What values of alpha make these curves tangent?

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In summary, the conversation discusses the values of alpha that would make two intersecting curves tangent. It is mentioned that two curves are tangent when they touch at a single point and have the same slope. The process of finding the value of alpha involves using the derivatives of the curves and solving for the satisfying value. It is also noted that there can be multiple values of alpha that make the curves tangent, which can have significance in understanding the relationship between the curves and solving optimization problems.
  • #1
Ackbach
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Here is this week's POTW:

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Find all values of $\alpha$ for which the curves $y = \alpha x^2 +\alpha x + \dfrac{1}{24}$ and $x = \alpha y^2 + \alpha y + \dfrac{1}{24}$ are tangent to each other.

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  • #2
Congratulations to Opalg and castor28 for their correct solutions to this week's POTW, which was Problem A-1 in the 2007 Putnam Archive. Opalg's solution follows:

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If the curves $$(1)\qquad y = \alpha x^2 + \alpha x + \tfrac1{24}$$ and $$(2)\qquad x = \alpha y^2 + \alpha y + \tfrac1{24}$$ touch at the point $(x,y)$ then they must have the same gradient at that point. So, differentiate the equations to get $y' = 2\alpha x + \alpha$ and $1 = (2\alpha y + \alpha)y'$. Therefore $$(3)\qquad 1 = \alpha^2(2x+1)(2y+1).$$ Next, subtract (2) from (1) to get $y-x = \alpha (x^2 - y^2) + \alpha(x-y),$ which factorises as $$(4)\qquad (x-y)\bigl(\alpha(x+y+1) + 1\bigr) = 0.$$ One of the factors in (4) must be zero, so there are two cases to look at.Case 1: $x-y=0$. In that case, (3) becomes $$(5)\qquad \alpha(2x+1) = \pm1.$$ Substitute that in (1) (and use the fact that $y=x$) to get $$x = \pm\frac{x^2+x}{2x+1} + \frac1{24},$$ $$ 2x^2 + x = \pm(x^2+x) + \tfrac1{24}(2x+1).$$ There are then two subcases, depending on which $\pm$ sign is used.

Subcase 1+. If $ 2x^2 + x = x^2+x + \tfrac1{24}(2x+1)$ then $x^2 - \frac1{12}x - \frac1{24} = 0$, a quadratic with solutions $x = \frac14$ and $x = -\frac16.$ The corresponding values for $\alpha$ are given from (5) as $\alpha = \frac23$ and $\alpha = \frac32$.

Subcase 1-. If $ 2x^2 + x = -(x^2+x) + \tfrac1{24}(2x+1)$ then $24(3x^2 + 2x) = 2x+1$, so $72x^2 + 46x - 1 = 0.$ This time the quadratic does not factor neatly. So you have to use the formula to get the solutions as $x = \dfrac{-23 \pm\sqrt{601}}{72}.$ The corresponding values for $\alpha$ are given from (5) as $\alpha = \dfrac{-36}{13 \pm\sqrt{601}}.$

The other possibility from (4) is

Case 2: $\alpha(x+y+1) = -1$. Square that equation, to get $\alpha^2(x+y+1)^2 = 1$. Comparing that with (3), you see that $(x+y+1)^2 = (2x+1)(2y+1).$ Therefore $$x^2 + 2xy + y^2 + 2x + 2y + 1 = 4xy + 2x + 2y + 1,$$ $$ x^2 - 2xy + y^2 = 0,$$ $$ (x-y)^2 = 0.$$ So $x=y$, and Case 2 reduces to Case 1.

In conclusion, there are four values of $\alpha$ for which the given curves are tangent to each other. Their (approximate) numerical values are $-0.9596$, $0.6667$, $1.5$, $3.1263$. You can see them geometrically by dragging the slider in this graphic: [DESMOS]advanced: {"version":5,"graph":{"squareAxes":false,"viewport":{"xmin":-2.5,"ymin":-2.5,"xmax":2.5,"ymax":2.5}},"expressions":{"list":[{"id":"graph1","type":"expression","latex":"y\\ =\\ ax^2\\ +\\ ax\\ +\\ \\frac{1}{24}","color":"#2d70b3"},{"id":"2","type":"expression","latex":"x\\ =\\ ay^2\\ +\\ ay\\ +\\ \\frac{1}{24}","color":"#388c46"},{"id":"3","type":"expression","latex":"a=3.14","hidden":true,"color":"#fa7e19","sliderMin":"-2","sliderMax":"4","sliderHardMin":true,"sliderHardMax":true,"sliderInterval":"0.01"}]}}[/DESMOS]
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FAQ: What values of alpha make these curves tangent?

What values of alpha make these curves tangent?

The question is referring to two intersecting curves and asking for the value of alpha that would make them tangent to each other.

What does it mean for two curves to be tangent?

When two curves are tangent, it means that they touch each other at a single point and have the same slope at that point.

How do you find the value of alpha that makes two curves tangent?

This can be done using the derivative of the curves. Set the derivatives equal to each other and solve for the value of alpha that satisfies the equation.

Can there be more than one value of alpha that makes two curves tangent?

Yes, it is possible for there to be multiple values of alpha that make two curves tangent. This would result in multiple points of tangency between the curves.

What is the significance of finding the value of alpha that makes two curves tangent?

It can provide useful information about the relationship between the two curves and can also be used to solve optimization problems in mathematics and physics.

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