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Rising water levels (Posted on 2006-06-26) Difficulty: 4 of 5
You have rectangular pot of water 10cm by 10cm at the bottom and 6 cm deep. It is filled to a depth of 3cm.

You have 7 solid steel shapes in front of you. The question is to find the new level of the water after each shape is put in the pot in the orientation described. The previous shape is removed before adding the next.

1. A cube 5cm on a side.

2. A prism in the shape of a right triangle with legs 5cm long. It is 5cm high but is placed on one of its legs with its hypotenuse sloping out of the water.

3. Another prism, this one having an equilateral triangle of sides 5cm and height 5cm. It is to be placed on its side with two faces sloping up out of the water.

4. A regular hexagonal prism. Each edge of the hexagon is 4cm and the height is 5cm. It is to be placed on its side.

5. A right square pyramid. Its base is 6cm on a side. It is 5cm high. It is to be placed base down.

6. A right cylinder of radius 3cm and length 5cm. It is to be placed on its side.

7a. A right cone of radius 3cm and height 5cm. It is to be placed base down.

7b. The same cone as 7a. This time placed on its side.

No Solution Yet Submitted by Jer    
Rating: 4.1667 (6 votes)

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re(2): Solution to 7b using numerical integration | Comment 25 of 26 |
(In reply to re: Solution to 7b using numerical integration by Mindrod)

In my diagram, I have the base of the cone on the left and the vertex on the right.

The line representing the base (which is seen edge-on) goes upward arctan(3/5) to the right of directly up (we'll see why later) and is 6 cm long.  The line representing one of the slant heights is horizontal (length sqrt(3^2 + 5^2)).  A line connects the vertex of the cone at the bottom-right to the top of the line representing the base.  Another line connects the vertex of the cone with the center of the base (midpoint of the line representing the base), and is the cone's axis, and the altitude of the isosceles triangle when you consider the 6-cm side of the isosceles triangle as the base, though it is not flat on the bottom of the diagram.  That altitude is 5 cm long and is also a bisector of the isosceles triangle's angle at the vertex on the bottom right.  Call the vertex of the cone, at the bottom right point A. Call the point of the isosceles triangle on the bottom left B, and the point at the top C.  Call the midpoint of BC, M.  AM = 5 and BM=3. So the angle of this right triangle (half the isosceles triangle) at A is arctan(3/5). The angle at B in degrees is 90-arctan(3/5); that's the angle the base plane makes with the bottom of the box.  The angle it makes with the vertical is arctan(3/5).  Since the half-angle at A is arctan(3/5) also, the whole angle of the isosceles triangle at A is 2 arctan(3/5).

Perhaps you were thinking is was referring to the angle that the base makes with the bottom of the pot as atan(3/5); that's not the case; atan(3/5) is the angle the base makes with the vertical and also the half-angle at the vertex on the bottom-right, as the isosceles triangle is divided into two contruent but oppositely handed right triangles.

  Posted by Charlie on 2006-07-04 00:50:44
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