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Is it possible to find the point which is marked by question mark ? we know that the s1(x)=s2(x) (the areas of the two triangles are equal)

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Also asked at stackoverflow.com/questions/12683048/… but I think it belongs here. –  Ross Millikan Oct 2 '12 at 4:18
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I don't think so. The point common to both triangles can be moved to the left, increasing the area of the blue triangle and making the line from $(x_2,y_2)$ through the common point miss the point ? entirely. Then the point ? can move down to add the required area. In essence we have four unknowns-the coordinates of the common point and the coordinates of ? but only three constraints-the area equality and the fact that the two line segments through the common point are straight.

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we know that the y coordinate of ? is equal to y1, so the point ? can not move down. –  ben Oct 2 '12 at 4:28
    
@ben: I don't gather that from the drawing-if it is true it should be specified. That would be a fourth constraint and should give a unique solution. It looks like messy algebra to find it-four equations, one of them quadratic (the area one). –  Ross Millikan Oct 2 '12 at 4:30
    
what do you mean by "one of them quadratic (the area one)?" –  ben Oct 2 '12 at 4:35
    
@ben: the equations that express the straightness of the lines are linear in the coordinates. One way to express the area constraint is that the product of the lengths of the blue lines to the meeting point equals the product of the lengths of the red lines to the meeting point. This is of second degree. Sometimes these problems have an easy solution, but I haven't found it. –  Ross Millikan Oct 2 '12 at 12:55
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