x 2 − 6 x + 1 3 + x 2 + y 2 + y 2 − 8 y + 1 7
Given that x > 0 and y > 0 are real numbers, if the minimum of the above equation can be expressed as A B , where A and B are positive integers, and B is square free, input the product A B as your answer.
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Imagine the following diagram in which A B = 2 , B D = 3 , D F = 4 , and F G = 1 , and a path from A to G that passes through B D at C and D F at E :
If x = C D and y = D E , then by the Pythagorean Theorem on the three right triangles, the length of the path is A C + C E + E G = 2 2 + ( 3 − x ) 2 + x 2 + y 2 + ( 4 − y ) 2 + 1 2 = x 2 − 6 x + 1 3 + x 2 + y 2 + y 2 − 8 y + 1 7 , the given equation.
This path is a minimum when it is a straight line:
Since A H = A B + B H = A B + F D = 2 + 4 = 6 , and since H G = H F + F G = B D + F G = 3 + 1 = 4 , by Pythagorean Theorem on △ A H G the minimum path will be A G = A H 2 + H G 2 = 6 2 + 4 2 = 2 1 3 .
Therefore, A = 2 , B = 1 3 , and A B = 2 6 .
Both Sathvik Acharya and David Vreken have posted remarkable solutions. To accentuate how incredible their solutions are, I will post an atrocious solution.
Let f ( x , y ) denote the value of the expression that we want to minimize. Clearly f ( x , y ) has no maximum value because we can set either x or y to be unboundedly large. In other words, if f ( x , y ) has a critical point, it must be a minimum value.
At its critical points, both its partial derivatives must be 0, ∂ x ∂ f ( x , y ) = 0 ⇒ x 2 + y 2 x + x 2 − 6 x + 1 3 x − 3 = 0 ( 1 )
∂ y ∂ f ( x , y ) = 0 ⇒ x 2 + y 2 y + y 2 − 4 y + 1 7 y − 4 = 0 ( 2 )
From ( 1 ) , we can set y > 0 as the subject (I've omitted the tedious algebra):
y = x 2 − 6 x + 9 2 x ( 3 )
Likewise, from ( 2 ) , set x > 0 as the subject: x = ± y − 4 y ( 4 )
Substitute ( 4 ) into ( 3 ) yields ( x , y ) = ( 3 7 , 7 ) , ( 5 , 5 ) , ( 3 5 , 2 5 ) , ( 7 , 2 7 ) .
Trial and error shows that among these 4 critical points, ( x , y ) = ( 3 5 , 2 5 ) gives a smallest value of f ( x , y ) .
Hence, min ( f ( x , y ) ) = f ( 3 5 , 2 5 ) = 2 1 3 ⇒ A × B = 2 × 1 3 = 2 6 .
I am not gonna put a full explanation and it’s gonna use geometry and light ( physics )
The sum required is the sum of the distances in green
Here’s a Desmos link which is a little interactive on what the green distances are
If you notice it seems like there is a light ray coming from the origin bouncing off the point and landing on x=3 and y=4
If u keep the x coordinate constant and change the y coordinate you will also notice that point at which the light ray hits y=4 is constant
We can use fermats principle and see that when x=a acting as a mirror (purple vertical line) the first two distances ( from origin to point and point to y=4 ) can be minimised for a given x=a
Vice versa with y=b
The equations formed from these two will be that ( using similarity of triangles )
When x=a the sum of first two distances is minimised when y = a + 1 4 a
When y=b the sum of the next two distances (from origin to point, point to x=3) is minimised when x = b + 2 3 b
Solving these two equations with each other ( putting the value of x in a ) , we get x , y = 3 5 , 2 5
Putting these values in the equation given gives 2 1 3
@Michael Huang You can add physics to the title too now
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Define vectors a 1 , a 2 , a 3 , a 4 as, a 1 a 2 a 3 a 4 = ( 3 − x ) i ^ + 2 j ^ = x i ^ + y j ^ = 1 i ^ + ( 4 − y ) j ^ = a 1 + a 2 + a 3 = ( 3 − x + x + 1 ) i ^ + ( 2 + y + 4 − y ) j ^ = 4 i ^ + 6 j ^ ⟹ ∣ a 1 ∣ = x 2 − 6 x + 1 3 ⟹ ∣ a 2 ∣ = x 2 + y 2 ⟹ ∣ a 3 ∣ = y 2 − 8 y + 1 7 In a general quadrilateral, the sum of the lengths of any three sides is greater than the length of the fourth side , so we have, ∣ a 1 ∣ + ∣ a 2 ∣ + ∣ a 3 ∣ ≥ ∣ a 4 ∣ ⟹ x 2 − 6 x + 1 3 + x 2 + y 2 + y 2 − 8 y + 1 7 ≥ 4 2 + 6 2 = 5 2 Therefore, the four vectors must lie on a straight line for equality to be achieved and the minimum value of the expression is 5 2 = 2 1 3 ⟹ A B = 2 6