2022 AMC 12A Problem 25

Problem 25 of 25HarderGeometry Number Theory

A circle with integer radius

r

is centered at

(r,r).

Distinct line segments of length

c_i

connect points

(0,a_i)

(b_i,0)

for

1\le i\le14

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

\dfrac{c_{14}}{c_1}

for the least possible value of

r?

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Solution

The circle centered

(r,r)

with radius

r

is tangent to both axes. A segment from

(0,a)

(b,0)

with

a^2+b^2=c^2

is tangent to it when

r

equals either the inradius

\tfrac{a+b-c}{2}

or the semiperimeter

\tfrac{a+b+c}{2}

of the right triangle with legs

a,b.

In the inradius case, put

x=a-2r

and

y=b-2r.

Then

xy=2r^2,

and every positive divisor

x

determines one oriented segment, with

a=x+2r,

b=\frac{2r^2}{x}+2r,

and

c=x+\frac{2r^2}{x}+2r.

Thus there are exactly

d(2r^2)

such segments. For

r=1,2,3,4,5,

these counts are

2,4,6,6,6,

and no semiperimeter case is possible because the smallest integer right triangle has semiperimeter

6.

r=6,

d(72)=12,

and the

3

4

5

triangle contributes two more oriented segments. Hence

6

is the least possible radius and gives exactly

14

segments. The two semiperimeter segments have

c=5.

In the inradius family,

c=x+\frac{72}{x}+12

is largest at

x=1

72,

giving

c=85.

Therefore

c_1=5,

c_{14}=85,

and

\frac{c_{14}}{c_1}=17.

Thus, the correct answer is E.