Given a triangle ABC, M is the midpoint of [AB] and I is the midpoint of [MC]. We need to construct a point K such that $\vec{CK} = \frac{1}{3}\vec{CB}$. Then, we need to prove that points A, I, and K are collinear.

GeometryVectorsCollinearityTriangle GeometryMidpoint
2025/3/9

1. Problem Description

Given a triangle ABC, M is the midpoint of [AB] and I is the midpoint of [MC]. We need to construct a point K such that CK=13CB\vec{CK} = \frac{1}{3}\vec{CB}. Then, we need to prove that points A, I, and K are collinear.

2. Solution Steps

First, express AI\vec{AI} in terms of AC\vec{AC} and AB\vec{AB}.
Since I is the midpoint of MC, we have AI=12(AM+AC)\vec{AI} = \frac{1}{2}(\vec{AM} + \vec{AC}).
Since M is the midpoint of AB, AM=12AB\vec{AM} = \frac{1}{2}\vec{AB}.
Therefore,
AI=12(12AB+AC)=14AB+12AC\vec{AI} = \frac{1}{2}(\frac{1}{2}\vec{AB} + \vec{AC}) = \frac{1}{4}\vec{AB} + \frac{1}{2}\vec{AC}.
Next, express AK\vec{AK} in terms of AC\vec{AC} and AB\vec{AB}.
AK=AC+CK=AC+13CB=AC+13(ABAC)=AC+13AB13AC=13AB+23AC\vec{AK} = \vec{AC} + \vec{CK} = \vec{AC} + \frac{1}{3}\vec{CB} = \vec{AC} + \frac{1}{3}(\vec{AB} - \vec{AC}) = \vec{AC} + \frac{1}{3}\vec{AB} - \frac{1}{3}\vec{AC} = \frac{1}{3}\vec{AB} + \frac{2}{3}\vec{AC}.
Now, we need to check if AI\vec{AI} and AK\vec{AK} are collinear. In other words, we need to find a scalar λ\lambda such that AK=λAI\vec{AK} = \lambda \vec{AI}.
If such a λ\lambda exists, then A, I, and K are collinear.
13AB+23AC=λ(14AB+12AC)\frac{1}{3}\vec{AB} + \frac{2}{3}\vec{AC} = \lambda(\frac{1}{4}\vec{AB} + \frac{1}{2}\vec{AC})
13AB+23AC=λ4AB+λ2AC\frac{1}{3}\vec{AB} + \frac{2}{3}\vec{AC} = \frac{\lambda}{4}\vec{AB} + \frac{\lambda}{2}\vec{AC}
Equating the coefficients of AB\vec{AB} and AC\vec{AC}, we get:
13=λ4\frac{1}{3} = \frac{\lambda}{4} and 23=λ2\frac{2}{3} = \frac{\lambda}{2}
From the first equation, λ=43\lambda = \frac{4}{3}.
From the second equation, λ=43\lambda = \frac{4}{3}.
Since both equations yield the same value for λ\lambda, we have λ=43\lambda = \frac{4}{3}.
Therefore, AK=43AI\vec{AK} = \frac{4}{3}\vec{AI}.
Since AK\vec{AK} is a scalar multiple of AI\vec{AI}, the points A, I, and K are collinear.

3. Final Answer

AK=43AI\vec{AK} = \frac{4}{3}\vec{AI}

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