# The square of minimum area with three vertices on a parabola

I was given this question by a friend and after working tirelessly on it I have not come up with anything substantial. I was hoping someone in the community could provide a pointer or possibly a solution. The person who gave me the question says he has the answer to it, but I am more curious about the working.

Here is the question:

Three points $(B, C, D)$ lie on parabola $y = x^2$ and a fourth point $A$ is placed such that $ABCD$ forms a square. Find the minimal area of $ABCD$.

So far my working out has involved a collection of distance/mid-point formula's and I haven't even reached the point where I differentiate to minimize the area.

Any pointers would be welcome.

### Edit

Apparently the correct answer is $\sqrt{2}$ units squared.

• Let $B(b,b^2),\,C(a,a^2),\,D(c,c^2),\ \vec{CB}.\vec{CD}=0,\ {\vec{CB}}^2={\vec{CD}}^2$.Trying to get explicit formulas I end up like this: $1+(b+a)(c+a)=0,\, 2a^2(c+b)+2a-(b^2+c^2+1)(b+c)=0$, $(a-b)^2-(a^2-b^2)^2\to \min$. Maybe Lagrange multipliers would help. – Alexey Burdin May 25 '15 at 3:22
• I haven't worked it out, but I bet it's easier to find the rectangle with the minimum area, and I would be unsurprised if it turns out to be square. – user14972 May 25 '15 at 8:07
• "Apparently the correct answer is $\sqrt 2$ units squared": That would usually be taken to mean $\sqrt 2$ (units squared)", not "($\sqrt 2$ units) squared". In which case it's wrong. The usual way to express this area is "2 square units". – TonyK Jun 1 '15 at 12:20

Let one vertex be $(x,x^2)$, the side length $s$ and the side orientation $\theta$.

Two other vertices are at coordinates $(x+s\cos\theta,x^2+s\sin\theta)$ and $(x+s\sin\theta,x^2-s\cos\theta)$. Let us express that they belong to the parabola:

$$(x+s\cos\theta)^2=x^2+s\sin\theta\implies2xs\cos\theta+s^2\cos^2\theta=s\sin\theta,\\ (x+s\sin\theta)^2=x^2-s\cos\theta\implies2xs\sin\theta+s^2\sin^2\theta=-s\cos\theta.$$

Eliminating $x$, we find $$s=\frac1{\cos\theta\sin\theta\,(\cos\theta-\sin\theta)}.$$

The area is minimized when the denominator is maximized, i.e. $\theta=\dfrac{3\pi}4$, then $s^2=\color{green}2$.

• A very nice solution. Monacraft should select this as the best answer. – wdacda May 26 '15 at 19:06
• @wdacda: thanks for the comment. I was myself amazed to see how it nicely simplified. – Yves Daoust May 27 '15 at 6:36

2 square units. B,C,D are (-1,1) (0,0) and (1,1) so A is (0,2)

This is the way to produce the right angle for two congruent sides.

I did not solve mathematically, with any equations. I pictured a square, say, 2 units by 2 units, and slid it down the side of the parabola until the opposite corner met.

Then came the visual shrink. Any smaller than 2x2, the third vertex doesn't ever touch the parabola, regardless of where you slide it. Keep in mind, the OP asked for a pointer or solution. I gave both, even though the rigorous steps are admittedly missing.

• Check my edit. Also could you please show some working? – Monacraft May 25 '15 at 2:08
• Can you show this is the minimal one? @JoeTaxpayer – Alexey Burdin May 25 '15 at 2:12
• Can we argue by symmetry this must be the answer? – James S. Cook May 25 '15 at 3:25
• Yes if you can explain where the symmetry arises? – Faraz Masroor May 25 '15 at 3:51

JoeTaxpayer is dead on with the visual, and I am not posting a 'duplicate' solution, so much as honoring the request for a more 'rigorous' treatment of Joe's visual.

Since you are given that ABCD forms a square, you know that the BCD segment must be a right angle (by virtue of it being half of a square). You can skirt calculus and geometry formulas by using basic linear algebra to use the given $y = x^2$ function which finds solutions of the 'endpoints' of B and D, whose dot product will be zero (since they are necessarily orthogonal). The endpoints of B are $(-x,y)$, though we know that $y = x^2$ tells us that the endpoints are $(-x,x^2)$. The same process applied to D tells us that the endpoint will be some $(x, x^2)$, as they must lie on the same vertical point on the parabola to have the same length.

It should be clear that since $y \geq 0$ that C must lie on $(0,0)$ (this should be 'visually' immediate as well as any other 'anchor' point on $y$ would be a positive number, whose orthogonal segments would be of different lengths if their endpoints also lie on the function $y = x^2$). Now, since (0,0) is the base point of your square, and we want the dot product of these two vectors to be $0$, we calculate the dot product of these two vectors in $\mathbb{R}^2$: $-x \cdot x + x^2 \cdot x^2$, i.e. $-x^2 + x^4 = 0 \implies x^2 = x^4$, which only has solutions for $x = -1, 0, 1$ which we can plug back in to our equations for B, C, and D to get $(-1, 1), (0, 0)$ and $(1,1)$ respectively. This gives you the three points of the 'square' formed along the lines of the parabola, and forces your choice of A. But since you can already calculate the area of a square using two of its sides, i.e. BC and CD, you only need to know the line segment length between $(0,0)$ and $(-1,1),(1,1)$. This is $\sqrt{2}$. Square this to get the area of ABCD and you get the desired answer: 2.

• Do I read right that there are no square, except $(0,2),\, (-1,1),\,(0,0),\,(1,1)$ ? You can check these x-coordinates: $$a=-\sqrt{{9\over 4}+{{3 \sqrt{3}}\over 2}}, b={1\over2}\left(1+\sqrt{3}+\sqrt{3+2 \sqrt{3}}\right), c={1\over2} \left(-1-\sqrt{3}+\sqrt{3+2 \sqrt{3}}\right)$$, these 3 also forms a square. – Alexey Burdin May 25 '15 at 4:43
• In fact, you can easily see that every vertex can be $C$, by drawing a pair of perpendicular lines meeting $C$ and rotating them until they cut out chords of the same length. WLOG, let $C$ be in the first quadrant. If you start out with a horizontal-vertical pair, then the one chord has finite length and the other is "infinite". As you rotate it clockwise, the length of infinite one decreases to zero, and the length of the other chord increases. – user14972 May 25 '15 at 8:05
• @Alexey/Hurkyl are correct. However, once we choose the basepoint to be $(0,0)$, the only solutions are $(-1,1)$ and $(1,1)$. There are an infinite number of solutions to be sure, but all of them offer a larger area for ABCD. @Yves Daoust's solution above is also very nice, and it eliminates the need to choose such a basepoint explicitly. As I referenced in my solution, I was showing rigorously why JoeTaxpayer's solution was correct, not using calculus to derive a formula for the area itself. For this, Yves's solution above is very nice. – Alex May 27 '15 at 22:49

There is no square with sides parallel to the axes having three vertices on the parabola $\gamma: \>y=x^2$. Therefore we have a lowest vertex $C:=(u,u^2)$. There is a side of slope $m>0$ emanating from $C$ to the right, and this side intersects $\gamma$ at $$D=\bigl((m-u),(m-u)^2\bigr)\ .$$ Replacing $m$ by $-{1\over m}$ here produces $B=\bigl(\bigl(-{1\over m}-u\bigr), \bigl({1\over m}+u\bigr)^2\bigr)$. We now consider $m$ as primary variable. Since we want $B$ to the left and $D$ to the right of $C$ the value of $u$ is then restricted to $$-{1\over 2m}<u<{m\over2}\ .\tag{1}$$ One computes $$|CD|^2=(1+m^2)(m-2u)^2,\qquad |BC|^2={1+m^2\over m^4}(1+2m u)^2\ .$$ The condition $|BC|^2=|CD|^2=:s^2$ then leads to $$(-1 + m^3 - 2 m u - 2 m^2 u) (1 + m^3 + 2 m u - 2 m^2 u)=0$$ with the solutions $$u_1={m^3-1\over2m(m+1)},\qquad u_2={m^3+1\over 2m(m-1)}\ .$$ We may assume that $m\geq1$, i.e., that the square is tilted to the left. Then only $u_1$ satisfies $(1)$. This leads to $$s^2={(m^2+1)^3\over m^2(m+1)^2}=2+{(m-1)^2(1+2m+4m^2+2m^3+m^4)\over m^2(m+1)^2}\ ,$$ which is smallest, namely $=2$, when $m=1$. This will result in $C=(0,0)$ and $A=(0,2)$.

Let $x_3 , x_2 , x_1$ - $x$ coordinates of the points of square and $x_2$ is between $x_1$ and $x_3$. Then sides of the square are: $$Y = (x_1 + x_2)X - x_1x_2$$ $$Y=(x_2+x_3)X - x_2x_3$$

Perpendicularity means: $$(x_1+x_2)(x_2+x_3) = -1$$

Solving it for $x_2$ gives: $$x_2 = -\frac{x_1+x_3}{2} \pm \frac{\sqrt{d^2-4}}{2}$$ Where $d = x_1-x_3$ . Above concludes that $d$ cannot be less than 2. Square area $S$ can be expressed in $x_1$ and $x_3$ as: $$2S = (x_1-x_3)^2+(x_1^2-x_3^2)^2=d^2+d^2(x_3+x_1)^2$$ Keeping $d$ at minimum = 2 $S$ is minimal only if $x_1 = -x_3$ That corresponds to square with main diagonal parallel to axes.

This concludes that minimum area = 2.

• "Solving all this" yields to i.imgur.com/4JZF300.gif or is there any nice way? – Alexey Burdin May 25 '15 at 4:33
• Since you assume $x_3 \lt x_2 \lt x_1$, what does "swapping $x_1$ and $x_3$ mean? – wdacda May 25 '15 at 18:50
• Corrected. The only I need is that they are pair-distinct and x2 is in the middle. Swapping x1,x3 still keep x2 in the middle. – hOff May 25 '15 at 19:30
• I don't understand neither of your claims "But swapping shall change position of $x_2$, if $x_2$ is not located on the symmetry line. In particular, due to symmetry $x_2$ changes its sign." You need to explain in detail what "swapping" means. Otherwise, I don't think that this is a solution. Although it is getting close. But the final argument is missing. Why a triple $x_1,x_2,x_3$ with $x_2 \gt 0$ cannot lead to the minimal area? – wdacda May 25 '15 at 20:53
• I simplified the solution. – hOff May 29 '15 at 12:34

For the point $(x+s\cos(\theta),x^2+s\sin(\theta))$ to be on the curve $y=x^2$, we need $$(x+s\cos(\theta))^2=x^2+s\sin(\theta)\tag{1}$$ Expanding the left side of $(1)$ and cancelling, we get $$s\cos(\theta)=\tan(\theta)-2x\tag{2}$$ Now $(2)$ must also remain true for $\theta+\frac\pi2$ so that the other corner adjacent to $(x,x^2)$ is on the parabola. Thus, we must also have $$s\sin(\theta)=\cot(\theta)+2x\tag{3}$$ Adding the squares of $(2)$ and $(3)$ and subtracting $2$ yields \begin{align} s^2-2 &=(\tan(\theta)-2x)^2+(\cot(\theta)+2x)^2-2\\ &=(\cot(\theta)-\tan(\theta))^2+4x(\cot(\theta)-\tan(\theta))+8x^2\\ &=\underbrace{\tfrac12(\cot(\theta)-\tan(\theta))^2}_{\text{0 when \theta=\frac\pi4}} +\underbrace{8\left(x+\tfrac14(\cot(\theta)-\tan(\theta))\right)^2}_{\text{0 when \theta=\frac\pi4 and x=0}}\tag{4} \end{align} Thus, $(4)$ says that $s^2-2$ is the sum of two squares, and therefore, $s^2-2\ge0$. If we set $\theta=\frac\pi4$ and $x=0$, we get that $s^2-2=0$.

Therefore, we can achieve the minimum of $s=\sqrt2$ with $x=0$ and $\theta=\frac\pi4$. This is the situation in Joe Taxpayer's image.

• Note that no calculus was needed to find the minimum; only algebra and trigonometry. – robjohn May 29 '15 at 0:39