Is this even possible?
Consider the
4
distinct points
(
a
,
a
1
)
,
(
b
,
b
1
)
,
(
c
,
c
1
)
,
(
d
,
d
1
)
lying on a circle, centered at the origin, of radius
4
.
Find a × b × c × d .
The answer is 1.
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6 solutions
This is such an elegant solution, Daniel. It gets my vote. :)
Each point is a solution of the equation x 2 + x 2 1 = 1 6 .
Rearranging gives us the equation x 4 − 1 6 ∗ x 2 + 1 = 0 . Since a , b , c and d are given as being distinct, we know that they represent the 4 roots of this equation. By Vieta's, we thus know that a ∗ b ∗ c ∗ d = 1 .
Before concluding, however, we should verify that these 4 roots are real. Now the equation can be treated as a quadratic in x 2 , giving us that
x 2 = ( 2 1 ) ∗ ( 1 6 ± 2 5 6 − 4 ) = 8 ± 3 7 .
Since 8 > 3 7 we will have 4 distinct real roots, and thus we can have 4 distinct points as described in the question.
The solution is thus 1 .
Same thing I did. Great problem!
What's wrong with my solution?
x 2 + x 2 1 = 1 6 ⟹ ( x + x 1 ) 2 = 1 6 + 2 = 1 8
\displaystyle \implies x+\frac{1}{x}=\pm 3\sqrt{2}\stackrel{\times x}\implies x^2\mp 3\sqrt{2}x+1=0
⟹ ( x ∓ 1 . 5 2 ) 2 = 3 . 5 ⟹ x ∓ 1 . 5 2 = ± 3 . 5
⟹ x = 0 . 5 ( ± 7 ± 3 2 )
The points are all distinct, hence we can only have:
a b c d = 0 . 0 6 2 5 ( 7 + 3 2 ) ( 7 − 3 2 ) ( − 7 + 3 2 ) ( − 7 − 3 2 ) = 6 0 . 0 6 2 5
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That looks good down to the 3rd line, where you should have
x ± 1 . 5 2 = ± 3 . 5 ,
making the fourth line
x = 2 1 ( ± 3 ± 7 ) .
You had me worried there for a moment. :)
I have a query in the roots step... Sorry, I do not have superscript enabled in my phone so I'll try my best to use brackets...
x^2= [ 16 (+/-) [ {(4)^4} - 4] ^(1/2)] ] /2 ..... x^2= [ 16 (+/-) [ {(4^3) (4-1)} ^(1/2)] ] /2 ...... x^2= [ 16 (+/-) [ { (4^3) (3) } ^(1/2)] ] /2 ...... x^2= 8 (+/-) 4 [3^(1/2)] ............... ............... In your value of root you get x^2= 8 (+/-) 3 [7^(1/2)].... ..... ..... I have to agree with your value as well, as it makes sense but mine does too..... I'm only a beginner can you help me find the discrepancy here?
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In your second step you factored ( 4 4 − 4 ) to be ( 4 3 ) ( 4 − 1 ) , when in fact it should be 4 ∗ ( 4 3 − 1 ) . This then equals 4 ∗ 6 3 = 4 ∗ 9 ∗ 7 = 3 6 ∗ 7 . The square root of this is 6 7 , and so
x 2 = ( 2 1 ) ∗ ( 1 6 ± 6 7 ) = 8 ± 3 7 .
Note that a , b , c , d are actually roots of the equation
x 2 + x 2 1 = 4 2 ....
This comes by equation of the circle which will be of the form ( x − a ) 2 + ( y − b ) 2 = r 2 , if ( a , b ) is it's center and r is radius.
Thus we have x 2 + x 2 1 = 1 6
and hence x 4 − 1 6 x 2 + 1 = 0 (obviously this has no negative root, giving no-nonreal solutions for x )
Product of all the roots of this equation is the constant term and it is 1
You should reference Pythag rather than ( x − a ) 2 + ( y − b ) 2 = r 2 , since there is no use for the equation of a circle when it's centered on the origin.
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Why no use, it's just this time a = b = 0 , so x 2 + y 2 = r 2 , the Pythagoras.... but I felt because we talk about circles, if at all there are some people who don't know this (lol, seems a joke !) , then they will come to know that the equation looks like it :P
Good point @Finn Hulse ;)
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Yeah, I agree. Although I do feel bad for the people who don't know the general form of a circle or who can't derive it on the spot. :P
Aditya, you should also mention that the quartic equation has an even degree of exponent or else you would have to negate the constant term/leading coefficient if it's odd power.
All the 4 points are of the form ( x , 1 / x ) . They have to satisfy the equation of the circle of radius r with origin as center i.e.
x 2 + y 2 = r 2 .
x 2 + ( 1 / x 2 ) = r 2
x 4 − r 2 . x 2 + 1 = 0
The above equation has four roots a,b,c and d. The product of the roots is nothing but the constant term of the equation which is 1. Therefore a b c d = 1
This is a great way of solving that!
We could infer the answer without being given the radius r = 4 .
Not necessarily true. If the radius of the circle is too small, then the 4 points of intersection would not exist. You could argue that the varying the value of the radius above a lower constraint ( > root 2) would not affect the solution, but some information about the radius would be necessary.
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Can one say, that even if there is no real intersection (because the circle is too small) there will still be intersection with complex coordinates? Because the complex roots to actually multiply out to '1'! So the circle radius seems immaterial. Just a thought . . .
x^2 + y^2 = R^2
y = 1/x R^2 = 16
x^2 + 1/x^2 = 16 x^4 - 16x^2 + 1 = 0 let x^2 = A A^2 - 16 A + 1 = 0 solving quadratic gives A = 8 ± 3√7 since x^2 = A ----> x = ±√(8 ± 3√7) meaning there are exactly 4 distinct such points. √(8 +3√7) √(8 - 3√7) (-√(8 +3√7)) (-√(8 - 3√7)) = (8 +3√7)(8 - 3√7) = 1
Our points are the intersection of the two graphs x 2 + y 2 = 1 6 and x y = 1 . Note that these two graphs are symmetric along the line y = x ; thus the intersections are symmetric along y = x , so ( a , a 1 ) = ( b 1 , b ) and ( c , c 1 ) = ( d 1 , d ) and we have a = b 1 and c = d 1 .
Thus, the product of the 4 variables is a ⋅ a 1 ⋅ c ⋅ c 1 = 1 .