Difference between revisions of "1954 AHSME Problems/Problem 26"

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==Problem 26==
 
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
 
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
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The straight line <math> \overline{AB}</math> is divided at <math> C</math> so that <math> AC\equal{}3CB</math>.  Circles are described on <math> \overline{AC}</math> and <math> \overline{CB}</math> as diameters and a common tangent meets <math> AB</math> produced at <math> D</math>.  Then <math> BD</math> equals:
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<math> \textbf{(A)}\ \text{diameter of the smaller circle} \
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\textbf{(B)}\ \text{radius of the smaller circle} \
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\textbf{(C)}\ \text{radius of the larger circle} \
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\textbf{(D)}\ CB\sqrt{3}\
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\textbf{(E)}\ \text{the difference of the two radii}</math>
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==Solution==
  
 
==See Also==
 
==See Also==

Revision as of 17:03, 22 April 2020

Problem 26

https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles The straight line $\overline{AB}$ is divided at $C$ so that $AC\equal{}3CB$ (Error compiling LaTeX. Unknown error_msg). Circles are described on $\overline{AC}$ and $\overline{CB}$ as diameters and a common tangent meets $AB$ produced at $D$. Then $BD$ equals:

$\textbf{(A)}\ \text{diameter of the smaller circle} \\ \textbf{(B)}\ \text{radius of the smaller circle} \\ \textbf{(C)}\ \text{radius of the larger circle} \\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii}$

Solution

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions


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