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Difference between revisions of "1966 AHSME Problems/Problem 40"

(Created page with "== Problem == In this figure <math>AB</math> is a diameter of a circle, centered at <math>O</math>, with radius <math>a</math>. A chord <math>AD</math> is drawn and extended to m...")
 
(→‎Problem)
 
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== Problem ==
 
== Problem ==
 +
<asy>
 +
draw(circle((0,0),10),black+linewidth(1));
 +
MP("O", (0,0), S);MP("A", (-10,0), W);MP("B", (10,0), E);MP("C", (10,10), E);MP("D", (6,8), N);
 +
MP("a", (-5,0), S);MP("E", (-6,3), N);
 +
dot((0,0));dot((-6,2));
 +
draw((-10,0)--(10,0),black+linewidth(1));
 +
draw((-10,0)--(10,10),black+linewidth(1));
 +
draw((-10,-12)--(-10,12),black+linewidth(1));
 +
draw((10,-12)--(10,12),black+linewidth(1));
 +
</asy>
 +
 
In this figure <math>AB</math> is a diameter of a circle, centered at <math>O</math>, with radius <math>a</math>. A chord <math>AD</math> is drawn and extended to meet the tangent to the circle at <math>B</math> in point <math>C</math>. Point <math>E</math> is taken on <math>AC</math> so the <math>AE=DC</math>. Denoting the distances of <math>E</math> from the tangent through <math>A</math> and from the diameter <math>AB</math> by <math>x</math> and <math>y</math>, respectively, we can deduce the relation:
 
In this figure <math>AB</math> is a diameter of a circle, centered at <math>O</math>, with radius <math>a</math>. A chord <math>AD</math> is drawn and extended to meet the tangent to the circle at <math>B</math> in point <math>C</math>. Point <math>E</math> is taken on <math>AC</math> so the <math>AE=DC</math>. Denoting the distances of <math>E</math> from the tangent through <math>A</math> and from the diameter <math>AB</math> by <math>x</math> and <math>y</math>, respectively, we can deduce the relation:
  
Line 5: Line 16:
  
 
== Solution ==
 
== Solution ==
−
 
+
<math>\fbox{A}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 20:43, 23 September 2014

Problem

[asy] draw(circle((0,0),10),black+linewidth(1)); MP("O", (0,0), S);MP("A", (-10,0), W);MP("B", (10,0), E);MP("C", (10,10), E);MP("D", (6,8), N); MP("a", (-5,0), S);MP("E", (-6,3), N); dot((0,0));dot((-6,2)); draw((-10,0)--(10,0),black+linewidth(1)); draw((-10,0)--(10,10),black+linewidth(1)); draw((-10,-12)--(-10,12),black+linewidth(1)); draw((10,-12)--(10,12),black+linewidth(1)); [/asy]

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

$\fbox{A}$

See also

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
All AHSME Problems and Solutions

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