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 "2022 AMC 12A Problems/Problem 25"

(Created page with "==Problem== A circle with integer radius <math>r</math> is centered at <math>(r, r)</math>. Distinct line segments of length <math>c_i</math> connect points <math>(0,a_i)</mat...")
 
(→‎Solution)
Line 10: Line 10:
  
 
We can easily prove that
 
We can easily prove that
−
\[
+
<cmath>
 
a + b - 2 r = c . \hspace{1cm} (1)
 
a + b - 2 r = c . \hspace{1cm} (1)
−
\]
+
</cmath>
  
 
Recall that <math>c = \sqrt{a^2 + b^2}</math>.
 
Recall that <math>c = \sqrt{a^2 + b^2}</math>.
  
 
Taking square of (1) and reorganizing all terms, (1) is converted as
 
Taking square of (1) and reorganizing all terms, (1) is converted as
−
\[
+
<cmath>
 
\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .
 
\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .
−
\]
+
</cmath>
  
  
Line 27: Line 27:
  
 
We can easily prove that
 
We can easily prove that
−
\[
+
<cmath>
 
2 r - a - b  = c . \hspace{1cm} (2)
 
2 r - a - b  = c . \hspace{1cm} (2)
−
\]
+
</cmath>
  
 
Recall that <math>c = \sqrt{a^2 + b^2}</math>.
 
Recall that <math>c = \sqrt{a^2 + b^2}</math>.
  
 
Taking square of (2) and reorganizing all terms, (2) is converted as
 
Taking square of (2) and reorganizing all terms, (2) is converted as
−
\[
+
<cmath>
 
\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .
 
\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .
−
\]
+
</cmath>
  
 
Putting both cases together, for given <math>r</math>, we look for solutions of <math>a</math> and <math>b</math> satisfying
 
Putting both cases together, for given <math>r</math>, we look for solutions of <math>a</math> and <math>b</math> satisfying
−
\[
+
<cmath>
 
\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 ,
 
\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 ,
−
\]
+
</cmath>
 
with either <math>a, b > 2r</math> or <math>0 < a, b < r</math>.
 
with either <math>a, b > 2r</math> or <math>0 < a, b < r</math>.
  
Line 47: Line 47:
  
 
For equation
 
For equation
−
\[
+
<cmath>
 
uv = 2 r^2 ,
 
uv = 2 r^2 ,
−
\]
+
</cmath>
 
we observe that the R.H.S. is a not a perfect square. Thus, the number of positive <math>(u, v)</math> is equal to the number of positive divisors of <math>2 r^2</math>.
 
we observe that the R.H.S. is a not a perfect square. Thus, the number of positive <math>(u, v)</math> is equal to the number of positive divisors of <math>2 r^2</math>.
  

Revision as of 20:55, 11 November 2022

Problem

A circle with integer radius $r$ is centered at $(r, r)$. Distinct line segments of length $c_i$ connect points $(0,a_i)$ to $(b_i, 0)$ for $1 \leq i \leq 14$ and are tangent to the circle, where $a_i$, $b_i$, and $c_i$ are all positive integers and $c_1 \leq c_2 \leq \cdots \leq c_{14}$. What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$?

Solution

Case 1: The tangent and the origin are on the opposite sides of the circle.

In this case, $a, b > 2r$.

We can easily prove that \[a + b - 2 r = c . \hspace{1cm} (1)\]

Recall that $c = \sqrt{a^2 + b^2}$.

Taking square of (1) and reorganizing all terms, (1) is converted as \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .\]


Case 2: The tangent and the origin are on the opposite sides of the circle.

In this case, $0 < a, b < r$.

We can easily prove that \[2 r - a - b = c . \hspace{1cm} (2)\]

Recall that $c = \sqrt{a^2 + b^2}$.

Taking square of (2) and reorganizing all terms, (2) is converted as \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .\]

Putting both cases together, for given $r$, we look for solutions of $a$ and $b$ satisfying \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 ,\] with either $a, b > 2r$ or $0 < a, b < r$.

Now, we need to find the smallest $r$, such that the number of feasible solutions of $(a, b)$ is at least 14.

For equation \[uv = 2 r^2 ,\] we observe that the R.H.S. is a not a perfect square. Thus, the number of positive $(u, v)$ is equal to the number of positive divisors of $2 r^2$.

Second, for each feasible positive solution $(u, v)$, its opposite $(-u, -v)$ is also a solution. However, $(u,v)$ corresponds to a feasible solution if $(a, b)$ with $a = u + 2r$ and $b = v + 2r$, but $(-u, -v)$ may not lead to a feasible solution if $(a, b)$ with $a = 2 r - u$ and $b = 2 r - v$.

Recall that we are looking for $r$ that leads to at least 14 solutions. Therefore, the above observations imply that we must have $r$, such that $2 r^2$ has least 7 positive divisors.

Following this guidance, we find the smallest $r$ is 6. This leads to the following solutions:

\begin{enumerate} \item $\left( a_1, b_1, c_1 \right) = \left( 3, 4, 5 \right)$, $\left( a_2, b_2, c_2 \right) = \left( 4, 3, 5 \right)$. \item $\left( a_3, b_3, c_3 \right) = \left( 20, 21, 29 \right)$, $\left( a_4, b_4, c_4 \right) = \left( 21, 20, 29 \right)$. \item $\left( a_5, b_5, c_5 \right) = \left( 18, 24, 30 \right)$, $\left( a_6, b_6, c_6 \right) = \left( 24, 18, 30 \right)$. \item $\left( a_7, b_7, c_7 \right) = \left( 16, 30, 34 \right)$, $\left( a_8, b_8, c_8 \right) = \left( 30, 16, 34 \right)$. \item $\left( a_9, b_9, c_9 \right) = \left( 15, 36, 39 \right)$, $\left( a_{10}, b_{10}, c_{10} \right) = \left( 36, 15, 39 \right)$. \item $\left( a_{11}, b_{11}, c_{11} \right) = \left( 14, 48, 50 \right)$, $\left( a_{12}, b_{12}, c_{12} \right) = \left( 48, 14, 50 \right)$. \item $\left( a_{13}, b_{13}, c_{13} \right) = \left( 13, 84, 85 \right)$, $\left( a_{14}, b_{14}, c_{14} \right) = \left( 84, 13, 85 \right)$. \end{enumerate}

Therefore, $\frac{c_{14}}{c_1} = \boxed{\textbf{(E) 17}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/6RfGCTNQ2Jw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_25&oldid=180728"