Difference between revisions of "2022 AMC 12A Problems/Problem 25"

Revision as of 21:04, 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 $1 \le i \le 14$ 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

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 excircle, 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

@@ Line 1: / Line 1: @@
 ==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)</math> to <math>(b_i, 0)</math> for <math>1 \leq i \leq 14</math> and are tangent to the circle, where <math>a_i</math>, <math>b_i</math>, and <math>c_i</math> are all positive integers and
+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)</math> to <math>(b_i, 0)</math> for <math>1 \le i \le 14</math> and are tangent to the circle, where <math>a_i</math>, <math>b_i</math>, and <math>c_i</math> are all positive integers and <math>c_1 \le c_2 \le \cdots \le c_{14}</math>. What is the ratio <math>\frac{c_{14}}{c_1}</math> for the least possible value of <math>r</math>?
-<math>c_1 \leq c_2 \leq \cdots \leq c_{14}</math>. What is the ratio <math>\frac{c_{14}}{c_1}</math> for the least possible value of <math>r</math>?
+<math>\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 </math>
 ==Solution==
+Suppose that with a pair <math>(a_i,b_i)</math> the circle is an excircle. Then notice that the hypotenuse must be <math>(r-x)+(r-y)</math>, so it must be the case that <cmath>a_i^2+b_i^2=(2r-a_i-b_i)^2.</cmath> Similarly, if with a pair <math>(a_i,b_i)</math> the circle is an excircle, the hypotenuse must be <math>(x-r)+(y-r)</math>, leading to the same equation.
-Case 1: The tangent and the origin are on the opposite sides of the circle.
+Notice that this equation can be simplified through SFFT to <cmath>(a_i-2r)(b_i-2r)=2r^2.</cmath> Thus, we want the smallest <math>r</math> such that this equation has at least <math>14</math> distinct pairs <math>(a_i,b_i)</math> for which this holds. The obvious choice to check is <math>r=6</math>. In this case, since <math>2r^2=2^3\cdot 3^2</math> has <math>12</math> positive factors, we get <math>12</math> pairs, and we get another two if the factors are <math>-8,-9</math> or vice versa. One can check that for smaller values of <math>r</math>, we do not even get close to <math>14</math> possible pairs.
-In this case, <math>a, b > 2r</math>.
-We can easily prove that
-<cmath>
-a + b - 2 r = c . \hspace{1cm} (1)
-</cmath>
-Recall that <math>c = \sqrt{a^2 + b^2}</math>.
-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 .
-</cmath>
-Case 2: The tangent and the origin are on the opposite sides of the circle.
-In this case, <math>0 < a, b < r</math>.
-We can easily prove that
-<cmath>
-r - a - b  = c . \hspace{1cm} (2)
-</cmath>
-Recall that <math>c = \sqrt{a^2 + b^2}</math>.
-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 .
-</cmath>
-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 ,
-</cmath>
-with either <math>a, b > 2r</math> or <math>0 < a, b < r</math>.
-Now, we need to find the smallest <math>r</math>, such that the number of feasible solutions of <math>(a, b)</math> is at least 14.
-For equation
-<cmath>
-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>.
-Second, for each feasible positive solution <math>(u, v)</math>, its opposite <math>(-u, -v)</math> is also a solution. However, <math>(u,v)</math> corresponds to a feasible solution if <math>(a, b)</math> with <math>a = u + 2r</math> and <math>b = v + 2r</math>, but <math>(-u, -v)</math> may not lead to a feasible solution if <math>(a, b)</math> with <math>a = 2 r - u</math> and <math>b = 2 r - v</math>.
-Recall that we are looking for <math>r</math> that leads to at least 14 solutions.
-Therefore, the above observations imply that we must have <math>r</math>, such that <math>2 r^2</math> has least 7 positive divisors.
-Following this guidance, we find the smallest <math>r</math> is 6. This leads to the following solutions:
-<math>\left( a_1, b_1, c_1 \right) = \left( 3, 4, 5 \right)</math>, <math>\left( a_2, b_2, c_2 \right) = \left( 4, 3, 5 \right)</math>.
-<math>\left( a_3, b_3, c_3 \right) = \left( 20, 21, 29 \right)</math>, <math>\left( a_4, b_4, c_4 \right) = \left( 21, 20, 29 \right)</math>.
-<math>\left( a_5, b_5, c_5 \right) = \left( 18, 24, 30 \right)</math>, <math>\left( a_6, b_6, c_6 \right) = \left( 24, 18, 30 \right)</math>.
-<math>\left( a_7, b_7, c_7 \right) = \left( 16, 30, 34 \right)</math>, <math>\left( a_8, b_8, c_8 \right) = \left( 30, 16, 34 \right)</math>.
-<math>\left( a_9, b_9, c_9 \right) = \left( 15, 36, 39 \right)</math>, <math>\left( a_{10}, b_{10}, c_{10} \right) = \left( 36, 15, 39 \right)</math>.
-<math>\left( a_{11}, b_{11}, c_{11} \right) = \left( 14, 48, 50 \right)</math>, <math>\left( a_{12}, b_{12}, c_{12} \right) = \left( 48, 14, 50 \right)</math>.
-<math>\left( a_{13}, b_{13}, c_{13} \right) = \left( 13, 84, 85 \right)</math>, <math>\left( a_{14}, b_{14}, c_{14} \right) = \left( 84, 13, 85 \right)</math>.
-Therefore, <math>\frac{c_{14}}{c_1} = \boxed{\textbf{(E) 17}}</math>.
-~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
-==Video Solution==
-https://youtu.be/6RfGCTNQ2Jw
+When <math>r=6</math>, the smallest possible <math>c</math>-value is clearly when the factors are negative. When this occurs, <math>a_i=4, b_i=3</math> (or vice versa), so the mimimal <math>c</math> is <math>5</math>. The largest possible <math>c</math>-value occurs when the largest of <math>a_i</math> and <math>b_i</math> are maximized. This occurs when the factors are <math>72</math> and <math>1</math>, leading to <math>a_i=84, b_i=13</math> (or vice-versa), leading to a maximal <math>c</math> of <math>85</math>.
-~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+Hence the answer is <math>\frac{85}5=\boxed{17}</math>.
-== See Also ==
+~ bluelinfish
+==See also==
 {{AMC12 box|year=2022|ab=A|num-b=24|after=Last Problem}}
 {{MAA Notice}}

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Difference between revisions of "2022 AMC 12A Problems/Problem 25"

Revision as of 21:04, 11 November 2022

Problem

Solution

See also