Difference between revisions of "2015 AMC 12B Problems/Problem 24"

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==Problem==
 
==Problem==
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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> ?
  
 
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<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 $A$, $B$, $C$, and $D$, and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$, and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$. Furthermore, $AB = CD = 39$ and $PQ = 48$. Let $R$ be the midpoint of $\overline{PQ}$. What is $AR+BR+CR+DR$ ?

$\textbf{(A)}\; ? \qquad\textbf{(B)}\; ? \qquad\textbf{(C)}\; ? \qquad\textbf{(D)}\; ? \qquad\textbf{(E)}\; ?$

Solution

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
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

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