AO_A$. Therefore, $A_3H = 2 * OO_A$, so $A_3OA$ meets $OH$ at $G$, as desired. $\blacksquare$
[asy][asy]
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(130), B = dir(210), C = dir(-30);
pair A2 = rotate(-90, A) * ((B + 2 * A)/3);
pair B1 = rotate(90, B) * ((A + 2 * B)/3);
pair B2 = rotate(-90, B) * ((C + 2 * B)/3);
pair C1 = rotate(90, C) * ((B + 2 * C)/3);
pair C2 = rotate(-90, C) * ((A + 2 * C)/3);
pair A1 = rotate(90, A) * ((C + 2 * A)/3);
pair A3 = A1 + A2 - A, B3 = B1 + B2 - B, C3 = C1 + C2 - C;
pair Oa = (B + C1)/2, Ob = (C + A1)/2, Oc = (A + B1)/2;
draw(A--A2--B1--B--cycle^^A2--A3^^B1--B3, blue+linewidth(1));
draw(B--B2--C1--C--cycle^^B2--B3^^C1--C3, deepcyan+linewidth(1));
draw(C--C2--A1--A--cycle^^C2--C3^^A1--A3, heavymagenta+linewidth(1));
draw(A3--Oa^^B3--Ob^^C3--Oc, grey+dashed);
draw(A3--B3--C3--cycle, black+dashed);
string[] names = {"$A$", "$B$", "$C$", "$A_1$", "$A_2$", "$B_1$", "$B_2$", "$C_1$", "$C_2$",
"$A_3$", "$B_3$", "$C_3$", "$O_A$", "$O_B$", "$O_C$"
};
pair[] points = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc};
pair[] ll = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc};
for (int i=0; i<names.length; ++i)
dot(names[i], points[i], dir(ll[i]));
[/asy][/asy]
The claim here is that $B_3, O_A,$ and $C_3$ are collinear so that the three lines will concur at the centroid of $\bigtriangleup A_3B_3C_3$.
One way to prove this is by noting that $BB_3C_1C_3$ is a parallelogram, since both $C_1C_3$ and $BB_3$ are perpendicular to $AC$ (proving $BB_3 \perp AC$ is already discussed in the first solution above), and $BB_3 = C_1C_3 = k \cdot AC$, where $k = \frac{AC}{CC_2}$. Therefore, $O_A$ is the midpoint of $B_3C_3$ as well, as desired. $\blacksquare$
[asy][asy]
size(8cm);
defaultpen(fontsize(9pt));
pair A = dir(130), B = dir(210), C = dir(-30);
pair A2 = rotate(-90, A) * ((B + A)/2);
pair B1 = rotate(90, B) * ((A + B)/2);
pair B2 = rotate(-90, B) * ((C + 2 * B)/3);
pair C1 = rotate(90, C) * ((B + 2 * C)/3);
pair C2 = rotate(-90, C) * ((A + 7 * C)/8);
pair A1 = rotate(90, A) * ((C + 7 * A)/8);
pair A3 = A1 + A2 - A, B3 = B1 + B2 - B, C3 = C1 + C2 - C;
pair Oa = (B + C1)/2, Ob = (C + A1)/2, Oc = (A + B1)/2;
draw(A--A2--B1--B--cycle^^A2--A3^^B1--B3, blue+linewidth(1));
draw(B--B2--C1--C--cycle^^B2--B3^^C1--C3, deepcyan+linewidth(1));
draw(C--C2--A1--A--cycle^^C2--C3^^A1--A3, heavymagenta+linewidth(1));
draw(A--B1^^A--C2^^Ob--Oc^^C2--B1^^C3--B3, black+dashed);
string[] names = {"$A$", "$B$", "$C$", "$A_1$", "$A_2$", "$B_1$", "$B_2$", "$C_1$", "$C_2$",
"$A_3$", "$B_3$", "$C_3$", "$O_A$", "$O_B$", "$O_C$"
};
pair[] points = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc};
pair[] ll = {A, B, C, A1, A2, B1, B2, C1, C2, A3, B3, C3, Oa, Ob, Oc};
for (int i=0; i<names.length; ++i)
dot(names[i], points[i], dir(ll[i]));
[/asy][/asy]
Using this solution, we can generalize the problem by removing condition (b). Here we will show $\bigtriangleup A_3B_3C_3$ and $\bigtriangleup O_AO_BO_C$ are homothetic, so the three lines concur at the centre of the homothety.
FIrst, $O_CO_B || B_1C_2$ since $O_CO_B$ is a midline of $\bigtriangleup AB_1C_2$.
Second, it's easy to see that $B_1C_2C_3B_3$ is a parallelogram. So, $B_1C_2 || B_3C_3$.
Therefore, $O_CO_B || B_3C_3$. Similarly, $O_AO_B || A_3B_3$ and $O_AO_C || A_3C_3$. Therefore the two triangles $\bigtriangleup A_3B_3C_3$ and $\bigtriangleup O_AO_BO_C$ are homothetic as desired. $\blacksquare$