2003 AIME II Problem 4 (part 2)

2003 AIME II Problem 4 (part 2)

Category:

Education

Tags:

• 2003
• AIME
• II
• Problem
• (part
• 2)
• Three
• Dimensional
• Geometry

[unnddd]

The last two ratios in my post should have been fractions ie. 3/1 and 27/1.

[unnddd]

This problem took me about 3 minutes to solve and my solution took up about a fifth of an A4 page. Best of all it involved no coordinate geometry.

The in-centre of a triangle divides the top and bottom in the ratio 2:1 - this is quite simple to find. It's clear from this that the ratio of the height of the big tetrahedron to the small tetrahedron is 3:1. Therefore the ratio of the volumes is 27:1. 27+1=28.

You guys should try some BMO questions. This is complete piss in comparison.

[Vindex27]

@mvszao It was tricky. First, one had to realize that the cube ratio of sides was all that was needed. Volume would have taken a while unless one remembered the formula. But it looks like a familiarity with coordinate geometry would have helped a lot!

[Vindex27]

@HighFive5555 So what are you drawing that O axis to get your Ozy plan? Which triangles are you specifically talking about and what angle do they share? Finally, how do you deduce from the these two that the smaller one is a third of the larger one?

[Vindex27]

@HighFive5555 Where's the O axis making the Ozy plan? Which angle is shared? And how do you deduce from these two that the smaller one's side are 1/3 of the larger one?

[abodyVIP]

form a triangle with two sides sqrt(3)/3 and one is unknown and an angle (x) between them.

Then by cosine rule we'll get that the small length is 3/2 and we can get the ration then.

Thanks!

[abodyVIP]

Nice Solution, now I know more about 3D coordinate geometry.

I've another solution without using it.

if the length of a side is 2 => we can figure out that h=2sqrt(2)/sqrt(3) by using the pythagorean theorem twice.

The last triangle we use have the sides sqrt(3), 2sqrt(2)/sqrt(3) and sqrt(3)/3

so we can conclude that Cos(x) ,where (x) is the angle between any two planes in this shape, is equal to 1/3 (a/hyp.)

Then, we take two altitudes of the triangles with the small shape side and form ---->

[PoleReseal]

nice problem, cool explanation. Cheers Sal :)

[tttybhgf]

@Zaibatsu4Life True !

[Zakamiro]

I bet Sal's wife is ridiculously hot.