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 $1 \leq i \leq 14$ 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:

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

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

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

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

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

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

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

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

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

Video Solution

https://youtu.be/6RfGCTNQ2Jw

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

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2022 AMC 12A Problems/Problem 25

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