# Thread: Tangent twice -- challenge problem

1. ## Tangent twice -- challenge problem

Find the equation of the line which is tangent to $\displaystyle y=-{x}^{4}+18\,{x}^{3}-97\,{x}^{2}+180\,x-52$ at exactly two distinct points. Also find the points of tangency. The graph, shown with no axes and different vertical scale than horizontal scale is shown below. Note: this is just for fun, I am not looking for help.

2. Well, I tried, and after some tedious calculations, I got this...

Equation of the line that contains these two points:

$\displaystyle y = 36.46062956x + 2.620763081$

Am I right?

EDIT:
Points of tangency:
$\displaystyle C_{1} = 1.450646899$
$\displaystyle C_{2} = 8.321649395$

3. Originally Posted by Chop Suey
Well, I tried, and after some tedious calculations, I got this...

Equation of the line that contains these two points:

$\displaystyle y = 36.46062956x + 2.620763081$

Am I right?

EDIT:
Points of tangency:
$\displaystyle C_{1} = 1.450646899$
$\displaystyle C_{2} = 8.321649395$
Actually, the numbers should work out nicely, in fact I have chosen coefficients so the coordinates of the points of tangency are integers. So, back to the drawing board

4. Just curious, but are you aware that this exact question is posted on the AoPs forum? That is why I cannot answer it, for I would be "cheating" in a sense, for after seeing the solution I cannot say how much I would have been able to do on my own.

5. Originally Posted by Mathstud28
Just curious, but are you aware that this exact question is posted on the AoPs forum? That is why I cannot answer it, for I would be "cheating" in a sense, for after seeing the solution I cannot say how much I would have been able to do on my own.
I posted the same question to yahoo answers forum, but the only answer there was very incomplete, so I thought some people here might enjoy it. I am not familiar with "AoPs forum" but if that is different and it appeared with the same coefficients they probably took it from my yahoo post. On the other hand the problem, probably with different coefficients has certainly been around for a while -- I gave a version to my students around the year 2000, but no one got it.

Thank you for "not cheating". BTW, the solution sketched on the yahoo answers forum was quite different than the way I did it myself, so you could try some method different than what you saw. Or try to figure out how to design the problem so it will have integer answers -- a little number theory to pick coefficients.

6. Originally Posted by jbuddenh
I posted the same question to yahoo answers forum, but the only answer there was very incomplete, so I thought some people here might enjoy it. I am not familiar with "AoPs forum" but if that is different and it appeared with the same coefficients they probably took it from my yahoo post. On the other hand the problem, probably with different coefficients has certainly been around for a while -- I gave a version to my students around the year 2000, but no one got it.

Thank you for "not cheating". BTW, the solution sketched on the yahoo answers forum was quite different than the way I did it myself, so you could try some method different than what you saw. Or try to figure out how to design the problem so it will have integer answers -- a little number theory to pick coefficients.
If I have time I will try . And as for the AoPs forum, it is probably the one with the most Mathematicians per capita. It's a good site.

7. Hello,

I didn't know it had integer answers... I did it the rough way because I had no idea where to start

$\displaystyle f'(a)(x-a)+f(a)=f'(b)(x-b)+f(b)$

$\displaystyle xf'(a)-af'(a)+f(a)=xf'(b)-bf'(b)+f(b)$

$\displaystyle \Rightarrow \left\{\begin{array}{ll} f'(a)=f'(b) & \text{that's logical, I know} \\ f(a)-f(b)=af'(a)-bf'(b) \end{array} \right.$

This is the first equation of the system, the second one is presumed to be nasty

8. Hello Moo,

That was certainly a good start. As you noted (b-a) factors out of your first equation. If you pursue the 'presumed nasty' second equation you will find that (b-a)^2 factors out of it. What results then is a system of two quadratic equations in a,b which in fact are intersecting ellipses. They are actually not too difficult to solve. Since nothing done so far distinguishes a from b, if (a,b) is a solution so is (b,a). Two of the four intersection points are extraneous -- for reasons that are not yet clear to me.

I did the problem by a different method, which avoids the extraneous solutions but still has some messy algebra -- I will give a hint for that method, if nobody posts a complete solution in a few more days.

9. Hello everybody!

This problem is really good an interesting as well.

Here's what I figured out:

let (x_1;f(x_1)) and (x_2,f(x_2)) are two distinct points of the graph of f(x)

the slope between them is defined as: m = [f(x_2) - f(x_1)]/[x_2 - x_1]

but it can be also defined via the 1st derivative: m = f(x_1) = f(x_2)

so the system: [f(x_2) - f(x_1)]/[x_2 - x_1] = f(x_1) = f(x_2)

should provide us with the correct answer

However the calculations and the algebra is terrible (though I haven't got time yet to try)

so, what do you think?

10. Originally Posted by Marine
Hello everybody!

This problem is really good an interesting as well.

Here's what I figured out:

let (x_1;f(x_1)) and (x_2,f(x_2)) are two distinct points of the graph of f(x)

the slope between them is defined as: m = [f(x_2) - f(x_1)]/[x_2 - x_1]

but it can be also defined via the 1st derivative: m = f(x_1) = f(x_2)

so the system: [f(x_2) - f(x_1)]/[x_2 - x_1] = f(x_1) = f(x_2)

should provide us with the correct answer

However the calculations and the algebra is terrible (though I haven't got time yet to try)

so, what do you think?
---------------------------------
Looks good. Go for it! Be aware that (x_2 - x_1) may be a factor of some of your polynomials, and presumably x_2 - x_1 is non-zero so the other factor is the relevant one. This approach is actually similar to what Moo did above. My method was different but I see no reason why your approach should not work.

11. I think this problem has been out here long enough. My approach was to consider a straight line intersecting a quartic, where one would normally expect four solutions, but to use the fact that in this case there are two double roots. The one is automatic if you start with a line tangent at one point. I'm not sure if that is a hint or not.

Tomorrow I will post a solution to the more general problem of finding a line that is tangent to $\displaystyle y=a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+e$ at two different points. This more general problem is really no harder and has the advantage that one gets as a by-product conditions for when this is possible.

12. Theorem: Let $\displaystyle p \left( x \right) =a{x}^{4}+b{x}^{3}+c{x}^{2}+dx+e$, with real coefficients and $\displaystyle a\neq 0$. There exists a line that is tangent to $\displaystyle y=p \left( x \right)$ at two distinct points if and only if $\displaystyle 8\,ac<3\,{b}^{2}$, in which case the $\displaystyle x$-coordinates of the two points of tangency are

$\displaystyle {\frac {-b+\sqrt {3\,{b}^{2}-8\,ac}}{4a}},$ and $\displaystyle {\frac {-b-\sqrt {3\,{b}^{2}-8\,ac}}{4a}}$.

Proof: Let $\displaystyle ({\it x_1},{\it y_1})$ be a point on the quartic, and equate the right hand sides of $\displaystyle y=p \left( x \right)$ and the tangent line $\displaystyle y=p \left( {\it x_1} \right) +p\,' \left( {\it x_1} \right) \left( x-{\it x_1} \right)$ at that point. This gives

(1) ......... $\displaystyle [p \left( x \right) -p \left( {\it x_1} \right) ] +p\,' \left( {\it x_1} \right) \left( x-{\it x_1} \right) =0$.

The polynomial in brackets is zero at $\displaystyle x={\it x_1}$ and so has $\displaystyle (x-{\it x_1})$ as a factor. Find the other factor by substituting for $\displaystyle p \left( x \right)$ and $\displaystyle p \left( x_1 \right)$ and organizing the terms as follows:

$\displaystyle [p \left( x \right) -p \left( {\it x_1} \right) ] = a({x}^{4}-{{\it x_1}}^{4})+b({x}^{3}-{{\it x_1}}^{3})+c({x}^{2}-{{\it x_1 }}^{2})+d(x-{\it x_1})$,

which allows $\displaystyle \left( x-{\it x_1} \right)$ to factor out from each term on the right and hence from the right side. After simplification the result is:

(2) ...... $\displaystyle [p \left( x \right) -p \left( {\it x_1} \right) ]=$ $\displaystyle \left( x-{\it x_1} \right) \left[ a{x}^{3}+ \left( a{\it x_1}+b \right) {x}^{2}+ \left( a{{\it x_1}}^{2}+b{\it x_1}+c \right) x+a{{\it x_1}}^{3 }+b{{\it x_1}}^{2}+c{\it x_1}+d \right]$

Now substitute (2) and the derivative $\displaystyle p '\left( x_1 \right) =4\,a{x_1}^{3}+3\,{x_1}^{2}b+2c\,x +d$ back into (1) and divide out the factor $\displaystyle \left( x-{\it x_1} \right)$, since we are interested in the other point of tangency and not the one we chose arbitrarily. This turns (1) into the following equation:

(3) ....... $\displaystyle a{x}^{3}+b{x}^{2}+a{x}^{2}{\it x_1}+cx+ax{{\it x_1}}^{2}+bx{\it x_1}-c{ \it x_1}\,-3\,a{{\it x_1}}^{3}-2\,b{{\it x_1}}^{2}=0$

Now, since points of tangency are double roots, the left side of (3) should contain $\displaystyle \left( x-{\it x_1} \right)$ as a factor again, and it is easy to check that substituting $\displaystyle x={\it x_1}$ makes the left side of (3) zero, so this must be the case. By dividing (3) by $\displaystyle \left( x-{\it x_1} \right)$ we find the factorization:

(3)' ........ $\displaystyle \left( x-{\it x_1} \right) \left( a{x}^{2}+ \left( 2\,a{\it x_1}+b \right) x+3\,a{{\it x_1}}^{2}+2\,b\,{\it x_1}\,+c \right) =0$

As before we are not interested in the current point of tangency, so we discard the factor $\displaystyle \left( x-{\it x_1} \right)$ and are left with the quadratic equation:

(4) ........ $\displaystyle a{x}^{2}+ \left( 2\,a{\it x_1}+b \right) x+3\,a{{\it x_1}}^{2}+2\,b{\it x_1} +c=0$

Solving (4) for $\displaystyle x$ will give the $\displaystyle x$-coordinates of the other two points where the tangent line at $\displaystyle ({\it x_1},{\it y_1})$ crosses the quartic curve. We want those to points to coincide, so the line will be tangent to the quartic at a second point. This means we need to force the quadratic equation (4) to have a double root. But this will happen if and only if the discriminant of the quadratic in (4) equals zero. So calculate that discriminant and set it to zero, and this gives the equation:

(5) ......... $\displaystyle 8\,{a}^{2}{{\it x_1}}^{2}+4\,ab{\it \,x_1}-{b}^{2}+4\,ac=0$

The two roots, $\displaystyle x_1$, of (5) are the $\displaystyle x$-coordinates of the two points at which the line (1) is tangent to the quartic. Solving (5) for $\displaystyle x_1$ gives these values of $\displaystyle x_1$ as:

$\displaystyle {\frac {-b+\sqrt {3\,{b}^{2}-8\,ac}}{4a}},$ and $\displaystyle {\frac {-b-\sqrt {3\,{b}^{2}-8\,ac}}{4a}}$

and the proof is complete.
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Note: The quartic given in the challenge problem was $\displaystyle p \left( x \right) =-{x}^{4}+18\,{x}^{3}-97\,{x}^{2}+180\,x-52$, so $\displaystyle a=-1$, $\displaystyle b=18$ and $\displaystyle c=-97$. Plugging these into the formulas above give the $\displaystyle x$-coordinates of the points of tangency as $\displaystyle x=1$ and $\displaystyle x=8$.

13. Originally Posted by Moo
Hello,

I didn't know it had integer answers... I did it the rough way because I had no idea where to start

$\displaystyle f'(a)(x-a)+f(a)=f'(b)(x-b)+f(b)$

$\displaystyle xf'(a)-af'(a)+f(a)=xf'(b)-bf'(b)+f(b)$

$\displaystyle \Rightarrow \left\{\begin{array}{ll} f'(a)=f'(b) & \text{that's logical, I know} \\ f(a)-f(b)=af'(a)-bf'(b) \end{array} \right.$

This is the first equation of the system, the second one is presumed to be nasty
Buwahahaha...I got something different...I probably screwed up on the algebra part...It was a ton of work though...

--Chris