Difference between revisions of "2022 AMC 12A Problems/Problem 25"
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<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_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>. | ||
Revision as of 21:56, 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
Case 1: The tangent and the origin are on the opposite sides of the circle.
In this case, .
We can easily prove that
Recall that .
Taking square of (1) and reorganizing all terms, (1) is converted as
Case 2: The tangent and the origin are on the opposite sides of the circle.
In this case, .
We can easily prove that
Recall that .
Taking square of (2) and reorganizing all terms, (2) is converted as
Putting both cases together, for given , we look for solutions of
and
satisfying
with either
or
.
Now, we need to find the smallest , such that the number of feasible solutions of
is at least 14.
For equation
we observe that the R.H.S. is a not a perfect square. Thus, the number of positive
is equal to the number of positive divisors of
.
Second, for each feasible positive solution , its opposite
is also a solution. However,
corresponds to a feasible solution if
with
and
, but
may not lead to a feasible solution if
with
and
.
Recall that we are looking for that leads to at least 14 solutions.
Therefore, the above observations imply that we must have
, such that
has least 7 positive divisors.
Following this guidance, we find the smallest is 6. This leads to the following solutions:
,
.
,
.
,
.
,
.
,
.
,
.
,
.
Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)