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 "2002 AIME I Problems/Problem 2"
m (→‎Solution 2)
 
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as <math>\dfrac{1}{2}(\sqrt{p}-q)</math> where <math>p</math> and <math>q</math> are positive integers. Find <math>p+q</math>.
+
The diagram shows twenty congruent [[circle]]s arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The [[ratio]] of the longer dimension of the rectangle to the shorter dimension can be written as <math>\dfrac{1}{2}(\sqrt{p}-q)</math> where <math>p</math> and <math>q</math> are positive integers. Find <math>p+q</math>.
  
 
<center>[[Image:AIME_2002I_Problem_02.png]]</center>
 
<center>[[Image:AIME_2002I_Problem_02.png]]</center>
  
 
== Solution ==
 
== Solution ==
−
Let the radius of the circles is <math>r</math>. The longer dimension can be written as <math>14r</math>, and by Pythagorean Theorem, we can find that the shorter dimension: <math>2r\sqrt{3}+2r</math>.
+
Let the [[radius]] of the circles be <math>r</math>. The longer dimension of the rectangle can be written as <math>14r</math>, and by the [[Pythagorean Theorem]], we find that the shorter dimension is <math>2r\left(\sqrt{3}+1\right)</math>.
  
−
Therefore, <math>\frac{14r}{2r\sqrt{3}+2r}=\frac{1}{2}(\sqrt{p}-q)</math>
+
Therefore, <math>\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)</math>. Thus we have <math>p=147</math> and <math>q=7</math>, so <math>p+q=\boxed{154}</math>.
  
−
After simplifying, we get <math>7\sqrt{3}-7=\sqrt{p}-q</math>, which gives <math>p=147</math> and <math>q=7</math>. So, <math>p+q=154</math>
+
== Solution 2 ==
 +
Since we only care about the ''ratio'' between the longer side and shorter side, we can set the longer side to <math>14</math>. So, this means that each of the radii is <math>1</math>. Now, we connect the radii of three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is <math>2\sqrt{3}</math>, and the shorter side of the triangle is therefore <math>2\sqrt{3}+2</math> and we use simplification similar to as showed above, and we reach the result <math>\frac{1}{2} \cdot (\sqrt{147}-7)</math> and the final answer is <math>147+7 = \boxed{154}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2002|n=I|num-b=1|num-a=3}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 19:21, 16 September 2024

Problem

The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$.

AIME 2002I Problem 02.png

Solution

Let the radius of the circles be $r$. The longer dimension of the rectangle can be written as $14r$, and by the Pythagorean Theorem, we find that the shorter dimension is $2r\left(\sqrt{3}+1\right)$.

Therefore, $\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)$. Thus we have $p=147$ and $q=7$, so $p+q=\boxed{154}$.

Solution 2

Since we only care about the ratio between the longer side and shorter side, we can set the longer side to $14$. So, this means that each of the radii is $1$. Now, we connect the radii of three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is $2\sqrt{3}$, and the shorter side of the triangle is therefore $2\sqrt{3}+2$ and we use simplification similar to as showed above, and we reach the result $\frac{1}{2} \cdot (\sqrt{147}-7)$ and the final answer is $147+7 = \boxed{154}$.

See also

2002 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME 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=2002_AIME_I_Problems/Problem_2&oldid=227346"