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3 distinct points are on the graph $y=4x^2$ The x coordinates form an arithmetic sequence while the y coordinates form a geometric sequence, what are the possible values of the common ratio?

Naming the points : $(A,4A^2), (B, 4B^2), (C, 4C^2)$

X = common difference r = common ratio

Equations: $A+x = B; B+x =C$

$A^2(r) = B^2; B^2(r)=C^2$

Solving for x

$A^2(r) = (A+x)^2$

I get $x=(-A\pm A\sqrt{r})/2$

then $B^2(r) = C^2$

$A^2(r^2)=(A+2x)^2$

$A^2(r^2)=(A-A\pm A\sqrt{r})^2$

which results in $r=1$ or $r=0$

Where did i go wrong?

SuperMage1
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2 Answers2

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$x=(-2A\pm 2A\sqrt r)/2=-A\pm A\sqrt r$, not $x=(-A\pm A\sqrt r)/2$.

Continuing from $A^2(r^2)=(A+2x)^2$,

$$A^2(r^2)=(A-2A\pm 2A\sqrt r)^2\\ A^2(r^2)=(-A\pm 2A\sqrt r)^2\\ A^2(r^2)=A^2+4A^2r\pm4A^2\sqrt r\\ r^2=1+4r\pm4\sqrt r\\ r^2-4r-1=\pm\sqrt{16r}\text.$$

Squaring both sides, $$r^4+16r^2+1-8r^3-2r^2+8r=16r\\ r^4-8r^3+14r^2-8r+1=0\\ (r-1)^2(r^2-6r+1)=0\\ r=1\lor r=3\pm2\sqrt2\text.$$

Consider that $A,B,C$ are distinct, so $r\ne1$. Therefore, $$r=3\pm2\sqrt2\text.$$

user_194421
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  • Of which $,r=1,$ is not eligible, since the three points are supposedly distinct. – dxiv Jan 19 '18 at 06:42
  • @dxiv Where in the question it says so? – user_194421 Jan 19 '18 at 06:44
  • You are right, it nowhere literally says so. But I thought that was what 3 points was supposed to mean in the context, so I took it as an implicit assumption. Otherwise, for example, one could choose all three points to be $,(0,0),$ and then the answer would be "any common ratio will work". – dxiv Jan 19 '18 at 06:48
  • sorry i forgot to include that part, it does say distinct points – SuperMage1 Jan 19 '18 at 11:05
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    @SuperMage1 Would you mind editing it into the question? – user_194421 Jan 19 '18 at 11:10
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Let the abscissae of the three points be $a,b,c$, then the conditions for AP and GP are:

$$ \begin{align} a+c &= 2b \tag{1} \\ (4a^2)\cdot(4c^2) &= (4b^2)^2 \tag{2} \end{align} $$

Substituting $(1)$ into $(2)$ gives:

$$ \begin{align} 16a^2c^2 = (a+c)^4 \;\;&\iff\;\; (a+c)^4 - 16a^2c^2 = 0 \\ &\iff\;\; \big((a+c)^2-4ac\big)\big((a+c)^2+4ac)=0 \\ &\iff\;\; (a-c)^2(a^2+6ac+c^2) = 0 \end{align} $$

The first factor cannot be $0$ since $a,b,c$ are distinct, which leaves $\,a^2+6ac+c^2=0\,$.

None of $a,b,c$ can be $0\,$, since the GP condition would imply that the other two are $0$ as well, and a quadratic cannot have three distinct roots. Then dividing by $a^2\ne0$ gives $\,\left(\dfrac{c}{a}\right)^2+6\left(\dfrac{c}{a}\right)+1=0\,$ with the roots $\dfrac{c}{a}=-3 \pm 2 \sqrt{2}\,$, and so the common ratio of the GP is $\,\pm\sqrt{\dfrac{4c^2}{4a^2}} = \pm(3 \pm 2 \sqrt{2})\,$. It still remains to be verified whether actual $\,a,b,c\,$ solutions do in fact exist in each of the four cases.

[ EDIT ] The common ratio of the GP must be positive since all $y$ coordinates are positive, as noted by @user_194421. This eliminates the negative solutions, and leaves only $\,3 \pm 2\sqrt{2}\,$ to be checked.
dxiv
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