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 #1
avatar+9488 
+2

Let's start by expanding both sides and combining like terms:

 

(a+1b)(b+1c)(c+1a)=(1+1a)(1+1b)(1+1c) (ab+ac+1+1bc)(c+1a)=(1+1b+1a+1ab)(1+1c) abc+b+a+1c+c+1a+1b+1abc=1+1c+1b+1bc+1a+1ac+1ab+1abc abc+b+a+1c+c+1a+1b+1abc=1+1c+1b+1bc+1a+1ac+1ab+1abc abc+b+a+c=1+1bc+1ac+1ab abc+a+b+c=1+aabc+babc+cabc abc+a+b+c=1+a+b+cabc

 

 

At this point, let's substitute   x   in for   a + b + c   and   13   in for   abc    so we want to solve this for  x:

 

13+x=1+x13 169+13x=13+x 13xx=13169 12x=156 x=13_

06.10.2019
 #2
avatar+9488 
+2

 

See:  https://www.desmos.com/calculator/sln5zuurfx

 

We know that

 

AP / PC  =  1 / 3

                              We are also given that  AP = 10   so we can substitute  10  in for  AP

10 / PC  =  1 / 3

                              If it isn't already clear that  PC  must be  30 , we can invert both sides

PC / 10  =  3 / 1

                              and then multiply both sides of the equation by  10

PC  =  3 / 1 * 10

 

PC  =  30

 

And now we can find the length of  AC

 

AC   =   AP + PC   =   10 + 30   =   40

 

The diagonals of a square are the same length and bisect each other so...

 

h   =   AC / 2   =   40 / 2   =   20       where  h  is the height of triangle BPC

 

Area of triangle BPC   =   ( 1/2 )( PC )( h )   =   ( 1/2 )( 30 )( 20 )   sq units   =   300   sq units

05.10.2019