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Difference between revisions of "1967 AHSME Problems/Problem 29"

(Created page with "== Problem == <math>\overline{AB}</math> is a diameter of a circle. Tangents <math>\overline{AD}</math> and <math>\overline{BC}</math> are drawn so that <math>\overline{AC}</mat...")
 
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== Problem ==
 
== Problem ==
−
<math>\overline{AB}</math> is a diameter of a circle.  Tangents <math>\overline{AD}</math> and <math>\overline{BC}</math> are drawn so that <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect in a point on the circle.  If <math>\overline{AD}=a</math> and <math>\overline{BD}=b</math>, <math>a \not= b</math>, the diameter of the circle is:
+
<math>\overline{AB}</math> is a diameter of a circle.  Tangents <math>\overline{AD}</math> and <math>\overline{BC}</math> are drawn so that <math>\overline{AC}</math> and <math>\overline{BD}</math> intersect in a point on the circle.  If <math>\overline{AD}=a</math> and <math>\overline{BC}=b</math>, <math>a \not= b</math>, the diameter of the circle is:
  
 
<math>\textbf{(A)}\ |a-b|\qquad
 
<math>\textbf{(A)}\ |a-b|\qquad

Revision as of 02:56, 23 February 2015

Problem

$\overline{AB}$ is a diameter of a circle. Tangents $\overline{AD}$ and $\overline{BC}$ are drawn so that $\overline{AC}$ and $\overline{BD}$ intersect in a point on the circle. If $\overline{AD}=a$ and $\overline{BC}=b$, $a \not= b$, the diameter of the circle is:

$\textbf{(A)}\ |a-b|\qquad \textbf{(B)}\ \frac{1}{2}(a+b)\qquad \textbf{(C)}\ \sqrt{ab} \qquad \textbf{(D)}\ \frac{ab}{a+b}\qquad \textbf{(E)}\ \frac{1}{2}\frac{ab}{a+b}$

Solution

$\fbox{C}$

See also

1967 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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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