Post mathematic related questions here

Niilesh

Niilesh

Padawan
Re: mathematic related questions

A question for you guys

Take 5 points in a plane so that no two of the straight lines joining them are parallel, perpendicular or coincident. From each point perpendeculars are drawn to all the lines joining the other 4 pts. Then find total no. of intersection of these perpendiculars.
 
Niilesh

Niilesh

Padawan
Re: mathematic related questions

I think chess board is a square
yup you are right it is not possible to divide it in 8 squares, I thought he ment 8*8=64 squares
 
V

Vignesh B

Youngling
Re: mathematic related questions

Niilesh said:
A question for you guys

Take 5 points in a plane so that no two of the straight lines joining them are parallel, perpendicular or coincident. From each point perpendeculars are drawn to all the lines joining the other 4 pts. Then find total no. of intersection of these perpendiculars.
Click to expand...
We can use nC2 right? So, is 10 the answer?
 
OP
rst

rst

Youngling
Re: mathematic related questions

Niilesh said:
A question for you guys

Take 5 points in a plane so that no two of the straight lines joining them are parallel, perpendicular or coincident. From each point perpendeculars are drawn to all the lines joining the other 4 pts. Then find total no. of intersection of these perpendiculars.
Click to expand...

315 point of intersection
 
V

Vignesh B

Youngling
Re: mathematic related questions

rst said:
there are 5 points
for drawing a line we require 2 points
so using nC2 , we can draw 10 lines

but it is not related with point of intersection



it requires lot of explanation

its really a tough question
Click to expand...
Ok, my bad.
Care to explain the answer in brief? :p

One from my side too -
How many pairs of integers are there such that a^2 + 6b^2 and b^2+6a^2 are perfect squares?
Source : I found it in one of my books.
 
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OP
rst

rst

Youngling
Re: mathematic related questions

Vignesh B said:
One from my side too -
How many pairs of integers are there such that a^2 + 6b^2 and b^2+6a^2 are perfect squares?
Source : I found it in one of my books.
Click to expand...

you all are asking tough and time consuming question
Is there any way, other than hit and trial method for this question

I think (0,0) is the only pair (by hit and trial method)
 
V

Vignesh B

Youngling
Re: mathematic related questions

rst said:
you all are asking tough and time consuming question
Is there any way, other than hit and trial method for this question

I think (0,0) is the only pair (by hit and trial method)
Click to expand...
Correct.
Only the trivial solution (0,0) works.
Suppose there is a nontrivial solution (a, b). Without loss of generality, we may assume it is a minimal solution, say in the sense that |a|+|b| is as small as possible.
The only squares mod 4 are 0 and 1.
If a is even and b is odd, then
a^2 + 6b^2 = 2 (mod 4),
so a^2 + 6b^2 is not a square. Similarly, if a is odd and b is even we arrive at a contradiction by using the assumption that b^2 + 6a^2 is a square.
If instead a is odd and b is odd, then
a^2 + 6b^2 +3 ( mod 4),
so again we obtain a contradiction.
The only possibility then is that both a & b are even. But if you look at a/2, b/2 then one easily sees that both
(a/2)^2 + 6(b/2)^2
(b/2)^2 + 6(a/2)^2
are squares (since the squares a^2 + 6b^2 and b^2 + 6a^2 are both even, hence divisible by 4). This contradicts the minimality of our original solution.
 
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OP
rst

rst

Youngling
Re: mathematic related questions

objective question
obj.PNG

Also plz show some work

(Also you can ask your mathematics related questions or queries
we will try to solve them)
 
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Niilesh

Niilesh

Padawan
Re: mathematic related questions

rst said:
objective question
View attachment 10057

Also plz show some work
Click to expand...

I am getting ans as 5 :|

can someone point out mistake in the following solution?
PQC is concurrent to PQA by SSS
ABQ is concurrent to CDP by AAS

this means that PC=AP=AQ=QC and PD=QB, now I equated the area and got PD=0=PD. so PQ will be the diagonal so it will be equal to five
 
OP
rst

rst

Youngling
Re: mathematic related questions

Niilesh said:
I am getting ans as 5 :|

can someone point out mistake in the following solution?
PQC is concurrent to PQA by SSS
ABQ is concurrent to CDP by AAS

this means that PC=AP=AQ=QC and PD=QB, now I equated the area and got PD=0=PD. so PQ will be the diagonal so it will be equal to five
Click to expand...

< B= <D (each 90)
AB=CD (opposite sides of rectangle)
what is another angle
 
Niilesh

Niilesh

Padawan
Re: mathematic related questions

rst said:
right (then it should be ASA)

I also don't understand I equated the area and got PD=0=PD
Click to expand...
I found the area of the 4 triangles and equated its sum to 4*3=12
 
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OP
rst

rst

Youngling
Re: mathematic related questions

Then by solving
you will get , 24=24
so by this method you will not be able to find anything
It doesnot mean PD=0
 
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