Difference between revisions of "2012 AMC 10A Problems/Problem 11"
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| − | == Video Solution == | + | == Video Solution by Calculocity (Easy Similar Triangles) == |
| + | https://youtu.be/YGu8AegJZ80 | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
https://youtu.be/NsQbhYfGh1Q?t=758 | https://youtu.be/NsQbhYfGh1Q?t=758 | ||
Latest revision as of 17:41, 22 April 2024
Contents
Problem
Externally tangent circles with centers at points 
and 
have radii of lengths 
and 
, respectively. A line externally tangent to both circles intersects ray 
at point 
. What is 
?
Solution
![[asy] unitsize(3.5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(8,0); pair C=(20,0); pair D=(1.25,-0.25sqrt(375)); pair E=(8.75,-0.15sqrt(375)); path a=Circle(A,5); path b=Circle(B,3); draw(a); draw(b); draw(C--D); draw(A--C); draw(A--D); draw(B--E); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,N); label("$D$",D,SE); label("$E$",E,SE); label("$5$",(A--D),SW); label("$3$",(B--E),SW); label("$8$",(A--B),N); label("$x$",(C--B),N); [/asy]](http://latex.artofproblemsolving.com/c/3/a/c3a9d5ccbf12a08c74fe1ebfe6d2f966349b51f6.png)
Let 
and 
be the points of tangency on circles and with line 
. 
. Also, let 
. As 
and 
are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share 
, 
. From this we can get a proportion.
Video Solution by Calculocity (Easy Similar Triangles)
Video Solution by OmegaLearn
https://youtu.be/NsQbhYfGh1Q?t=758
~ pi_is_3.14
See Also
| 2012 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
