# Equilateral triangle and area

I need help with the following problem: Given an equilateral triangle $\triangle ABC$ with $AB=1$, $M,\ N$ and $Q$ are points on the sides $AB,\ BC$ and $AC$ such that the lines $AN,\ BQ$ and $CM$ divide the triangle into $4$ triangles and $3$ quadrilaterals. We color the triangles in two colors (black and blue) in such way that any two triangle with common vertex are colored with different colors. If the area colored in black is equal to that colored in blue, find the sum $AM+BN+CQ$.

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The diagram is shown above. Let $a,b,c,d,e,f,g$ be the areas of each region as shown. Then $a+b+f+g=CN\frac {\sqrt 3}2, d+e+f+g=AQ\frac {\sqrt 3}2, b+c+d+g=BM\frac {\sqrt 3}2$. The condition on coloring says $a+c+e=g$. Adding the first three gives $(a+c+e)+2(b+d+f)+3g=(CN+AQ+BM)\frac {\sqrt 3}2$. Using the coloring, we get the left side is twice the area of the triangle: $2\frac {\sqrt 3}2=(CN+AQ+BM)\frac {\sqrt 3}2$ or $CN+AQ+BM=2$

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let the triangle be EDF.

• now [EDF]=[ABC]$-$( [AMC]+[ABN]+[BQC]$-$[ECQ]$-$[ADM]$-$[BFN] )
• SO [EDF]=[ABC]$-$( [AMC]+[ABN]+[BQC] )+[EQC]+[ADM]+[BFN]
• so we have [ABC]=[AMC]+[ABN]+[BQC]
• IN $\bigtriangleup$ABC each hight is $\sqrt[2]{3}$.$\frac{1}{2}$
• so we have $\frac{a}{4}$=$\frac{1}{2}$.$\frac{a}{2}$.AM +$\frac{1}{2}$.$\frac{a}{2}$.BN +$\frac{1}{2}$.$\frac{a}{2}$.CQ [ WHERE a=$\sqrt[2]{3}$ ]
• from this we get AM +BN +CQ=2
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