# 1969 AHSME Problems/Problem 26

## Problem $[asy] draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); draw((-1.7,0)--(0,0)--(1.7,0),dot); draw((0,0)--(0,.98),dot); MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); [/asy]$

A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$ inches from the center $M$ is: $\text{(A) } 1\quad \text{(B) } 15\quad \text{(C) } 15\tfrac{1}{3}\quad \text{(D) } 15\tfrac{1}{2}\quad \text{(E) } 15\tfrac{3}{4}$

## Solution

Because the arch has a height of $16$ inches, an equation that models the arch is $y = ax^2 + 16$, where $x$ is the horizontal distance from the center and $y$ is the height. The arch has a span of $40$ inches, so the arch meets the ground $20$ inches from the center. That means $0 = 400a + 16$, so $a = -\frac{1}{25}$.

Thus, the equation that models height based on distance from the center $y = -\frac{1}{25}x^2 + 16$, so the height of the arch $5$ inches from the center is $-\frac{1}{25} \cdot 5^2 + 16 = \boxed{\textbf{(B) } 15}$ inches.

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