1
$\begingroup$

A circle rests in the interior of the parabola with equation $y=x^2$ so that it is tangent to the parabola at two points. How much higher is the center of the circle than the points of tangency ?

I've seen the solution of the problem here on Aops site and it left me a bit skeptical.

I am not referring to the validity of the solution,which is correct, but rather to the fact that the problem has been solved using basic calculus.

Now, I came across this problem on Aops 2 volume book which doesn't teach any calculus,so it's clear that there's a simpler solution to the problem that relies only on basic geometry;however I've not been able to find it...

So I am asking if someone knows a solution which involves only basic geometry.

$\endgroup$
5
$\begingroup$

You don't need calculus for this one. Let the points of tangency be $(\pm a, a^2)$ and the center of the circle be $(0,c)$. Then the radius of the circle, $r$, is given by the Pythagorean theorem: $$r^2 = (c-a^2)^2 + a^2.$$ We can now plug this into the equation for the circle, $x^2 + (y-c)^2 = r^2$, to get: $$x^2 + y^2 - 2cy + c^2 - (c-a^2)^2 - a^2 = 0$$ but then we have $y=x^2$ so we get $$x^4 + (1-2c)x^2 + c^2 - a^2 - (c - a^2)^2 = 0.$$ We want the circle to only have two points of tangecy with the parabola, so we want the discriminant to $=0$, i.e. $$1 - 4c + 4c^2 - 8a^2c + 4a^4 + 4a^2 = 0$$ so that $$c = a^2 + \frac{1}{2}.$$ So now the answer to the question is simply $$a^2 + \frac{1}{2} - a^2 = \frac{1}{2} .$$

$\endgroup$
  • $\begingroup$ +1 but a typo in $r^2=a^4+(1-2c)a^2+c$, it should be $r^2=a^4+(1-2c)a^2+c^2$ $\endgroup$ – Ekaveera Kumar Sharma Dec 4 '15 at 13:33
  • $\begingroup$ Sorry about the typos, I think it's all cleaned up now. $\endgroup$ – TrivialCase Dec 4 '15 at 14:39
1
$\begingroup$

This problem has a name -- "The Quadratic Circle".

To solve it, we need knowledge slightly beyond simple geometry. They are :- (1) The theory of quadratic equation; and (2) Analytic geometry on circles.

enter image description here $y = x^2$ .............(1)

From given, $K$, the center of the unit circle must lie on the y-axis. Thus, $K= (0, k)$ for some $k$.

Then, the equation of that circle is $x^2 + (y – k)^2 = 1^2$ .............(2)

Putting (1) in (2), we have $y + (y – k)^2 = 1$

: :

$y^2 + (1 – 2k)y + [k^2 – 1] = 0$ ..........(3)

Let the points of intersection of the two curves be at $(x_1, y_1)$ and $(x_2, y_2)$.

By symmetry again, $y_1 = y_2$.

That means (3) has equal roots.

∴ $(1 – 2k)^2 – 4(1)[k^2 – 1] = 0$

: :

∴ $k = \dfrac {5}{4}$

$\endgroup$
  • $\begingroup$ Thanks for your answer @Mick $\endgroup$ – Mr. Y Dec 4 '15 at 19:38
  • $\begingroup$ @Mr.Y Hope that helps. $\endgroup$ – Mick Dec 5 '15 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.