The problem statement is problem 15. Given a triangle ABC, locate points I, J, and K such that $\vec{BI} = \frac{3}{2} \vec{BC}$, $\vec{CJ} = \frac{1}{3} \vec{CA}$, and $\vec{AK} = \frac{2}{5} \vec{AB}$. We need to prove that the points I, J, and K are aligned (collinear).

1. Problem Description

The problem statement is problem

1

5. Given a triangle ABC, locate points I, J, and K such that $\vec{BI} = \frac{3}{2} \vec{BC}$, $\vec{CJ} = \frac{1}{3} \vec{CA}$, and $\vec{AK} = \frac{2}{5} \vec{AB}$.

We need to prove that the points I, J, and K are aligned (collinear).

2. Solution Steps

To prove that the points I, J, and K are aligned, we need to show that the vectors $\vec{IJ}$ and $\vec{IK}$ are collinear, which means that there exists a scalar $t$ such that $\vec{IJ} = t \vec{IK}$.

First, express $\vec{IJ}$ and $\vec{IK}$ in terms of $\vec{AB}$ and $\vec{AC}$.

$\vec{IJ} = \vec{IB} + \vec{BC} + \vec{CJ} = -\vec{BI} + \vec{BC} + \vec{CJ} = -\frac{3}{2} \vec{BC} + \vec{BC} + \frac{1}{3} \vec{CA} = -\frac{1}{2} \vec{BC} - \frac{1}{3} \vec{AC}$

Since $\vec{BC} = \vec{AC} - \vec{AB}$,

$\vec{IJ} = -\frac{1}{2} (\vec{AC} - \vec{AB}) - \frac{1}{3} \vec{AC} = \frac{1}{2} \vec{AB} - \frac{1}{2} \vec{AC} - \frac{1}{3} \vec{AC} = \frac{1}{2} \vec{AB} - \frac{5}{6} \vec{AC}$

$\vec{IK} = \vec{IB} + \vec{BA} + \vec{AK} = -\vec{BI} - \vec{AB} + \vec{AK} = -\frac{3}{2} \vec{BC} - \vec{AB} + \frac{2}{5} \vec{AB} = -\frac{3}{2} (\vec{AC} - \vec{AB}) - \frac{3}{5} \vec{AB} = -\frac{3}{2} \vec{AC} + \frac{3}{2} \vec{AB} - \frac{3}{5} \vec{AB} = \frac{9}{10} \vec{AB} - \frac{3}{2} \vec{AC}$

Now we want to find a scalar $t$ such that $\vec{IJ} = t \vec{IK}$:

$\frac{1}{2} \vec{AB} - \frac{5}{6} \vec{AC} = t (\frac{9}{10} \vec{AB} - \frac{3}{2} \vec{AC})$

$\frac{1}{2} \vec{AB} - \frac{5}{6} \vec{AC} = \frac{9}{10} t \vec{AB} - \frac{3}{2} t \vec{AC}$

Equating the coefficients of $\vec{AB}$ and $\vec{AC}$:

$\frac{1}{2} = \frac{9}{10} t \implies t = \frac{1}{2} \cdot \frac{10}{9} = \frac{5}{9}$

$-\frac{5}{6} = -\frac{3}{2} t \implies t = \frac{5}{6} \cdot \frac{2}{3} = \frac{5}{9}$

Since the same value of $t = \frac{5}{9}$ works for both equations, $\vec{IJ} = \frac{5}{9} \vec{IK}$.

3. Final Answer

$\vec{IJ} = \frac{5}{9}\vec{IK}$