In the triangle $ABC$, the medians $AA_1$, $BB_1$, and $CC_1$ are concurrent at a point $T$ such that $BA_1=TA_1$. The points $C_2$ and $B_2$ are chosen on the extensions of $CC_1$ and $BB_2$, respectively, such that $$C_1C_2 = \frac{CC_1}{3} \quad \text{and} \quad B_1B_2 = \frac{BB_1}{3}.$$Show that $TB_2AC_2$ is a rectangle.
Problem
Source: Macedonia JBMO TST 2017, Problem 2
Tags: geometry, rectangle
Devastator
26.06.2018 05:21
Here's my sol
Notice that $TA_1=BA_1=CA_1$, so the circumcenter of triangle $BTC$ lies on $BC$, so $BTC$ is a right angle. Hence, angle $B_2TC_2$ is a right angle, and we are left to prove that $AB_2TC_2$ is a parallelogram.
But note that since $AC_1=BC_1$and $C_1C_2=\frac{1}{3}CC_1=TC_1$, we get that $ATBC_2$ is a parallelogram, so $AC_2$ is parallel to BT, and thus $AC_2$ and $B_2T$ are parallel. Similarly, $AB_2$ and $C_2T$ are parallel
It follows that $AB_2TC_2$ is a parallelogram with a right angle, so it's a rectangle.
$QED$
$BTC$ is a right triangle
although it does not use $B_2$ and $C_2$) With the same conditions, prove that $$AB^2+AC^2=5BC^2$$
socr4tes
27.06.2018 13:18
Amir Hossein wrote: In the triangle $ABC$, the medians $AA_1$, $BB_1$, and $CC_1$ are concurrent at a point $T$ such that $BA_1=TA_1$. The points $C_2$ and $B_2$ are chosen on the extensions of $CC_1$ and $BB_2$, respectively, such that $$C_1C_2 = \frac{CC_1}{3} \quad \text{and} \quad B_1B_2 = \frac{BB_1}{3}.$$Show that $TB_2AC_2$ is a rectangle. $C_2C_1=C_1T=\frac{C_1C}{3}$ , $BC_1=C_1A \Rightarrow C_2ATB$ is a parallelogram. Similarly $AB_2CT$ is a parallelogram so $C_2AB_2T$ is a parallelogram.$C_2T=TC=\frac{C_2C}{2}$, $BT=B_2T=\frac{BB_2}{2}$ so $C_2B_2\parallel BC$ and $AT=\frac{2AA_1}{3}$, $C_2B_2=2*BA_1=2*TA_1=\frac{2AA_1}{3}$, $AT=C_2B_2 \Rightarrow C_2ATB_2$ rectangle.
zuss77
28.06.2018 14:46
Let's construct point $A_2$ similarly. Now, gomothety with center $T$ and factor $k=-1$ sends quadrilateral $BTCA_2$ to $B_2TC_2A$. (wee know how T divides medians). Now it only remains to notice that $BTCA_2$ is a rectangle. ($BA_1=TA_1=CA_1=A_2A_1$).
DensSv
20.10.2024 00:10
Since $TA_{1}=BA_{1}=CA_{1}$ this means that $TB_{1}C_{1}$ is a right triangle in $T$. Also, $T$ is the centroid of $ABC$ thus $C_{1}C_{2}=\frac{CC_{1}}{3}= C_{1}T$ This means that $T$ is the midpoint of $CC_{2}$ but also $B_{1}\perp CC_{2}$ hence $B_{1}C_{2}=B_{1}C_{1}$ similarly $B_{2}C_{2}=B_{2}C_{1}$, so $ B_{2}C_{2}B_{1}C_{1}$ is a rhombus with its side lengths $=2TA_{1}=AT$. Also $TA_{1}$ is a midline in $B_{2}B_{1}C_{1}$ thus $AT \stackrel{||}{=}B_{2}C_{1}$ so $ AB_{2}C_{1}T$ is a parallelogram $\Rightarrow AB_{2}\stackrel{||}{=}TC_{1} \Rightarrow AB_{2}\stackrel{||}{=}TC_{2} $ hence $TB_{2}AC_{2}$ is a parallelogram with equal diagonals so $TB_{2}AC_{2}$ is a rectangle q.e.d.
Ianis
20.10.2024 00:44
Devastator wrote: With the same conditions, prove that $$AB^2+AC^2=5BC^2$$ This is a problem from Ireland 2001.