Let $ABC$ be a triangle. Let $M$ be the midpoint of segment $AB$, and let $I$ be the midpoint of segment $MC$. Construct the point $K$ such that $\vec{CK} = \frac{1}{3} \vec{CB}$. Prove that the points $A$, $I$, and $K$ are collinear.

GeometryVector GeometryCollinearityTriangle GeometryMidpoint
2025/3/9

1. Problem Description

Let ABCABC be a triangle. Let MM be the midpoint of segment ABAB, and let II be the midpoint of segment MCMC. Construct the point KK such that CK=13CB\vec{CK} = \frac{1}{3} \vec{CB}. Prove that the points AA, II, and KK are collinear.

2. Solution Steps

Since MM is the midpoint of ABAB, we have AM=12AB\vec{AM} = \frac{1}{2} \vec{AB}.
Since II is the midpoint of MCMC, we have MI=12MC\vec{MI} = \frac{1}{2} \vec{MC}.
We want to show that AA, II, and KK are collinear, which means that AI=λAK\vec{AI} = \lambda \vec{AK} for some scalar λ\lambda.
We can write
AI=AM+MI=12AB+12MC\vec{AI} = \vec{AM} + \vec{MI} = \frac{1}{2} \vec{AB} + \frac{1}{2} \vec{MC}
Since MC=ACAM=AC12AB\vec{MC} = \vec{AC} - \vec{AM} = \vec{AC} - \frac{1}{2} \vec{AB},
AI=12AB+12(AC12AB)=12AB+12AC14AB=14AB+12AC\vec{AI} = \frac{1}{2} \vec{AB} + \frac{1}{2} (\vec{AC} - \frac{1}{2} \vec{AB}) = \frac{1}{2} \vec{AB} + \frac{1}{2} \vec{AC} - \frac{1}{4} \vec{AB} = \frac{1}{4} \vec{AB} + \frac{1}{2} \vec{AC}
Now, 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}.
We want to find λ\lambda such that AI=λAK\vec{AI} = \lambda \vec{AK}. Thus,
14AB+12AC=λ(13AB+23AC)=λ3AB+2λ3AC\frac{1}{4} \vec{AB} + \frac{1}{2} \vec{AC} = \lambda (\frac{1}{3} \vec{AB} + \frac{2}{3} \vec{AC}) = \frac{\lambda}{3} \vec{AB} + \frac{2\lambda}{3} \vec{AC}.
Equating the coefficients, we have
14=λ3\frac{1}{4} = \frac{\lambda}{3} and 12=2λ3\frac{1}{2} = \frac{2\lambda}{3}.
From the first equation, λ=34\lambda = \frac{3}{4}.
From the second equation, λ=34\lambda = \frac{3}{4}.
Thus, AI=34AK\vec{AI} = \frac{3}{4} \vec{AK}, which means AA, II, and KK are collinear.

3. Final Answer

AI=34AK\vec{AI} = \frac{3}{4} \vec{AK}

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