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

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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:
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<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

Latest 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 (ProblemsAnswer KeyResources)
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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