a has an equation: 3 x + 4 y + 5 z = 1 A triangle has as its vertices points of intersection of the plane a with the axes of the coordinate system. Incircle of that triangle is projected onto the X Y plane and the projected shape has an area A . Give answer as ⌊ A × 1 0 0 0 0 ⌋
A plane
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S c i r c l e cos θ = S e l l i p s e , where θ - is angle between plane a and plane z = 0 and cos θ = 3 2 1 + 4 2 1 + 5 2 1 5 1
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Vertices of the triangle are A ( 3 , 0 , 0 ) , B ( 0 , 4 , 0 ) and C ( 0 , 0 , 5 ) . Sides can be easily calculated using Pythagorean theorem: 5 , 3 4 , 4 1 . The radius of the incircle can be calculated using the formula: r = s △ A B C where s is a semi-perimeter. Area of △ A B C can be found using De Gua's theorem: ( △ A B C ) 2 = ( ( 3 × 4 ) 2 + ( 3 × 5 ) 2 + ( 4 × 5 ) 2 ) / 4 ⇒ △ A B C = 2 7 6 9 The projected incircle is ellipses with major axis = 2 r . Minor axis can be obtained by multiplying length of major axis with "the shortening factor". That shortening factor is 1 − flattening factor of the the ellipse g . It is really cosine of the angle between plane XY and plane a. It is also a ratio of original segment lengths to the projected ones, perpendicular to the intersection line between the two planes.
The shortening factor is of the △ A B C can be obtained as △ A B C △ A B O = 7 6 9 3 × 4 where O is the point of origin. The area of the ellipse will be: π r 2 7 6 9 1 2 = 6 7 6 9 π / ( 3 4 4 1 + 5 3 4 + 5 4 1 + 5 0 )