+0  
 
+5
2987
9
avatar+1807 

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?

 Apr 18, 2015

Best Answer 

 #3
avatar+111438 
+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%

 

  

 Apr 18, 2015
 #1
avatar+1807 
+3

A point $P$ is chosen at random in the interior of equilateral triangle $ABC$. What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP$?

 Apr 18, 2015
 #2
avatar+111438 
+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 exactly on EF, which would make both triangles equal in area.......thus the need for the " ≈  "    symbol  !!!  ]

 

  

 Apr 18, 2015
 #3
avatar+111438 
+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
avatar+111438 
+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.

 

 

  

 Apr 18, 2015
 #5
avatar+110208 
+5

That is brilliant logic Chris.   :))

 Apr 19, 2015
 #6
avatar+111438 
+5

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

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

 

  

 Apr 19, 2015
 #7
avatar+110208 
0

Well, it sounds good to me anyway :/

 Apr 19, 2015
 #8
avatar+4695 
+10

Same here!

 

I don't understand it. (Well just. A bit)

 

But every time you commented a new try I'd be staring.

 

That is some good logic

 Apr 19, 2015
 #9
avatar+110208 
+15

Thanks MG     (on behalf of Chris)

 Apr 19, 2015

12 Online Users

avatar
avatar
avatar