Difference between revisions of "2015 AMC 12B Problems/Problem 24"
Flamedragon (talk | contribs) (Created page with "==Problem== ==Solution== ==See Also== {{AMC12 box|year=2015|ab=B|num-a=25|num-b=23}} {{MAA Notice}}") |
Pi over two (talk | contribs) (→Problem) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| + | Four circles, no two of which are congruent, have centers at <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>, and points <math>P</math> and <math>Q</math> lie on all four circles. The radius of circle <math>A</math> is <math>\tfrac{5}{8}</math> times the radius of circle <math>B</math>, and the radius of circle <math>C</math> is <math>\tfrac{5}{8}</math> times the radius of circle <math>D</math>. Furthermore, <math>AB = CD = 39</math> and <math>PQ = 48</math>. Let <math>R</math> be the midpoint of <math>\overline{PQ}</math>. What is <math>AR+BR+CR+DR</math> ? | ||
| − | + | <math>\textbf{(A)}\; ? \qquad\textbf{(B)}\; ? \qquad\textbf{(C)}\; ? \qquad\textbf{(D)}\; ? \qquad\textbf{(E)}\; ?</math> | |
==Solution== | ==Solution== | ||
Revision as of 13:30, 3 March 2015
Problem
Four circles, no two of which are congruent, have centers at 
, 
, 
, and 
, and points 
and 
lie on all four circles. The radius of circle is 
times the radius of circle , and the radius of circle is times the radius of circle . Furthermore, 
and 
. Let 
be the midpoint of 
. What is 
?
Solution
See Also
| 2015 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
