Functions $y = x^2 + x - 1$ and $y = x^3 + 2x^2 + (a + b\sqrt{3})x - 3$ have three common points $A, B, C$ such that the circumradius is $R = 3$.

Question

Consider two functions $y = x^2 + x - 1$ and $y = x^3 + 2x^2 + (a + b\sqrt{3})x - 3$ with $a$ and $b$ being two rational numbers such that the graphs of the aforementioned functions share three common points $A, B, C$ such that the radius of the circumscribed circle of $\triangle ABC$ is $R = 3$. Calculate the value of $a + b$.

[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)

By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]

It really was one run-on sentence, wasn't it? The kids aren't happy with this one. (Is this okay to call my friends—me included—kids? [nervous laughing])

For sake of inconvenience, I'll just set $c = a + b\sqrt 3$.

In order for the graphs of $y = x^2 + x - 1$ and $y = x^3 + 2x^2 + cx - 3$ to have been common points, the equation $f(x) = x^3 + x^2 + (c - 1)x - 2 = 0$ needs to have three pairwise distinct roots.

Now, I could just use the discriminant of the polynomial itself, $\Delta = -4c^3 + 13c^2 - 50c - 59$, to determine the condition that $c$ has to satisfy in order for the polynomial to have the pairwise distinct roots. But since that seems too cumbersome (and since our modern education system seems to have forgotten about the discriminant of cubic equations anyway), so I'm going to take the derivative instead, $f'(x) = 3x^2 + 2x + (c - 1)$, (to literally arrive at the same conclusion).

The equation $f'(x) = 0$ needs to have two roots $x_1 \ne x_2$ such that $f(x_1)f(x_2) < 0$.

First of all, $\Delta > 0 \implies 4(4c - 3) > 0 \iff c < \dfrac{3}{4}$.

Furthermore, since $f(x) = \dfrac{(3x + 1)f'(x) + [(6c - 8)x - (c + 17)]}{9} \implies \left\{ \begin{aligned} f(x_1) &= \dfrac{(6c - 8)x_1 - (c + 17)}{9}\\ f(x_2) &= \dfrac{(6c - 8)x_2 - (c + 17)}{9} \end{aligned} \right.$

$$\implies f(x_1)f(x_2) = \dfrac{(36c^2 - 96c + 64)x_1x_2 - (6c^2 + 94c - 136)(x_1 + x_2) + (c^2 + 34c + 289)}{81}$$

According to Vieta's formulas, as $x_1$ and $x_2$ are roots of the equation $3x^2 + 2x + (c - 1) = 0$, $x_1 + x_2 = -\dfrac{2}{3}$ and $x_1x_2 = \dfrac{c - 1}{3}$, which means $f(x_1)f(x_2) = \dfrac{4c^3 - 13c^2 + 50c + 59}{27}$. And eventually, $$f(x_1)f(x_2) < 0 \iff 4c^3 - 13c^2 + 50c + 59 < 0 \iff c < \dfrac{1}{12}\left(z + 13 - \dfrac{431}{z}\right)$$, where $z = \sqrt[3]{1320\sqrt{330} - 22247}$. (You know where this result came from.)

But was all of this necessary? Hmmm~

Let $(m; m^2 + n - 1), (n; n^2 + n - 1), (p; p^2 + p - 1)$ be the coordinates of points $A, B, C$ respectively, where $(n - p)(p - m)(m - n) \ne 0$. The perpendicular bisectors of line segments $AB$ and $BC$ are $$\left\{ \begin{aligned} (d_1)\colon \left(x - \dfrac{m + n}{2}\right) &+ (m + n + 1)\left(y - \dfrac{m^2 + n^2 + m + n + 2}{2}\right) = 0\\ (d_2)\colon \left(x - \dfrac{n + p}{2}\right) &+ (n + p + 1)\left(y - \dfrac{n^2 + p^2 + n + p + 2}{2}\right) = 0 \end{aligned} \right.$$

For another sake of inconvenience, let $m + n + p = M, np + pm + mn = N$ and $mnp = P$.

Solving the system of equations, we can obtain that $$\left\{ \begin{aligned} x &= -\dfrac{M^2 + MN + 2M + N - P + 2}{2}\\ y &= \dfrac{M^2 + 2M - N + 4}{2} \end{aligned} \right.$$

This is quite confusing. Although, I have another idea.

Let $A' (m; 0), B' (n; 0), C' (p; 0)$ and $I' (q; 0)$ be the projections of points $A, B, C$ and $I$ on the $x$-axis in that order. It can be incurred that $A', B', C'$ are the three common points between the graph of $y = x^3 + x^2 + (c - 1)x - 3$ and the $x$-axis. Additionally, $$\dfrac{|q - m|}{\cos(IA, Ox)} = \dfrac{|q - n|}{\cos(IB, Ox)} = \dfrac{|q - p|}{\cos(IC, Ox)} = 3$$

There might be too many points...

Anyhow, as always, thanks for reading, (and even more if you could help~) By the way, the options for this problem were $-1, 1, 3$ and $6$.

Answer 1

I do not claim that this is the simplest solution. I didn't use the calculator while solving this problem, so there might be a simpler solution using one. (I come from a country that thinks calculators are bad for children's math improvement) Anyways, my solution goes as follows:

Let $f(x)=x^2+x-1$, and $g(x)=x^3+2x^2+cx-3$, and denote $\alpha, \beta, \gamma$ be three solutions of $f(x)=g(x)$, i.e $$x^3+x^2+(c-1)x-2=0$$ I will assume that there indeed exists three (distinct) solutions. Then, we have $\alpha+\beta+\gamma=-1$, $\alpha\beta\gamma=2$ by Vieta's formula. Take the points $A(\alpha, \alpha^2+\alpha-1), B(\beta, \beta^2+\beta-1), C(\gamma, \gamma^2+\gamma-1)$. We will mainly use the formula $S_{\triangle ABC}=\frac{abc}{4R}$ where $a=\overline{BC}, b=\overline{AC}, c=\overline{AB}$.(This formula comes from the formula $S=\frac{1}{2}ab\sin C$ and the sine law.) We find the area of the triangle $ABC$ by the shoelace formula. Then, we have $$S_{\triangle ABC}=\frac{1}{2}\begin{vmatrix} \alpha&\beta&\gamma&\alpha\\ \alpha^2+\alpha-1&\beta^2+\beta-1&\gamma^2+\gamma-1&\alpha^2+\alpha-1 \end{vmatrix}$$ After some tedious computation(I don't know whether a calculator can do this calculation simpler, but well..), I got: $$S_{\triangle ABC}=-\frac{1}{2}(\alpha-\beta)(\beta-\gamma)(\alpha-\gamma)$$ We know that $R=3$, and so we get $$-\frac{1}{2}(\alpha-\beta)(\beta-\gamma)(\alpha-\gamma)=\frac{abc}{12}\cdots \star$$ Now, we figure out $a$. We get $$a^2=(\gamma-\beta)^2+((\gamma^2+\gamma-1)-(\beta^2+\beta-1))^2=(\gamma-\beta)^2(1+\alpha^2)$$ by using the identity $\alpha+\beta+\gamma=-1$. So, we get: $$a=(\gamma-\beta)\sqrt{1+\alpha^2}$$ $$b=(\gamma-\alpha)\sqrt{1+\beta^2}$$ $$c=(\beta-\alpha)\sqrt{1+\gamma^2}$$ by symmetry, and assuming $\alpha<\beta<\gamma$. Now plug all of this back into $\star$, then we get $$\frac{1}{2}(\gamma-\beta)(\gamma-\alpha)(\beta-\alpha)=\frac{(\gamma-\beta)(\gamma-\alpha)(\beta-\alpha)}{12}\sqrt{(\alpha^2+1)(\beta^2+1)(\gamma^2+1)}$$ We have assumed that $\alpha<\beta<\gamma$, so we can divide each side by $(\gamma-\beta)(\gamma-\alpha)(\beta-\alpha)$, and by squaring each side, we finally get: $$(\alpha^2+1)(\beta^2+1)(\gamma^2+1)=36$$ Now, we reformulate the given equation as: $$(\alpha\beta\gamma)^2+(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)+(\alpha^2+\beta^2+\gamma^2)+1=36$$ We know that $\alpha+\beta+\gamma=-1$, $\alpha\beta\gamma=2$, and $\alpha\beta+\beta\gamma+\gamma\alpha=c-1$. So, $$\alpha^2\beta^2\gamma^2=(\alpha\beta\gamma)^2=4$$ $$(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)=(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)=(c-1)^2+4$$ $$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=1-2(c-1)$$ Plugging all of these, we get: $$4+(c-1)^2+4+1-2(c-1)+1=36$$ Viewing this as a quadratic polynomial, we get $(c-1)=1\pm 3\sqrt{3}$, or $c=2\pm 3\sqrt{3}$. As you have showed that $c<\frac{3}{4}$, we conclude that the solution is $c=2-3\sqrt{3}$, and $a=2$, $b=-3$, so $a+b=-1$.

If there are some errors, I am happy to accept them. But I highly doubt if these are meant to solved in such short time.