Post mathematic related questions here

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

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
 

Vignesh B

Youngling
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.
We can use nC2 right? So, is 10 the answer?
 
OP
rst

rst

Youngling
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.

315 point of intersection
 

Vignesh B

Youngling
Re: mathematic related questions

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
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

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.

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)
 

Vignesh B

Youngling
Re: mathematic related questions

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)
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

Padawan
Re: mathematic related questions

objective question
View attachment 10057

Also plz show some work

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

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

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

Niilesh

Padawan
Re: mathematic related questions

right (then it should be ASA)

I also don't understand I equated the area and got PD=0=PD
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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