+0

# done

0
105
2
+82

done

Mar 19, 2020
edited by rubikx2910  Apr 15, 2020

#1
+24925
+3

Let O denote the centroid of triangle ABC.

Let M and N be points on sides AB and AC, respectively, so that M, O, and N are collinear, and AM/MB = 5/2.

Find AN/NC.

$$\begin{array}{|rcll|} \hline \text{Let } A=(x_A,\ y_A)&=&(0,\ 0) \\ \text{Let } B=(x_B,\ y_B)&=&(7,\ 0) \\ \text{Let } C=(x_C,\ y_C) \\ \text{Let } M=(x_M,\ y_M)&=&(5,\ 0) \\ \text{Let } N=(x_N,\ y_N) \\ \text{Let } O=(x_O,\ y_O)&=&\left(\dfrac{x_A+x_B+x_C}{3},\ \dfrac{y_A+y_B+y_C}{3} \right) \\ &=&\left(\dfrac{0+7+x_C}{3},\ \dfrac{0+0+y_C}{3} \right) \\ &=&\left(\dfrac{7+x_C}{3},\ \dfrac{y_C}{3} \right) \\ \hline \end{array}$$

M, O, and N are collinear

$$\begin{array}{|rcll|} \hline \mathbf{\dfrac{y_O-y_M}{x_O-x_M}} &=& \mathbf{\dfrac{y_N-y_O}{x_N-x_O}} \\\\ (x_N-x_O)(y_O-y_M) &=& (y_N-y_O)(x_O-x_M) \quad | \quad y_M=0,\ x_M=5 \\ (x_N-x_O)(y_O-0) &=& (y_N-y_O)(x_O-5) \\ (x_N-x_O)y_O &=& (y_N-y_O)(x_O-5) \\ x_Ny_O-x_Oy_O &=& y_Nx_O-5y_N-y_Ox_O+5y_O \\ x_Ny_O &=& y_Nx_O-5y_N+5y_O \quad | \quad y_O=\dfrac{y_C}{3},\ x_O=\dfrac{7+x_C}{3} \\ \dfrac{x_Ny_C}{3} &=& \dfrac{y_N(7+x_C)}{3}-5y_N+\dfrac{5y_C}{3} \\ \dfrac{x_Ny_C}{3} &=& \dfrac{7y_N}{3}+\dfrac{y_Nx_C}{3}-5y_N+\dfrac{5y_C}{3} \\ && \boxed{ \dfrac{y_N}{x_N}= \dfrac{y_C}{x_C}\quad \text{ or } \quad y_Nx_C=x_Ny_C } \\ 0 &=& \dfrac{7y_N}{3}-5y_N+\dfrac{5y_C}{3} \quad | \quad * 3 \\ 0 &=& 7y_N -15y_N+ 5y_C \\ 8y_N &=& 5y_C \\\\ \mathbf{\dfrac{y_N}{y_C}} &=& \mathbf{\dfrac{5}{8}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\dfrac{AN}{y_N}} &=& \mathbf{\dfrac{AN+NC}{y_C}} \\\\ \dfrac{AN}{AN+NC} &=& \dfrac{y_N}{y_C} \quad | \quad \mathbf{\dfrac{y_N}{y_C}=\dfrac{5}{8}} \\\\ \dfrac{AN}{AN+NC} &=& \dfrac{5}{8} \\\\ 8AN &=& 5(AN+NC) \\ 8AN &=& 5AN+5NC \\ 3AN &=& 5NC \\\\ \mathbf{\dfrac{AN}{NC}} &=& \mathbf{\dfrac{5}{3}} \\ \hline \end{array}$$

Mar 19, 2020
#2
+82
+1

done

rubikx2910  Mar 20, 2020
edited by rubikx2910  Apr 15, 2020