What values of alpha make these curves tangent?

[Ackbach]

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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[Ackbach]

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:

[sp]
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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