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Unravelling an inequality problem

Let \(a\) and \(b\) be positive reals.

Consider all pairs of positive reals \( (x,y) \) subject to

\[ \frac{ a}{x} + \frac{b}{y} = 1. \]

What is the minimum value of

\[ x + y + \sqrt{ x^2 + y^2 } ? \]


Some of you might recognize this problem, with the specific case of \( a = 1, b = 8 \).

The purpose of this note is to unravel this question, working from what has been given to try and find a logical path.


Observation 1: Recognize the final expression of \( x + y + \sqrt{ x^2 + y^2 } \) as the perimeter of a right angled triangle with legs \(x\) and \(y\). Let's consider a right triangle with vertices \( O = (0,0), X = (x,0), Y = (o,y) \).

Observation 2: The condition that \( \frac{a}{x} + \frac{b}{y} = 1 \) implies that the point \( P = (a,b) \) lies on the line \( XY \). This is the "intercept-intercept" equation of a line, which is often not taught explicitly.

Hence, we can rephrase the question as follows:

Consider all right triangles with legs on the x and y axis, and whose hypotenuse passes through the point \(P = (a,b) \). Which of these triangle has the smallest perimeter?

Consider a circle \(\Gamma\) that lies that passes through the point \(P\), and is tangential to the x-axis and the y-axis. Specifically, if the circle has radius \(r\), it has center \(T = (r,r) \), and length \(TP \) gives us

\[ (r-a) ^2 + (r-b)^2 = r^2 \Rightarrow r^2 - (2a+2b)r + (a^2+b^2) = 0. \]

This has solutions \( r = (a+b) \pm \sqrt{ 2ab } \). The solution corresponding to a minus sign would correspond to a circle whose center is "between" \(O\) and \(P\). The solution corresponding to the plus sign would correspond to the circle "on the other side". Let the larger circle be \( \Gamma \) (see diagram below).

Claim: The triangle with minimal perimeter is tangential to \(\Gamma\) at \(P\).

Let's study this triangle. In the diagram, this triangle is denoted by \( O A B \) (blue line). Observe that

\[ OA + AB + BO = OA + AP + PB + BO = OA + AQ + RB + BO = OQ + OR = 2r. \]

image

image

Proof of claim: Take any other line \(\ell\) (yellow line above) through \(P\), which cuts the axis at \( x,y\). Then, it is not tangential to \(\Gamma\), hence cuts \(\Gamma\) again.

Take a parallel line to \(\ell\), which is tangential to \(\Gamma\) at \(S\), which is contained beneath \( xy \). This line cuts the axis at \( x^*, y^* \). Then, perimeter of \(Oxy \, \) is greater than perimeter of \( O x^* y^* \) (by scaling), which is equal to \( OQ + OR \) (similar argument as perimeter of \(OAB\)), which is equal to perimeter of \( O A B \).

Hence, the minimal perimeter occurs in triangle \(OAB \). \(_ \square \)

In conclusion, the minimal value is \( 2[ a + b + \sqrt{2ab}] \). In the case where \(a=1, b=8 \), we get \( 2 [ 1 + 8 + \sqrt{16} ] = 26 \).

Note by Calvin Lin
2 years, 10 months ago

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Follow up question:

Let \(a\) and \(b\) be positive reals.

Consider all pairs of positive reals \( (x,y) \) subject to

\[ \frac{ a}{x} + \frac{b}{y} = 1. \]

What is the range of values of

\[ x + y - \sqrt{ x^2 + y^2 } ? \] Calvin Lin Staff · 2 years, 10 months ago

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@Calvin Lin is it always greater than 0 ? without an upper bound ? ....... maybe its the variant of triangle inequality. Abhinav Raichur · 2 years, 7 months ago

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@Calvin Lin How do you justify that the circle must be tangent to the line at \((a,b)\) and not any other point [since while calculating the value of \(r\) you have taken the distance from the point \((a,b)\)] ?

Also, I have figured out something that might be useful to solve this and your follow up question and can be thought about independently too. However, can you help me since I'm stuck.

Problem 1
Again, we set up the coordinates as done by Sir Calvin and mark all points on the Cartesian Plane.
Now, consider the formula for ex-radius of a triangle,

\[R_{A}=\frac{\Delta}{s-a}\]
(where \(R_{A}\) is the ex-radius opposite to vertex A, \(\Delta\) is the area of the triangle and \(s\) is its semi perimeter.)

also, given expression to minimize is

\(x+y+\sqrt{x^2+y^2}\)

\(=\frac{2xy}{x+y-\sqrt{x^2+y^2}}\)

\(=2R_{o}\)

Now the question is to minimize the ex-radius of the triangle. Clearly, its center must be at \((R_{o},R_{o})\), so that it is tangent to the coordinate axes (by definition of the ex-circle). It also has to be tangent to the line (from outside the triangle) again by the definition of the ex-circle.

However, I can't reason out why the extremum will be achieved when the ex-circle is tangent at the point \((a,b)\) and why will it be the minimum value ?

Problem 2
Again, we set up the coordinates as done by Sir Calvin and mark all points on the Cartesian Plane.
Now, consider the formula for in-radius of a triangle,

\[r=\frac{\Delta}{s}\]

also, given expression is

\(x+y-\sqrt{x^2+y^2}\)

\(=\frac{2xy}{x+y+\sqrt{x^2+y^2}}\)

\(=2r\)

Now the question is to find the range of the in-radius of the triangle. Clearly, its center must be at \((r,r)\), so that it is tangent to the coordinate axes (by definition of the in-circle). It also has to be tangent to the line (from inside the triangle) again by the definition of the in-circle.

And again I face the same problem.

I'll be glad if you'd help me out. Thanks! Ishan Singh · 2 years, 10 months ago

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@Ishan Singh I've edited the claim to clarify that it is tangential at \(P\). The proof of the claim is below the image.

Nice way to recognize that the relevant quantity is the inradius. Now, you will need to relate it to a similar claim (albeit for a maximum and for a minimum). Calvin Lin Staff · 2 years, 10 months ago

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Beautiful. Adrian Neacșu · 2 years, 10 months ago

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Good Observation and Solution. I never thought this way. Shreyash S · 2 years, 10 months ago

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Could you please explain why minus sign ( from the value of r that we get from solving quadratic) didn't satisfy this problem ? Ty Sg · 2 years, 9 months ago

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@Ty Sg The line \( y = x \) is the angle bisector of \(BOA\). There are 2 circles which satisfy the conditions of tangency, namely the incircle and the excircle opposite \(O\). The smaller radius corresponds to the incircle, which we do not want, because it didn't satisfy the condition of being outside / on the other side. Calvin Lin Staff · 2 years, 9 months ago

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And of course, this problem was set up knowing these facts, and then engineering it to obscure the setup. Calvin Lin Staff · 2 years, 10 months ago

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Am I wrong if I say \( \dfrac{X}{a} + \dfrac{Y}{b}=1\) with a, the x-intersect,..b, the y-intersect ?
OR is it \( \dfrac{a}{X} + \dfrac{b}{Y}=1\) ? Niranjan Khanderia · 2 years, 10 months ago

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@Niranjan Khanderia Note that the notation used in the question is not the standard variables that you use in an equation.

In particular, the point that lies on the line is \( (a,b) \) (and not \((x,y)\) ), and the x-intercept is \(x\) (and not \(a\)), and the y-intercept is \(y\) (and not \(b\)).

So, you have the right idea, but just need to check the variables accordingly. Calvin Lin Staff · 2 years, 10 months ago

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@Calvin Lin Thanks. I was not able to understand your notation. Niranjan Khanderia · 2 years, 10 months ago

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