1966 AHSME Problems/Problem 40

Problem

In this figure $AB$ is a diameter of a circle, centered at $O$ , with radius $a$ . A chord $AD$ is drawn and extended to meet the tangent to the circle at $B$ in point $C$ . Point $E$ is taken on $AC$ so the $AE=DC$ . Denoting the distances of $E$ from the tangent through $A$ and from the diameter $AB$ by $x$ and $y$ , respectively, we can deduce the relation:

$\text{(A) } y^2=\frac{x^3}{2a-x} \quad \text{(B) } y^2=\frac{x^3}{2a+x} \quad \text{(C) } y^4=\frac{x^2}{2a-x} \\ \text{(D) } x^2=\frac{y^2}{2a-x} \quad \text{(E) } x^2=\frac{y^2}{2a+x}$

Solution

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Problem 40
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
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1966 AHSME Problems/Problem 40

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