 Posers and Puzzles

1. 24 Sep '11 23:54
An iceberg floats in water so that 90% of its volume is underwater. The iceberg is shaped like a cone, with peak pointing down, base up, and the base rises one foot above the surface of the water. If the iceberg tilts so that base is down and peak up, how high from the water surface will the peak rise?
2. 25 Sep '11 17:59
9 feet???
3. 26 Sep '11 04:47
I get the height to be

100^(1/3)
4. 27 Sep '11 03:02
Originally posted by joe shmo
I get the height to be

100^(1/3)
nevermind, thats not correct...looks a bit more complicated at this time of day.

my latest result is

approx

13.45 ft
5. 28 Sep '11 00:34
Is your silence an indicator that I should try again?
6. 28 Sep '11 03:331 edit
Originally posted by joe shmo
nevermind, thats not correct...looks a bit more complicated at this time of day.

my latest result is

approx

13.45 ft
This is what i get.

The total height of the cone is
H = 10 + 3 * 30^(1/3) + 30^(2/3) ~= 28.977
7. 28 Sep '11 07:191 edit
Nice work. =)

SOLUTION

When the peak of the iceberg points down, the underwater

part and the entire iceberg have the same shape.

Let h be the height of the iceberg.

Proportions in objects of the same shape are the same,

except that they are squared for areas and cubed for volumes,

which gives

((h-1) : h)^3 = 9 : 10

from which h = 1 ft / (1 - (9 : 10)^(1/3)) , or about 29 feet.

When the iceberg keels over, the part above water is of

the same shape as the entire iceberg, with volume ratios

1 : 10. Let x be the height of the visible part of the iceberg.

(x : h)^3 = 1 : 10

from which

x = (1 : 10)^(1/3) h

x = 1 ft * (1 : 10)^(1/3) / (1 - (9 : 10)^(1/3))