MATHCOUNTS Mini #72 - Three Dimensional Geometry

This video focuses on three dimensional Geometry.

Video by Art of Problem Solving's Richard Rusczyk, a MATHCOUNTS alum. Visit Art of Problem Solving for many more educational resources.

[amaarquadri]

I found an easier way to solve the second problem. Since each of the 2 triangles have a base of PQ and a height that is equal to the altitude from A and B respectively to the line 4x+3y=19, their area doesn't depend on where PQ actually is (as long as it on the line). Thus, we can slide PQ along the line resulting in a family of quadrilaterals that all have the same area. Lets call the new P and Q: P' and Q' respectively. If we slide PQ such that Q' is on the x axis, then the quadrilateral becomes one big triangle. The vertices of the new large triangle are A, B, and P'. Now we just have to find the height of this triangle, which can be done since we know the slope and length of P'Q'.
Picture the right triangle formed by P', Q', and the point on the x axis directly below P'. Lets call its height (which is equal to the height of the large triangle) y, and its base x. Thus by the Pythagorean theorem:
x^2 + y^2 = P'Q'^2
But since we know the slope: x = 3/4y. Thus,
(3/4y)^2 + y^2 = P'Q'^2
Solving for y gives: y = 12/5.
Now the area of the large triangle is 0.5 * AB * y = 0.5* 8 * 12/5 = 48/5

[keshavb]

for 3d problems, use a 2d case and that is called induction.

[elrichardo]

Oh god... 2016 State Sprint #30 gives me nightmares...

[michaelempeigne]

Team A has 4 girls and 5 boys. Team A takes 5 days to paint 4 rooms.
Team B has 7 girls and 10 boys. Team B takes 4 days to paint 5 rooms.
Team C has 8 girls and 5 boys. How long will it take Team C to paint 6 rooms ?

Assume every girls works at a constant rate, and every boy works at a constant rate, and every room is identical.