+0  
 
+5
1391
9
avatar+1760 

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

Best Answer 

 #3
avatar+78620 
+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
Sort: 

9+0 Answers

 #1
avatar+1760 
+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 $\triangle BCP$?

Mellie  Apr 18, 2015
 #2
avatar+78620 
+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  !!!  ]

 

  

CPhill  Apr 18, 2015
 #3
avatar+78620 
+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+78620 
+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
 #5
avatar+91024 
+5

That is brilliant logic Chris.   :))

Melody  Apr 19, 2015
 #6
avatar+78620 
+5

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

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

 

  

CPhill  Apr 19, 2015
 #7
avatar+91024 
0

Well, it sounds good to me anyway :/

Melody  Apr 19, 2015
 #8
avatar+4663 
+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

MathsGod1  Apr 19, 2015
 #9
avatar+91024 
+15

Thanks MG     (on behalf of Chris)

Melody  Apr 19, 2015

27 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details