Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1954 AHSME Problems/Problem 26"
m
 
(4 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==Problem 26==
 
==Problem 26==
−
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
+
 
−
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:
+
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:
  
 
<math> \textbf{(A)}\ \text{diameter of the smaller circle} \\
 
<math> \textbf{(A)}\ \text{diameter of the smaller circle} \\
Line 10: Line 10:
  
 
==Solution==
 
==Solution==
 +
 +
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>.
 +
 +
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
 +
 +
==Video Solution==
 +
 +
https://youtu.be/cbU3DxJKB8U
 +
 +
~Lucas
  
 
==See Also==
 
==See Also==

Latest revision as of 15:39, 7 October 2024

Problem 26

The straight line $\overline{AB}$ is divided at $C$ so that $AC=3CB$. 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

Let $x=\overline{BD}$ and let $r$ 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, $\frac{x+r}{r}=\frac{x+5r}{3r} \implies x=r$, or $\boxed{\textbf{(B)}}$.

https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles

Video Solution

https://youtu.be/cbU3DxJKB8U

~Lucas

See Also

1954 AHSC (Problems • Answer Key • Resources)
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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_26&oldid=229233"