1. Point P is chosen inside square ABCD. What is the probability that triangle ABP has a greater area than triangle CDP?

2. Point P is chosen inside square ABCD. What is the probability that triangle ABP has a greater area than each of triangles BCP, CDP, and DAP?

3. Point P is chosen inside square ABCD. What is the probability that ABP has a greater area than each of triangles BCP and CDP?

Mellie Apr 18, 2015

#3**+23 **

I'll also take a stab at (2)

Consider square ABCD below:

This square is divided into 8 equal areas. It is clear that P cannot fall anywhere into rectangle ABFE because, although triangle APB and and triangle DPC would have equal bases, DPC would always have a greater height, and thus, a greater area.

It's also clear that P cannot fall into area IFC......To see this, note that if P fell on AC, its distance from AD and AB would be exactly the same. But, if P falls in area IFC, its distance from AD would always be greater than its distance from AB. Thus, triangles APB and APD formed by having P fall into this region would have equal bases, but APD would always have a greater height than triangle APB.

For the same reason, P cannot fall into area EID, because all triangles BPC would have greater heights than triangles APB.

So, P can only fall into region CID. Here triangles APB, BPC, CPD and DPA will all have the same bases, but APB will have a height greater than the other three.

And since CID is 25% of the area of the square, the probabilty that APB is greater than the other three mentioned triangles is ≈ 25%

CPhill Apr 18, 2015

#1**+3 **

A point is chosen at random in the interior of equilateral triangle . What is the probability that has a greater area than each of and ?

Mellie Apr 18, 2015

#2**+21 **

I'll attempt to answer (1)

Shown below is square ABCD

Notice that if point P falls anywhere within rectangle EFCD, then the triangle APB will have a greater area than triangle DPC because each will have an equal base, but APB will have a greater height than DPC. And since rectangle EFCD is 1/2 of the area of the square, then the probability of P falling into this area is ≈ 1/2. So, the probability that APB > DPC ≈ 1/2. [Of course, there is the * slight* possibility that P falls

CPhill Apr 18, 2015

#3**+23 **

Best Answer

I'll also take a stab at (2)

Consider square ABCD below:

This square is divided into 8 equal areas. It is clear that P cannot fall anywhere into rectangle ABFE because, although triangle APB and and triangle DPC would have equal bases, DPC would always have a greater height, and thus, a greater area.

It's also clear that P cannot fall into area IFC......To see this, note that if P fell on AC, its distance from AD and AB would be exactly the same. But, if P falls in area IFC, its distance from AD would always be greater than its distance from AB. Thus, triangles APB and APD formed by having P fall into this region would have equal bases, but APD would always have a greater height than triangle APB.

For the same reason, P cannot fall into area EID, because all triangles BPC would have greater heights than triangles APB.

So, P can only fall into region CID. Here triangles APB, BPC, CPD and DPA will all have the same bases, but APB will have a height greater than the other three.

And since CID is 25% of the area of the square, the probabilty that APB is greater than the other three mentioned triangles is ≈ 25%

CPhill Apr 18, 2015

#4**+16 **

I'll try (3), too

Here's the same figure as in the answer to (2)

Using reasoning from the other two answers, P cannot fall into area ABFE, because then, any triangle DPC is greater than any triangle APB. P also cannot fall into area EID because all triangles BPC would be greater than all triangles APB.

If P falls into any of the regions IFC, ICG of IGD, all triangles APB will have greater areas than either triangles BPC or CPD. And these areas are 3/8 of the square ABCD = 37.5%.....so that's the probabilty that tiangle APB will have a greater area than either BPC or CPD.

CPhill Apr 18, 2015

#6**+5 **

Heck....I don't even know if it's correct , or not.......LOL!!!!

Thanks for the confindence boost, though.......

CPhill Apr 19, 2015