Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0323bbf84409a1adeb34g-39.jpg?height=321&width=517&top_left_y=195&top_left_x=468)
Answer: -6.

![](https://cdn.mathpix.com/cropped/2024_05_06_0323bbf84409a1adeb34g-39.jpg?height=359&width=614&top_left_y=600&top_left_x=420)

Fig. 11: to the solution of problem 10.7

Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the condition, it is clear that $x_{1} < 0$ and $x_{2} > 0$. Since $x_{1}$ and $x_{2}$ are the roots of the quadratic trinomial $f(x)$, by Vieta's theorem, we have $x_{1} \cdot x_{2} = 12b$, from which we get $b = \frac{x_{1} \cdot x_{2}}{12} < 0$.

Let $H$ be the point with coordinates $(3 ; 0)$ (Fig. 11). Clearly, in the isosceles triangle $A T C$, the segment $T H$ is the height, and therefore it is also the median. Thus, $3 - x_{1} = A H = H C = x_{2} - 3$, from which we get $x_{1} = 6 - x_{2}$.

Let $M$ be the point with coordinates $(0, 3)$. Since $T H = T M = 3$ and $T A = T B$, the right triangles $A T H$ and $B T M$ are equal by the leg and hypotenuse. Therefore, $3 - x_{1} = H A = M B = 3 - b$, that is, $x_{1} = b = \frac{x_{1} \cdot x_{2}}{12}$ (by Vieta's theorem), from which we find $x_{2} = 12$. Finally, $x_{1} = 6 - x_{2} = 6 - 12 = -6$ and $b = x_{1} = -6$.

Another solution. As in the previous solution, let the abscissas of points $A$ and $C$ be $x_{1}$ and $x_{2}$, respectively; we will also use the fact that point $B$ has coordinates $(0 ; b)$. Immediately, we understand that $O A = |x_{1}| = -x_{1}$, $O C = |x_{2}| = x_{2}$, and $O B = |b| = -b$.

Let's find the second intersection of the circle with the y-axis, let this be point $D$ with coordinates $(0 ; d)$ (Fig. 12). The chords $A C$ and $B D$ of the circle intersect at the origin $O$; from the properties of the circle, we know that $O A \cdot O C = O B \cdot O D$. We get $-x_{1} \cdot x_{2} = -b \cdot d$, from which, replacing $x_{1} \cdot x_{2}$ with $12b$ by Vieta's theorem, we get $d = 12$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0323bbf84409a1adeb34g-40.jpg?height=521&width=571&top_left_y=83&top_left_x=437)

Fig. 12: to the solution of problem 10.7

It remains to note that triangle $BTD$ is isosceles, and the midpoint of its base, point $M$, has coordinates $(0 ; 3)$. Reflecting point $D(0 ; 12)$ relative to it, we get $B(0 ; -6)$.