Calculus, Geometry or Algebra?

Geometry Level 4

x 2 6 x + 13 + x 2 + y 2 + y 2 8 y + 17 \sqrt{x^2 - 6x + 13} + \sqrt{x^2 + y^2} + \sqrt{y^2 - 8y + 17}

Given that x > 0 x > 0 and y > 0 y > 0 are real numbers, if the minimum of the above equation can be expressed as A B A \sqrt{B} , where A A and B B are positive integers, and B B is square free, input the product A B AB as your answer.


The answer is 26.

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4 solutions

Sathvik Acharya
Jan 26, 2021

Define vectors a 1 , a 2 , a 3 , a 4 \overrightarrow{a_1},\;\overrightarrow{a_2},\;\overrightarrow{a_3},\;\overrightarrow{a_4} as, a 1 = ( 3 x ) i ^ + 2 j ^ a 1 = x 2 6 x + 13 a 2 = x i ^ + y j ^ a 2 = x 2 + y 2 a 3 = 1 i ^ + ( 4 y ) j ^ a 3 = y 2 8 y + 17 a 4 = a 1 + a 2 + a 3 = ( 3 x + x + 1 ) i ^ + ( 2 + y + 4 y ) j ^ = 4 i ^ + 6 j ^ \begin{aligned} \overrightarrow{a_1}&=(3-x)\hat{i}+2\hat{j}&&\implies \mid \overrightarrow{a_1}\mid =\sqrt{x^2-6x+13} \\ \\ \overrightarrow{a_2}&=x\hat{i}+y\hat{j}&&\implies \mid \overrightarrow{a_2}\mid =\sqrt{x^2+y^2} \\ \\ \overrightarrow{a_3}&=1\hat{i}+(4-y)\hat{j} &&\implies \mid \overrightarrow{a_3}\mid =\sqrt{y^2-8y+17} \\ \\ \overrightarrow{a_4}&=\overrightarrow{a_1}+\overrightarrow{a_2}+\overrightarrow{a_3} \\ &=(3-x+x+1)\hat{i}+(2+y+4-y)\hat{j} \\ &=4\hat{i}+6\hat{j} \end{aligned} 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 \mid \overrightarrow{a_1}\mid+\mid \overrightarrow{a_2}\mid+\mid \overrightarrow{a_3}\mid\;\ge\; \mid \overrightarrow{a_4}\mid x 2 6 x + 13 + x 2 + y 2 + y 2 8 y + 17 4 2 + 6 2 = 52 \implies \sqrt{x^2-6x+13}+\sqrt{x^2+y^2}+\sqrt{y^2-8y+17}\ge \sqrt{4^2+6^2}=\sqrt{52} Therefore, the four vectors must lie on a straight line for equality to be achieved and the minimum value of the expression is 52 = 2 13 A B = 26 \sqrt{52}=2\sqrt{13}\implies AB=\boxed{26}

David Vreken
Jan 26, 2021

Imagine the following diagram in which A B = 2 AB = 2 , B D = 3 BD = 3 , D F = 4 DF = 4 , and F G = 1 FG = 1 , and a path from A A to G G that passes through B D BD at C C and D F DF at E E :

If x = C D x = CD and y = D E y = DE , 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 + 13 + x 2 + y 2 + y 2 8 y + 17 AC + CE + EG = \sqrt{2^2 + (3 - x)^2} + \sqrt{x^2 + y^2} + \sqrt{(4 - y)^2 + 1^2} = \sqrt{x^2 - 6x + 13} + \sqrt{x^2 + y^2} + \sqrt{y^2 - 8y + 17} , 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 AH = AB + BH = AB + FD = 2 + 4 = 6 , and since H G = H F + F G = B D + F G = 3 + 1 = 4 HG = HF + FG = BD + FG = 3 + 1 = 4 , by Pythagorean Theorem on A H G \triangle AHG the minimum path will be A G = A H 2 + H G 2 = 6 2 + 4 2 = 2 13 AG = \sqrt{AH^2 + HG^2} = \sqrt{6^2 + 4^2} = 2\sqrt{13} .

Therefore, A = 2 A = 2 , B = 13 B = 13 , and A B = 26 AB = \boxed{26} .

Pi Han Goh
Jan 27, 2021

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 ) f(x,y) denote the value of the expression that we want to minimize. Clearly f ( x , y ) f(x,y) has no maximum value because we can set either x x or y y to be unboundedly large. In other words, if f ( x , y ) 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 x 2 + y 2 + x 3 x 2 6 x + 13 = 0 (1) \dfrac{\partial }{\partial x} f(x,y) = 0 \quad \Rightarrow\quad \dfrac x{\sqrt{x^2 + y^2}} + \dfrac{x-3}{\sqrt{x^2-6x+13}} = 0 \tag1

y f ( x , y ) = 0 y x 2 + y 2 + y 4 y 2 4 y + 17 = 0 (2) \dfrac{\partial }{\partial y} f(x,y) = 0 \quad \Rightarrow\quad \dfrac y{\sqrt{x^2 + y^2}} + \dfrac{y-4}{\sqrt{y^2-4y+17}} = 0 \tag2

From ( 1 ) (1) , we can set y > 0 y>0 as the subject (I've omitted the tedious algebra):

y = 2 x x 2 6 x + 9 (3) y = \dfrac {2x}{\sqrt{x^2-6x+9}} \tag3

Likewise, from ( 2 ) (2) , set x > 0 x>0 as the subject: x = ± y y 4 (4) x = \pm \dfrac y{y-4} \tag4

Substitute ( 4 ) (4) into ( 3 ) (3) yields ( x , y ) = ( 7 3 , 7 ) , ( 5 , 5 ) , ( 5 3 , 5 2 ) , ( 7 , 7 2 ) . (x,y) = (\tfrac73, 7), (5,5), (\tfrac53, \tfrac52), (7, \tfrac72) .

Trial and error shows that among these 4 critical points, ( x , y ) = ( 5 3 , 5 2 ) (x,y) =(\tfrac53, \tfrac52) gives a smallest value of f ( x , y ) f(x,y) .

Hence, min ( f ( x , y ) ) = f ( 5 3 , 5 2 ) = 2 13 A × B = 2 × 13 = 26 . \min(f(x,y)) = f(\tfrac53, \tfrac52) = 2\sqrt{13} \Rightarrow A \times B = 2\times13=\boxed{26}.

Jason Gomez
Jan 27, 2021

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 = y= 4 a a + 1 \frac{4a}{a+1}

When y=b the sum of the next two distances (from origin to point, point to x=3) is minimised when x = x= 3 b b + 2 \frac{3b}{b+2}

Solving these two equations with each other ( putting the value of x in a ) , we get x , y = x,y= 5 3 \frac{5}{3} , 5 2 \frac{5}{2}

Putting these values in the equation given gives 2 13 2\sqrt{13}

@Michael Huang You can add physics to the title too now

Jason Gomez - 4 months, 2 weeks ago

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Intriguing insight, using physics! ^.^

Michael Huang - 4 months, 2 weeks ago

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