Question

Extra Credit Geometry Problem

Answer 1

Draw PB. You know triangle ADP and DBP have area in ratio of 2/3 because area ratio will match ratio of bases AD/DB. So call those areas 2k and 3k.

We know triangle ABE to triangle AEC areas have ratio 1:4. Also triangle PBE to triangle PEC areas have ratio 1:4. So the differences ABE - PBE and AEC - PEC also have ratio 1:4.

But those differences are triangle APB and triangle APC.

Since triangle APB has area 5x, that means triangle APC must have area 20x to maintain that 1:4 ratio.

Now you know area ADP = 2x and area APC = 20x. So the ratio of PC to DP is 20x/2x, which is 10:1.

Answer 2

I believe you can also use vectors and Archimedes' Law of the Lever to get the answer. Archimedes says that the weight times the distance from the point of rotation gives the torque (rotational force) about the point.

Think about it in the following way. Our triangle is made out of a stiff and weightless piece of sheet metal. We have a knife edge running from C to D and we have to place masses at A and D so the triangle balances.

If the point D was in the middle of AB then it would balance. But D cuts the segment into unequal segments. We are given that AD / DB = 2 / 3. To make matters simple suppose AD has length 2 and DB has length 3. (We can scale later if need be.) To counteract the unequal segments we should put a mass of 3 at A and a mass of 2 at B to create balance. This is because the length of AD is 2 and we multiply by the mass 3 at A to get a torque of 6. Similarly the length of DB is 3 and the mass at B is 2 and we multiply them together to get a torque of 6 in the opposite direction.

Similarly with BC. If BE / EC = 1 / 4 then we need a mass of 4 at B and a mass of 1 at C to create balance.

To create harmony in the two situations we can scale the AB situation by 2 and in both cases the mass at B will be 4. This tells us we can proceed because the two masses we want to place at B are the same.

So we have a mass 6 at A, a mass 4 at B, and a mass 1 at C. We now have complete information.

Since D is the point of balance of AB we can treat the mass at A and the mass at B as being concentrated at D with a total mass of 10.

So the mass at D is 10 and the mass of C is 1. To get P to be a point of balance between the two masses we need CP to be 10 times the length of PD. This is because the length of 1 times the mass of 10 counterbalances the length of 10 times the mass of 1.

So CP / PD = 10 / 1

Edit: spelling errors

Answer 3

I was curious where I begin on this problem.

Answer 4

Hi,

I can think of two methods here.

Method 1. In the non-Cartesian coordinate system in which B = (0,0), C = (1,0), A = (0,1), write down the equations of the lines CD and AE. Solve the system of equations to find the coordinates of P.

Method 2. Consider the pencil of four lines consisting of AB, AE, AC, and the line through A parallel to CD. Then the cross-ratio of their respective points of intersection with BC is equal to the cross-ratio of their respective points of intersection with CD (the last of these being the point at infinity, since the fourth line is parallel to CD).

Answer 5

Here is an intuition: all the numbers are dimensionless. In fact they are all ratio of length of parallel lines. You can check that the hypothesis and the answer to the problem are all invariant under scaling in any directions, and shear in any directions. Over all, they are invariant under affine transformations.

So by scaling and shearing, you can transform the 3 points A,B,C into any coordinates you want, then you can just compute.

Another way of solving this problem is to just use barycentric coordinate.