Let $ABC$ be a triangle. Construct rectangles $BA_1A_2C$, $CB_1B_2A$, and $AC_1C_2B$ outside $ABC$ such that $\angle BCA_1=\angle CAB_1=\angle ABC_1$. Let $A_1B_2$ and $A_2C_1$ intersect at $A'$ and define $B',C'$ similarly. Prove that line $AA'$ bisects $B'C'$. Linus Tang
Problem
Source: ELMO Shortlist 2024 G7
Tags: geometry, rectangle
22.06.2024 18:59
23.06.2024 02:08
Let $k$ be imaginary, and $a_1=b+(c-b)k$, $a_2=c+(c-b)k$, $b_1=c+(a-c)k$, $b_2=a+(a-c)k$, $c_1=a+(b-a)k$, $c_2=b+(b-a)k$. Since $b_2-a_1=a-b+(a+b-2c)k$, triangle $A'B'C'$ is homothetic to the triangle $-a+(c-b)k$, $-b+(a-c)k$, $-c+(b-a)k$, so the median has slope $2(-a+(c-b)k)-(-b+(a-c)k)-(-c+(b-a)k)=-2a+b+c+3(c-b)k$. We will show that the intersection of the line through $a$ and $a+(-2a+b+c+3(c-b)k)$ intersects $A_2C_1$ at a point that is the same after swapping $b$ and $c$ and multiplying $k$ by $-1$. Assume without loss of generality $a=0$, so we need to find the intersection of the line through $0$ and $b+c+3(c-b)k$ and the line through $c+(c-b)k$ and $bk$. The intersection is \begin{align*} \frac{(\overline a(b+c+3(c-b)k)-a\overline{(b+c+3(c-b)k)})(a_2-c_1)-(a-(b+c+3(c-b)k))(\overline a_2c_1-a_2\overline c_1)}{(\overline a-\overline{(b+c+3(c-b)k)})(a_2-c_1)-(a-(b+c+3(c-b)k))(\overline a_2-\overline c_1)}&=\frac{(b+c+3(c-b)k)(\overline{(c+(c-b)k)}bk-(c+(c-b)k)\overline{bk})}{-\overline{(b+c+3(c-b)k)}(c+(c-2b)k)-(b+c+3(c-b)k)(\overline c-(\overline c-2\overline b)k)}\\ &=\frac{(b+c+3(c-b)k)\left((\overline c-(\overline c-\overline b)k)bk+(c+(c-b)k)\overline bk\right)}{-(\overline b+\overline c-3(\overline c-\overline b)k)(c+(c-2b)k)+(b+c+3(c-b)k)(\overline c-(\overline c-2\overline b)k)}\\ &=\frac{(b+c+3(c-b)k)k(k(-b\overline c+\overline bc)+b\overline c+\overline bc)}{k^2(3\overline bc-3b\overline c)+k(4c\overline c-2\overline bc-2b\overline c+4b\overline b)-\overline bc+b\overline c}. \end{align*} This is the same after swapping $b$ and $c$ and multiplying $k$ by $-1$, which implies that this point is $A'$ and lies on the line through $A$ parallel to the line through $A'$ and the midpoint of $B'C'$, so $AA'$ bisects $B'C'$.
23.06.2024 02:30
We define a notion of "distance" in the direction of a line as follows: for any non-parallel lines $\ell_1$ and $\ell_2$ and a point $P$, let $f(P,\ell_1,\ell_2)$ denote $\overrightarrow{PX}$, where $X$ is the point on $\ell_1$ such that $\overline{PX} \parallel \ell_2$. Let $O_A$, $O_B$, and $O_C$ be the centers of $BA_1A_2C$, $CB_1B_2A$, and $AC_1C_2B$, respectively. Notice that a homothety of scale factor $2$ at $A$ sends $\overline{O_BO_C}$ to $\overline{B_1C_2}$, and symmetric statements apply to $B$ and $C$. It suffices to show $f(A,\overline{A_1B_2},\overline{O_BO_C})=-f(A,\overline{A_2C_1},\overline{O_BO_C})$, as this shows $\overline{A'A}$ is the $A'$-median of $A'B'C'$. Since $f(A,\overline{O_AO_C},\overline{O_BO_C})=-f(B,\overline{O_AO_C},\overline{O_BO_C})$ and similar, it suffices to show \[f(A,\overline{O_AO_C},\overline{O_BO_C})+f(B,\overline{O_AO_C},\overline{O_BO_C})=-f(A,\overline{O_AO_B},\overline{O_BO_C})-f(C,\overline{O_AO_B},\overline{O_BO_C})\]Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively. Then, it suffices to show \[f(M,\overline{O_AO_C},\overline{O_BO_C})=-f(N,\overline{O_AO_B},\overline{O_BO_C}).\]Let the lines parallel to $\overline{O_BO_C}$ through $M$ and $N$ intersect $\overline{O_AO_C}$ and $\overline{O_AO_B}$ at $P$ and $Q$, respectively, and let $O$ be the circumcenter of $ABC$. Then, we have \begin{align*} \frac{MP}{NQ}&=\frac{MO_C \cdot \frac{\sin \angle MO_CP}{\sin \angle MPO_C}}{NO_B \cdot \frac{\sin \angle NO_BQ}{\sin \angle NQO_B}}=\frac{AB \cdot \frac{\sin \angle OO_CO_A}{\sin \angle O_AO_CO_B}}{AC \cdot \frac{\sin \angle OO_BO_A}{\sin \angle O_AO_BO_C}}=\frac{\sin \angle ACB}{\sin \angle ABC} \cdot \frac{\sin \angle OO_CO_A}{\sin \angle OO_BO_A} \cdot \frac{O_AO_C}{O_AO_B} \\ &=\frac{\sin \angle O_AOO_B}{\sin \angle O_AOO_C} \cdot \frac{\sin \angle OO_CO_A}{\sin \angle OO_BO_A} \cdot \frac{O_AO_C}{O_AO_B}=\frac{\frac{\sin \angle O_AOO_B}{O_AO_B \sin \angle OO_BO_A}}{\frac{\sin \angle O_AOO_C}{O_AO_C \sin \angle OO_CO_A}}=\frac{OO_A}{OO_A}=1, \end{align*}as desired. $\square$
12.11.2024 17:44
This problem was proposed by Kostas Vittas since 2011 for Mathley of Vietnam, see here.