Recent content by Charles Link

I would like to give a big thank you to @anuttarasammyak , @pasmith , @Gavran , and @DaveE for their inputs to this thread. I really think many could benefit by seeing how each of them solved this problem and reading through their solutions. Otherwise, I urge the reader to try and see if you...

@Gavran Very good. I think your Vieta calculation above might have come from Wolfram. My apologies if I am incorrect. When I recommended others to try it by hand above, (post 16), the version of the cubic I had, ## x^3-\frac{3x}{2} -\frac{3}{4}=0 ##, (from post 12 above), didn't need the...

@DaveE Excellent. In hindsight, your solution where you pointed out the inverted and translated symmetry of the two essentially identical parabolas is almost obvious, but I failed to recognize it. Thank you very much for your input. <3 <3 and your solution is in complete agreement with...

Please see the two Edits added to post 16 above. I think they may be useful.

@Gavran Very good=Excellent. :) You have an interesting approach where you translated the first curve by ## (a,b) ## until it made contact with the second curve at ## (x_1+a,y_1+b) ##. (To get this to occur you set the discriminant to zero=very clever). I checked over your calculations, and...

Just a couple additional comments: I thought I did pretty well by getting a numerical solution by writing out the partial derivative expressions and solving for ##x_1 ## and ## x_2 ## to first order about ## x=3/2 ##. The exact solutions given by @anuttarasammyak and @pasmith are really a...

@pasmith I think it should read ##-2(x_1+x_2-3)=0 ##. Otherwise, yes, very good=excellent. :) and I think I see where he got the wrong sign: Right before that he has ## y_2=(x+3)^2 ## where it should be a minus sign on the 3.

Yes, sometime I'm going to need to try Vieta's substitution on this cubic and see if I can get the same answers that Wolfram got. :) and perhaps @anuttarasammyak already identified the cubic in post 4 that needs to be solved. It may be simply solving ## 4a^3-6a-3=0 ## to get the ## x_1 ##...

For the calculation with the slopes mentioned in post 8, we have ## m=\frac{2-(4-a^2)}{3/2-a}=\frac{(3-a-3)^2-2}{3-a-3/2}## and these are both ## \frac{a^2-2}{3/2-a}=\frac{1}{2a} ## because of what @anuttarasammyak had previously mentioned in post 4 that comes from the first partial derivative...

@anuttarasammyak Perhaps you also noticed this already, but I find it interesting that with your exact solution, the slopes of the curves are precisely parallel, as they should be, at ## x_1= a ## and ## x_2=3-a ##, with slopes of ## -2 x_1 =-2a ## and ## 2(x_2-3)=-2a ##. Edit: I also believe...

@anuttarasammyak Thank you very much for your solution above. It is excellent. <3 <3

@anuttarasammyak We are going to need to look over your two partial derivative expressions carefully, because your solution is incorrect and not even in the right ballpark. I got, doing it all by hand with no computer that ## (x_1,y_1)=(1.43,1.96)##, and ##(x_2,y_2)=(1.57,2.04) ## for a...

@anuttarasammyak That's one part of what I did to solve it. Those two partial derivative expressions are really too clumsy to solve exactly. I used them though by getting an approximate ## x ## for where the two ## x ## values occur by first solving for the ## x ## where the vertical distance...

The other day a friend of mine gave me the above problem which I found rather interesting. I was able to get a numerical solution. I'll post how I solved it later in the thread. Others may want to try it and see what they come up with.

The above is a photo of a 3'0" snowman that I made today, Monday 3-16-26 in Chicago. It was several degrees below freezing so packing was very difficult but made possible by the warm ground below the snow where the temperature got up to 63 degrees yesterday.