Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.7 AIME, harder Find the answer

One, (20 points) (1) Table 1 shows the data measured in a projectile experiment of an object.

Assuming the trajectory equation of the object is $y=a x^{2}$ , find the value of the real number $a$ so that the sum of the squares of the distances from each measurement point $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3, 4)$ to the corresponding point $A_{i}\left(x_{i}, a x_{i}^{2}\right)$ on the trajectory curve $\sum_{i=1}^{4} P_{i} A_{i}^{2}$ is minimized.
(2) Under the requirement of (1), for the data in Table 2, find the equation of the trajectory curve $y=a x^{2}$ (the coefficient $a$ should be accurate to $0.01$ ).

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

(1) From the problem, we know
$\begin{array}{l} \sum_{i=1}^{4} P_{i} A_{i}^{2}=\sum_{i=1}^{4}\left(a x_{i}^{2}-y_{i}\right)^{2} \\ =a^{2} \sum_{i=1}^{4} x_{i}^{4}-2 a \sum_{i=1}^{4} x_{i}^{2} y_{i}+\sum_{i=1}^{4} y_{i}^{2} \\ =\sum_{i=1}^{4} x_{i}^{4}\left[a^{2}-\frac{2 \sum_{i=1}^{4} x_{i}^{2} y_{i}}{\sum_{i=1}^{4} x_{i}^{4}} a+\left(\frac{\sum_{i=1}^{4} x_{i}^{2} y_{i}}{\sum_{i=1}^{4} x_{i}^{4}}\right)^{2}+\right. \\ \sum_{i=1}^{4} y_{i}^{2}-\left(\frac{\sum_{i=1}^{4} x_{i}^{2} y_{i}}{\sum_{i=1}^{4} x_{i}^{4}}\right)^{2} \\ =\sum_{i=1}^{4} x_{i}^{4}\left(a-\frac{\sum_{i=1}^{4} x_{i}^{2} y_{i}}{\sum_{i=1}^{4} x_{i}^{4}}\right)^{2}+\sum_{i=1}^{4} y_{i}^{2}-\left(\frac{\sum_{i=1}^{4} x_{i}^{2} y_{i}}{\sum_{i=1}^{4} x_{i}^{4}}\right)^{2} \\ \text { Therefore, when } a=\frac{\sum_{i=1}^{4} x_{i}^{2} y_{i}}{\sum_{i=1}^{4} x_{i}^{4}}, \sum_{i=1}^{4} P_{i} A_{i}^{2} \text { reaches its minimum value } \\ \sum_{i=1}^{4} y_{i}^{2}-\left(\frac{\sum_{i=1}^{4} x_{i}^{2} y_{i}}{\sum_{i=1}^{4}} x_{i}^{4}\right. \end{array}$

Therefore, when $a=\frac{\sum_{i=1}^{4} x_{i}^{2} y_{i}}{\sum_{i=1}^{4} x_{i}^{4}}$ , $\sum_{i=1}^{4} P_{i} A_{i}^{2}$ reaches its minimum value.
(2) From the data in Table 2, we get
$\begin{array}{l} \sum_{i=1}^{4} x_{i}^{4}=1^{4}+2^{4}+3^{4}+4^{4}=354, \\ \sum_{i=1}^{4} x_{i}^{2} y_{i}=1^{2} \times(-2.6)+2^{2} \times(-9.9)+ \\ \quad 3^{2} \times(-22.6)+4^{2} \times(-39.8) \\ =-882.4 . \end{array}$

Thus, from (1), we have $a=\frac{-882.4}{354} \approx-2.49$ .
Therefore, the desired orbital curve is $y=-2.49 x^{2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.