2022 AMC 12A Problems/Problem 25

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 $1 \le i \le 14$ and are tangent to the circle, where $a_i$ , $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$ ?

$\textbf{(A)} ~\frac{21}{5} \qquad\textbf{(B)} ~\frac{85}{13} \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~\frac{39}{5} \qquad\textbf{(E)} ~17$

Solution 1

Suppose that with a pair $(a_i,b_i)$ the circle is an excircle. Then notice that the hypotenuse must be $(r-x)+(r-y)$ , so it must be the case that $a_i^2+b_i^2=(2r-a_i-b_i)^2.$ Similarly, if with a pair $(a_i,b_i)$ the circle is an incircle, the hypotenuse must be $(x-r)+(y-r)$ , leading to the same equation.

Notice that this equation can be simplified through SFFT to $(a_i-2r)(b_i-2r)=2r^2.$ Thus, we want the smallest $r$ such that this equation has at least $14$ distinct pairs $(a_i,b_i)$ for which this holds. The obvious choice to check is $r=6$ . In this case, since $2r^2=2^3\cdot 3^2$ has $12$ positive factors, we get $12$ pairs, and we get another two if the factors are $-8,-9$ or vice versa. One can check that for smaller values of $r$ , we do not even get close to $14$ possible pairs.

When $r=6$ , the smallest possible $c$ -value is clearly when the factors are negative. When this occurs, $a_i=4, b_i=3$ (or vice versa), so the mimimal $c$ is $5$ . The largest possible $c$ -value occurs when the largest of $a_i$ and $b_i$ are maximized. This occurs when the factors are $72$ and $1$ , leading to $a_i=84, b_i=13$ (or vice-versa), leading to a maximal $c$ of $85$ .

Hence the answer is $\frac{85}5=\boxed{17}$ .

~ bluelinfish

Solution 2

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 same 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:

$\left( a_1, b_1, c_1 \right) = \left( 3, 4, 5 \right)$ , $\left( a_2, b_2, c_2 \right) = \left( 4, 3, 5 \right)$ .

$\left( a_3, b_3, c_3 \right) = \left( 20, 21, 29 \right)$ , $\left( a_4, b_4, c_4 \right) = \left( 21, 20, 29 \right)$ .

$\left( a_5, b_5, c_5 \right) = \left( 18, 24, 30 \right)$ , $\left( a_6, b_6, c_6 \right) = \left( 24, 18, 30 \right)$ .

$\left( a_7, b_7, c_7 \right) = \left( 16, 30, 34 \right)$ , $\left( a_8, b_8, c_8 \right) = \left( 30, 16, 34 \right)$ .

$\left( a_9, b_9, c_9 \right) = \left( 15, 36, 39 \right)$ , $\left( a_{10}, b_{10}, c_{10} \right) = \left( 36, 15, 39 \right)$ .

$\left( a_{11}, b_{11}, c_{11} \right) = \left( 14, 48, 50 \right)$ , $\left( a_{12}, b_{12}, c_{12} \right) = \left( 48, 14, 50 \right)$ .

$\left( a_{13}, b_{13}, c_{13} \right) = \left( 13, 84, 85 \right)$ , $\left( a_{14}, b_{14}, c_{14} \right) = \left( 84, 13, 85 \right)$ .

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

Solution 3

As $c_i$ is the length of the segment $(0,a_i)$ and $(b_i,0)$ ， $a_i^2+b_i^2=c_i^2$ . The equation for the line that passes through $(0,a_i)$ and $(b_i,0)$ is $a_ix+b_iy-a_ib_i=0$ .

By the point-line distance formula from point $(r,r)$ to line $a_ix+b_iy-a_ib_i=0$

$r = \frac{ |a_ir+b_ir-a_ib_i| }{ \sqrt{a_i^2+b_i^2} }, \quad c_i = \frac{ |a_ir+b_ir-a_ib_i| }{r} = |a_i+b_i- \frac{a_ib_i}{r}|$

$\because c_i \quad \text{is an integer}, \quad \therefore r|a_ib_i$

To find $c_1$ and $c_{14}$ for the smallest $r$ , we will list Pythagorean triples $(a_i,b_i,c_i)$ with relatively prime elements from the smallest. Notice that we only need to find $7$ triples with $a_i<b_i$ , the $7$ other triples will simply be $(b_i,a_i,c_i)$ . $a_i$ will not equal $b_i$ because then $c_i = a_i \cdot \sqrt{2}$ , meaning that $a_i$ , $b_i$ , and $c_i$ cannot all be integers.

The $7$ smallest triples are: $(3,4,5)$ , $(5,12,13)$ , $(7,24,25)$ , $(8,15,17)$ , $(11,60,61)$ , $(12,35,37)$ , $(13,84,85)$

Therefore, $\frac{c_{14}}{c_1} = \frac{85}{5} = \boxed{\textbf{(E) 17}}$ .

~isabelchen

Video Solution by Math-X (Let's understand the question first)

https://youtu.be/7yAh4MtJ8a8?si=2dTtBDEZLIjN3YWS&t=9728 ~Math-X

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

Video Solution by MOP 2024

https://youtu.be/5jO__fUbgs8

~r00tsOfUnity

Video Solution by Steven Chen

https://youtu.be/6RfGCTNQ2Jw

Video Solution (5 Minutes)

https://youtu.be/79CLeiJrPvs

~MathProblemSolvingSkills.com

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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

Page

Toolbox

Search