Projected Incircle

Geometry Level 5

A plane a a has an equation: x 3 + y 4 + z 5 = 1 \frac{x}{3} + \frac{y}{4} + \frac{z}{5}=1 A triangle has as its vertices points of intersection of the plane a a with the axes of the coordinate system. Incircle of that triangle is projected onto the X Y XY plane and the projected shape has an area A A . Give answer as A × 10000 \lfloor A \times 10000 \rfloor


The answer is 35198.

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2 solutions

Maria Kozlowska
Jul 23, 2015

Vertices of the triangle are A ( 3 , 0 , 0 ) A(3,0,0) , B ( 0 , 4 , 0 ) B(0,4,0) and C ( 0 , 0 , 5 ) C(0,0,5) . Sides can be easily calculated using Pythagorean theorem: 5 , 34 , 41 5, \sqrt{34}, \sqrt{41} . The radius of the incircle can be calculated using the formula: r = A B C s r=\frac{\triangle ABC}{s} where s is a semi-perimeter. Area of A B C \triangle ABC 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 = 769 2 (\triangle ABC )^2 = ((3 \times 4)^2 + (3 \times 5)^2 + (4 \times 5)^2)/4 \Rightarrow \triangle ABC=\frac{\sqrt{769}}{2} The projected incircle is ellipses with major axis = 2 r = 2r . Minor axis can be obtained by multiplying length of major axis with "the shortening factor". That shortening factor is 1 1- flattening factor of the the ellipse g 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 \triangle ABC can be obtained as A B O A B C = 3 × 4 769 \frac{\triangle ABO } {\triangle ABC}=\frac{3\times 4}{\sqrt{769}} where O O is the point of origin. The area of the ellipse will be: π r 2 12 769 = 6 769 π / ( 34 41 + 5 34 + 5 41 + 50 ) \pi r^2 \frac{12}{\sqrt{769}}=6\sqrt{769} π / (\sqrt{34} \sqrt{41} + 5 \sqrt{34} + 5\sqrt{41} + 50)

Yuriy Kazakov
May 30, 2020

S c i r c l e cos θ = S e l l i p s e S_{circle}\cos\theta=S_{ellipse} , where θ \theta - is angle between plane a a and plane z = 0 z=0 and cos θ = 1 5 1 3 2 + 1 4 2 + 1 5 2 \displaystyle \cos\theta = \frac{\frac{1}{5}}{ \sqrt {\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}}}

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