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1988 AMC 12 Problem 29

Attempt Problem 29 of the 1988 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AMC 12 solutions, or check the answer key.

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29.

You plot weight (y)(y) against height (x)(x) for three of your friends and obtain the points (x1,y1),(x_1,y_1), (x2,y2),(x_2,y_2), (x3,y3).(x_3,y_3). If x1<x2<x3,x3−x2=x2−x1, \begin{aligned} x_1&\lt x_2\lt x_3,\\ x_3-x_2&=x_2-x_1, \end{aligned} which of the following is necessarily the slope of the line which best fits the data? “Best fits” means that the sum of the squares of the vertical distances from the data points to the line is smaller than for any other line.

y3−y1x3−x1\frac{y_3-y_1}{x_3-x_1}

(y2−y1)−(y3−y2)x3−x1\frac{(y_2-y_1)-(y_3-y_2)}{x_3-x_1}

2y3−y1−y22x3−x1−x2\frac{2y_3-y_1-y_2}{2x_3-x_1-x_2}

y2−y1x2−x1+y3−y2x3−x2\frac{y_2-y_1}{x_2-x_1}+\frac{y_3-y_2}{x_3-x_2}

none of these

Answer: A
Concepts:least squaresslopesymmetric coordinates
Difficulty rating: 2480
Small Hint:

Translate and scale the xx-coordinates to −1,0,1-1,0,1

Big Hint:

For a least-squares line through symmetric xx-values, compute the slope from the covariance numerator

Solution:

Translate and scale so the xx-coordinates are −d,0,d.-d,0,d. Their mean is 0,0, so the least-squares slope is (−d)y1+0y2+dy3d2+0+d2=y3−y12d=y3−y1x3−x1. \begin{aligned} \frac{(-d)y_1+0y_2+dy_3}{d^2+0+d^2} &=\frac{y_3-y_1}{2d}\\ &=\frac{y_3-y_1}{x_3-x_1}. \end{aligned}

Thus the correct answer is A.

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