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
MarkBcc168
22.06.2024 18:59
Let $D, E, F$ be the centers of the three rectangles, so $\triangle BDC$, $\triangle CEA$, $\triangle AFB$ are similar isosceles triangles. Notice $EF\parallel B_1C_2$ and similar, so the problem is equivalent to $AA'$ is parallel to median of $\triangle DEF$.
Let $O$, $G$, $H$ be the circumcenter, centroid, and orthocenter. Recall that $AD$, $BE$, $CF$ concur at point $P$ on the Kierpert's Hyperbola of $\triangle ABC$, which is the (rectangular) hyperbola $\mathcal H$ passing through $A$, $B$, $C$, $H$, $G$. We have the following lemma.
Lemma: The perpendiculars from $A$, $B$, $C$ to $EF$, $FD$, $DE$ concur at point $T\in\mathcal H$. Moreover, $O\in TP$.
Proof. The part that $O\in TP$ follows from Sondat's theorem. For the first part, note that $\mathcal H$ can be classified as the locus of point $U$ such that lines from $D$, $E$, $F$ parallel to $AU$, $BU$, $CU$ are concurrent. (This is also the conic used to prove Sondat's theorem). $\blacksquare$
Now, let $X$ be on orthocenter of $\triangle BTC$, so $X\in\mathcal H$ since $\mathcal H$ is rectangular. Note $BX\parallel DE$ and $CX\parallel DF$. Pascal on $GXTPAH$ gives $D\in GX$. Moreover, recall a well-known fact that $\triangle ABC$ and $\triangle DEF$ share the same centroid, so $DX$ bisects $EF$.
Finally, it suffices to show that $DX\parallel AA'$. To that end, we let $M$ be the midpoint of $BC$. Let $X'$ and $D'$ be the reflections of $X$ and $D$ across $M$. Since lines $BX', DF, A_2C_1$ are equally spaced and lines $CX', DE, A_1B_2$ are equally spaced, we have that $D$ is the midpoint of $A'X'$. Now, observe that
$X'D'\parallel DX$
midpoint of $D'A$ lies on $DX$ (recall: $AFD'E$ is parallelogram)
midpoint of $A'X'$ is $D$.
These three items give that $AA'$, $XD$, $D'X'$ are parallel and equally spaced, so we are done.
Let $\omega_A$ be the circumcircle of $BA_1A_2C$, and define $\omega_B$ and $\omega_C$ similarly. Let $R$ be the radical center of $\omega_A$, $\omega_B$, and $\omega_C$. First, we claim that $AR\perp B'C'$. Let $\omega_B$ and $\omega_C$ meet again at $X\neq A$. Then $\angle AXB_1=\angle AXC_2=90^\circ$, so $X$ is the foot from $A$ to $B_1C_2$. But both $B',C'$ lie on $B_1C_2$ and $AX$ is the radical axis of $\omega_B, \omega_C$, so $R$ lies on $AX$, proving the claim. Note that the same logic works for $BR$ and $CR$.
[asy][asy]
size(7cm);
defaultpen(fontsize(10pt));
import olympiad;
import geometry;
real t = 0.3;
real t = 0.3;
pair A = (1,4);
pair B = (0,0);
pair C = (4,0);
pair A_1 = B + t*dir(-90)*(C-B);
pair A_2 = C + t*dir(90)*(B-C);
pair B_1 = C + t*dir(-90)*(A-C);
pair B_2 = A + t*dir(90)*(C-A);
pair C_1 = A + t*dir(-90)*(B-A);
pair C_2 = B + t*dir(90)*(A-B);
pair A_prime = extension(A_1,B_2,A_2,C_1);
pair B_prime = extension(B_1,C_2,B_2,A_1);
pair C_prime = extension(C_1,A_2,C_2,B_1);
pair Y = extension(A,A+B_prime-C_prime,A_prime,B_prime);
pair Z = extension(A,Y,A_prime,C_prime);
pair R = extension(A,foot(A,B_1,C_2),B,foot(B,A_2,C_1));
fill(B--A_1--A_2--C--cycle, mediumgray);
fill(C--B_1--B_2--A--cycle, mediumgray);
fill(A--C_1--C_2--B--cycle, mediumgray);
draw(A--B--C--cycle, linewidth(1));
draw(A_1--B_2, black);
draw(B_1--C_2, black);
draw(C_1--A_2, black);
draw(Z--A, black, EndArrow(position=0.5));
draw(A--Y, black);
draw(B_prime--C_prime, black, EndArrow(position=0.5));
draw(B--A_1--A_2--C--cycle, black);
draw(C--B_1--B_2--A--cycle, black);
draw(A--C_1--C_2--B--cycle, black);
draw(A--R, gray);
draw(B--R, gray);
draw(C--R, gray);
dot("$A$", A, dir(101));
dot("$B$", B, dir(-142));
dot("$C$", C, dir(-27));
dot("$A_1$", A_1, dir(-136));
dot("$A_2$", A_2, dir(-45));
dot("$B_1$", B_1, dir(-9));
dot("$B_2$", B_2, dir(81));
dot("$C_1$", C_1, dir(121));
dot("$C_2$", C_2, dir(-149));
dot("$A'$", A_prime, dir(-172));
dot("$B'$", B_prime, dir(-53));
dot("$C'$", C_prime, dir(66));
dot("$Y$", Y, dir(-20));
dot("$Z$", Z, dir(-143));
dot("$R$", R, dir(45));
[/asy][/asy]
Now let the line through $A$ parallel to $B'C'$ meet $A_1B_2$ and $A_2C_1$ at $Y,Z$ respectively. Then the problem is equivalent to showing that $A$ is the midpoint of $YZ$. But note that $AB_2\perp AC$, $B_2Y=A'B'\perp CR$, and $AY\perp AR$, so $\triangle AYB_2\sim \triangle ARC$ and $AY=AR\cdot\tfrac{AB_2}{AC}$. Similarly, we get $AZ=AR\cdot\tfrac{AC_1}{AB}$. But the angle conditions imply rectangles $CB_1B_2A$ and $AC_1C_2B$ are similar, so $\tfrac{AB_2}{AC}=\tfrac{AC_1}{AB}$. Thus, $AY=AZ$, so $A$ is the midpoint of $YZ$, as desired.
The second solution only requires that rectangles $CB_1B_2A$ and $AC_1C_2B$ are similar. Rectangle $BA_1A_2C$ can be arbitrarily chosen and the problem is still true.
DottedCaculator
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 $B'C'$.
CyclicISLscelesTrapezoid
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. $\square$
buratinogigle
12.11.2024 17:44
This problem was proposed by Kostas Vittas since 2011 for Mathley of Vietnam, see here.