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
26.06.2018 05:21
Here's my sol
although it does not use $B_2$ and $C_2$) With the same conditions, prove that $$AB^2+AC^2=5BC^2$$
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.
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. ($BA_1=TA_1=CA_1=A_2A_1$).
20.10.2024 00:10
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.