Difference between revisions of "1954 AHSME Problems/Problem 26"
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==Problem 26== | ==Problem 26== | ||
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The straight line <math> \overline{AB}</math> is divided at <math> C</math> so that <math> AC=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: | The straight line <math> \overline{AB}</math> is divided at <math> C</math> so that <math> AC=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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==Solution== | ==Solution== | ||
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+ | Let <math> x=\overline{BD}</math> and let <math> r</math> be the radius of the small circle. Draw the line from the center of each of the circles to the point of contact of the tangent of the circle. By similar triangles, <math> \frac{x+r}{r}=\frac{x+5r}{3r} \implies x=r</math>, or <math> \boxed{\textbf{(B)}}</math>. | ||
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+ | https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles | ||
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+ | ==Video Solution== | ||
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+ | https://youtu.be/cbU3DxJKB8U | ||
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+ | ~Lucas | ||
==See Also== | ==See Also== |
Latest revision as of 15:39, 7 October 2024
Contents
Problem 26
The straight line is divided at so that . Circles are described on and as diameters and a common tangent meets produced at . Then equals:
Solution
Let and let be the radius of the small circle. Draw the line from the center of each of the circles to the point of contact of the tangent of the circle. By similar triangles, , or .
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
Video Solution
~Lucas
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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All AHSME Problems and Solutions |
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