Oh yes; championmathgirl gave that hint in post 8. I hadn't explored it because I had my solution, but let's do so now.
S lies in the plane of ABCD and in the plane of PQRC so CP extended must go through S
Triangles SAP and SDC are similar. Triangles SAQ and SDR are similar.
From the first pair SA/SD = PA/DC = 2/5
So SA/(SA+5) = 2/5 => 5xSA = 2xSA + 10 => SA = 10/3
From the second pair SA/SD = AQ/RD
=> RD = 1 x (5+10/3)/(10/3) = (15+10)/10 = 25/10 = 2.5
I've distorted the diagram slightly so that points don't end up looking like they are on top of each other.
Original tetrahedron ABCD. New tetrahedron EFGH.
I, J and K are the midpoints of BC, CD and BD.
So DI, AJ, AI and AK are medians. The centres are 2/3 of the way down from the top of the median.
Consider triangle IJK. Because it is formed from the midpoints it will be similar to DBC and half its size.
Triangle HFG is also similar to IJK and 2/3 of its size.
So you can work out the ratio BC:GF. The volume ratio will be the cube of the length ratio.
Welcome to the forum.
I think you'll always find this type of 'division' tricky unless you can do 'long division' with ordinary arithmetic first. How are you with long division? eg 3266 ÷ 23
If the answer is 'not so good' then it'll be worth looking at that first. The rules of algebra are just the same as the rules of arithmetic, so it's worth making sure of that first.
This comes from one of the angle properties of a circle. Have a look at post 7 here:
This relies on the other properties also proved in that thread.
I have also made a contents post for other geometry proofs which you will find here:
Hope that helps,
Here's a view of the cylinder on its side. The coloured region shows the oil.
You know enough measurements to calculate the angle ACB and hence the area of the sector from A through D to B (angle/360 times pi r^2). Also the area of the triangle ABC. Hence the area of the coloured segment.
So this will allow you to calculate the volume of oil. When the cylinder is upright that volume will be h times pi r^2 where h is the required height.
Hope that helps,
I've included a diagram:
Along the x axis slice solid into sections. Approximate each section to be a cylinder. The red line indicates the radius of one slice shaded orange. The 'height' (thickness)of a cylinder would depend on how far apart you make the slices. Let's say the radius of a slice is 'r' and the thickness 't'.
The area of the curved surface of the slice is 2.pi.r.t (think of a marrow; you peel of the surface making a strip 't' wide and circumference long.)
You'd have to make a separate calculation for every slice as the radii will vary.
The slices aren't really cylinders so the calculation is only approximate. If you have lots more slices by making 't' smaller, the calculation becomes more accurate.
I think the point of this question is just to get you thinking about how you can use what you already know to calculate tricky shaped solids.
You could imagine that the ball is sliced up into a series of disks along its major axis. Each disk has a radius which you'd have to measure and a thickness. The area of one disk would be the curved surface area of a cylinder which I think you know. Do each calculation and add them up.
This won't be wonderfully accurate ** but you can say that, and add that to improve the accuracy, I would take more but thinner disks. This is a method that is used to get the area of tricky to measure shapes.
** Because the strange shape means these aren't exactly disks.
Or you could get some squared paper and cut it up into squares (say 1 cm or 1 inch square) Then stick these all over the ball trying to cover as much area as possible. Where you have smaller uncovered bits, cut the squares in half to cover those bits .......
Sorry I've got to go now but I'll look in again later in case you want more help.
You could use vectors:
Take A as the origin and AB = p as one vector and AD = q as another.
Express BD, DC, AE and AC in terms of p and q.
Write a vector equation for the line BE and hence an expression with lambda for BX. Hence AX.
Similarly a vector equation for AC and an expression for AX with a 'mu'.
Equate the expressions and solve for lambda and mu. Show vector BE = 3EX.
and we're using a funny shaped ball not a UK soccer ball? That's round.
I think you're only expected to think of a sensible method ... it doesn't have to use advanced formulas ... especially if you have no idea where they come from.
For a volume I'd put the ball in a large tub with some water in and see how much the water goes up. If you can calculate that using formulas you do know that should be acceptable. So start with a cylindrical tub and measure its radius. Put in enough water to totally submerge the ball. Before you put the ball in mark where the water level is. Then push in the ball and mark again where the water is. Calculate the gain the height of the water ... use the volume of a cylinder formula to calculate the volume of water displaced ... that must equal the volume of the ball.
I'll post this whilst thinking about the area.
I'm using the following:
Sketchpad to make drawing.
tan and cot as calculated by this software.
If A, B and C are in GP then A/B - B/C = zero
If D, E and F are in AP then D -2E + F = zero.
I then constructed a diagram and measured as necessary. I calculated as above. Then I used trial and improvement to juggle the points into the best positions. Here is my best effort:
I cannot improve on these values because the minimum point movement is one pixel and any further movement moves me away from the optimum values.
My algebraic approach is to use the above formulas with distances DB, DC, DE etc substituted in. CBE, and ABE are not in right angled triangles but I have been able to eliminate them using compound angle formulas for tan. I also have (Pythag) BE^2 = 100 = (DB+DE)(BD-DE) so, in theory I can eliminate DE as well.
That would leave an equation only containing DB and DC, so I was hoping to reduce that to the required area. Unfortunately the expression is too complicated (five lines of my sheet of A4) so I've stuck at that.
Tried to find simple formulas connecting the tan and cot expressions but no luck there either.
These problems all require the use of the angle properties of a circle together with Pythagoras' theorem. You'll find a thread that deals with these here:
Have a look and see if that's enough to help you out with these. If you post your answers to those you can do, that'll give me a better idea of where you need help.
Cuts have got to leave congruent pieces so four cuts at right angles to the axis and equally spaced would be one way. This would make 5 equal shorter cylinders.
Or you could make the cuts through a diameter of the circular ends and along the axis. These cuts would have to be 45 degrees apart.
And there are several ways that mix these two types of cut. I'll leave you to find them.