# Equilateral triangle ABC

In the equilateral triangle ABC, K is the center of the AB side, the L point lies on one-third of the BC side near the point C, and the point M lies in the one-third of the side of the AC side closer to the point A. Find what part of the ABC triangle contains the triangle KLM.

Result

x =  0.278

#### Solution:

$a=1 \ \\ h=\sqrt{ a^2 - (a/2)^2 }=\sqrt{ 1^2 - (1/2)^2 } \doteq 0.866 \ \\ \ \\ S=\dfrac{ a \cdot \ h }{ 2 }=\dfrac{ 1 \cdot \ 0.866 }{ 2 } \doteq 0.433 \ \\ \ \\ S_{ 1 }=\dfrac{ \dfrac{ h }{ 3 } \cdot \ \dfrac{ a }{ 2 } }{ 2 }=\dfrac{ \dfrac{ 0.866 }{ 3 } \cdot \ \dfrac{ 1 }{ 2 } }{ 2 } \doteq 0.0722 \ \\ S_{ 2 }=\dfrac{ \dfrac{ 2 \cdot \ h }{ 3 } \cdot \ \dfrac{ a }{ 2 } }{ 2 }=\dfrac{ \dfrac{ 2 \cdot \ 0.866 }{ 3 } \cdot \ \dfrac{ 1 }{ 2 } }{ 2 } \doteq 0.1443 \ \\ S_{ 3 }=\dfrac{ \dfrac{ h }{ 3 } \cdot \ \dfrac{ 2 \cdot \ a }{ 3 } }{ 2 }=\dfrac{ \dfrac{ 0.866 }{ 3 } \cdot \ \dfrac{ 2 \cdot \ 1 }{ 3 } }{ 2 } \doteq 0.0962 \ \\ \ \\ S_{ 4 }=S - (S_{ 1 }+S_{ 2 }+S_{ 3 })=0.433 - (0.0722+0.1443+0.0962) \doteq 0.1203 \ \\ \ \\ x=\dfrac{ S_{ 4 } }{ S }=\dfrac{ 0.1203 }{ 0.433 } \doteq \dfrac{ 5 }{ 18 } \doteq 0.2778 \doteq 0.278$

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