Shadow of a Pole

Geometry Level 3

A vertical pole extends from the origin ( 0 , 0 , 0 ) (0, 0, 0) to the point ( 0 , 0 , 1 ) (0, 0, 1) . A point light source is located at the point ( 1 , 2 , 4 ) (1, 2, 4) .

What is the length of the shadow of the pole in the x y xy -plane ?

1 + 2 1 + \sqrt {2} 2 2 1 2 \sqrt {2} - 1 5 / 2 5 / \sqrt{2} 5 / 3 \sqrt {5} / 3

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1 solution

Unstable Chickoy
Jun 9, 2014

Bear with the notations.

O O (0, 0, 0) - origin.

P P (0, 0, 1) - highest point of the pole.

L L (1, 2, 4) - light source.

M M - a point directly below the light source L L perpendicular to the point P P .

S S - point from the tip of the shadow, perpendicular to the origin, lies in the X Y XY plane.

L P M \triangle{LPM} is similar to O S P \triangle{OSP}

O S O P = M P M L \frac{\overline{OS}}{\overline{OP}} = \frac{\overline{MP}}{\overline{ML}}

O S 1 = 1 2 + 2 2 4 1 \frac{\overline{OS}}{1} = \frac{\sqrt{1^2 + 2^2}}{4 - 1}

O S = 5 3 \overline{OS} = \boxed{\frac{\sqrt{5}}{3}}

Good use of similar triangles. Even though we were given 3-D coordinates, it ended up just being a 2-D problem.

Being able to reduce the dimension of the problem makes it much easier to approach.

Calvin Lin Staff - 4 years, 7 months ago

Thank you. I already forgot this website quite a while ago :) . I wish I could get back my spare time, same as the old days

Unstable Chickoy - 4 years, 5 months ago

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You should :)

Brilliant's community works by having numerous people contribute wherever they can. A problem here, a solution there, it all adds up!

Calvin Lin Staff - 4 years, 5 months ago

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