Junior 2009/5
In a $N\times N$ grid,Abir picks three lattice points as vertexes of a triangle.Surprisingly,he always chooses the (o,o) point.What is the largest area of the triangle Abir can draw.
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Re: Junior 2009/5
The largest triangle Abir can draw has the area of (N x N)/2
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Re: Junior 2009/5
If the base (as in fig 1) is parallel to the y axis, it is easily seen that the largest area is achieved when we move the base to the end of the line and the base is largest. In such situations we can get atmost $\frac {1}{2}n^2$.
Now consider that the base isn't parallel to the axis. By flipping, we can assume without the loss of generality that among the two chosen points, the one with the smaller $x$- coordinate has the larger $y$ coordinate. Drop perpendiculars as in figure 2. Let $A(x_1,y_1), B(x_2,y_2)$. Now $(OABO)=(OACO)+(CABDC)-(OBDO)$=$\frac {1}{2}[x_1y_1+(x_2-x_1)(y_1+y_2)-x_2y_2]$= $\frac{1}{2}[x_1y_1+x_2y_1+x_2y_2-x_1y_1-x_1y_2-x_2y_2]=\frac{1}{2}[x_2y_1-x_1y_2]$ We see that the largest value this fraction can get is $\frac{1}{2}[n^2-0]=\frac{1}{2}n^2$.
Now consider that the base isn't parallel to the axis. By flipping, we can assume without the loss of generality that among the two chosen points, the one with the smaller $x$- coordinate has the larger $y$ coordinate. Drop perpendiculars as in figure 2. Let $A(x_1,y_1), B(x_2,y_2)$. Now $(OABO)=(OACO)+(CABDC)-(OBDO)$=$\frac {1}{2}[x_1y_1+(x_2-x_1)(y_1+y_2)-x_2y_2]$= $\frac{1}{2}[x_1y_1+x_2y_1+x_2y_2-x_1y_1-x_1y_2-x_2y_2]=\frac{1}{2}[x_2y_1-x_1y_2]$ We see that the largest value this fraction can get is $\frac{1}{2}[n^2-0]=\frac{1}{2}n^2$.
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