Difference between revisions of "2022 AMC 12A Problems/Problem 25"
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==Problem== | ==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 \ | + | 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 \ | + | |
+ | <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== | ==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. | ||
− | + | 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. | |
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− | + | 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>. | |
− | + | Hence the answer is <math>\frac{85}5=\boxed{17}</math>. | |
− | + | ~ bluelinfish | |
+ | ==See also== | ||
{{AMC12 box|year=2022|ab=A|num-b=24|after=Last Problem}} | {{AMC12 box|year=2022|ab=A|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 22:04, 11 November 2022
Problem
A circle with integer radius is centered at
. Distinct line segments of length
connect points
to
for
and are tangent to the circle, where
,
, and
are all positive integers and
. What is the ratio
for the least possible value of
?
Solution
Suppose that with a pair the circle is an excircle. Then notice that the hypotenuse must be
, so it must be the case that
Similarly, if with a pair
the circle is an excircle, the hypotenuse must be
, leading to the same equation.
Notice that this equation can be simplified through SFFT to Thus, we want the smallest
such that this equation has at least
distinct pairs
for which this holds. The obvious choice to check is
. In this case, since
has
positive factors, we get
pairs, and we get another two if the factors are
or vice versa. One can check that for smaller values of
, we do not even get close to
possible pairs.
When , the smallest possible
-value is clearly when the factors are negative. When this occurs,
(or vice versa), so the mimimal
is
. The largest possible
-value occurs when the largest of
and
are maximized. This occurs when the factors are
and
, leading to
(or vice-versa), leading to a maximal
of
.
Hence the answer is .
~ bluelinfish
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.