Difference between revisions of "2023 AIME II Problems/Problem 12"
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Plugging <math>x</math> into the line equation gets you <math>y=\frac{147}{37}</math>. The distance between this point and <math>A</math>, which is <math>(5,12)</math> is <math>\sqrt{\frac{9801}{148}}</math>, or simplified to <math>\frac{99}{\sqrt{148}}\Longrightarrow99+148=\boxed{247}</math> | Plugging <math>x</math> into the line equation gets you <math>y=\frac{147}{37}</math>. The distance between this point and <math>A</math>, which is <math>(5,12)</math> is <math>\sqrt{\frac{9801}{148}}</math>, or simplified to <math>\frac{99}{\sqrt{148}}\Longrightarrow99+148=\boxed{247}</math> | ||
− | ~dragoon (<math>LaTeX</math> fixes by | + | ~dragoon (minor <math>LaTeX</math> fixes by rhydon516) |
==Solution 5 (similar to 3)== | ==Solution 5 (similar to 3)== |
Revision as of 13:52, 22 March 2023
Contents
Problem
In with side lengths
and
let
be the midpoint of
Let
be the point on the circumcircle of
such that
is on
There exists a unique point
on segment
such that
Then
can be written as
where
and
are relatively prime positive integers. Find
Solution
Because is the midpoint of
, following from the Steward's theorem,
.
Because ,
,
, and
are concyclic,
,
.
Denote .
In , following from the law of sines,
Thus,
In , following from the law of sines,
Thus,
Taking , we get
In , following from the law of sines,
Thus, Equations (2) and (3) imply
Next, we compute and
.
We have
We have
Taking (5) and (6) into (4), we get
Therefore, the answer is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
Define to be the foot of the altitude from
to
. Furthermore, define
to be the foot of the altitude from
to
. From here, one can find
, either using the 13-14-15 triangle or by calculating the area of
two ways. Then, we find
and
using Pythagorean theorem. Let
. By AA similarity,
and
are similar. By similarity ratios,
Thus,
. Similarly,
. Now, we angle chase from our requirement to obtain new information.
Take the tangent of both sides to obtain
By the definition of the tangent function on right triangles, we have
,
, and
. By abusing the tangent angle addition formula, we can find that
By substituting
,
and using tangent angle subtraction formula we find that
Finally, using similarity formulas, we can find
. Plugging in
and
, we find that
Thus, our final answer is
.
~sigma
Solution 3 (simplest)
It is clear that is a parallelogram. By Stewart's Theorem,
, POP tells
As leads to
~bluesoul
Solution 4 (LOS+ coordbash)
First, note that by Law of Sines, and that
. Equating the 2 expressions, you get that
. Now drop the altitude from
to
. As it is commonly known that the dropped altitude forms a
and a
triangle, you get the measures of
and
respectively, which are
and
. However, by the inscribed angle theorem, you get that
and that
, respectively. Therefore, by Law of Sines (as previously stated)
.
Now commence coordbashing. Let be the origin, and
be the point
. As
passes through
, which is
, and
, which is
, it has the equation
, so therefore a point on this line can be written as
. As we have the ratio of the lengths, which prompts us to write the lengths in terms of the distance formula, we can just plug and chug it in to get the ratio
. This can be squared to get
. This can be solved to get a solution of
, and an extraneous solution of
which obviously doesn’t work.
Plugging into the line equation gets you
. The distance between this point and
, which is
is
, or simplified to
~dragoon (minor fixes by rhydon516)
Solution 5 (similar to 3)
We use the law of Cosine and get
We use the power of point
with respect circumcircle
and get
It is clear that if
then
if is simmetric to
with respect
then
There exists a unique point on segment
vladimir.shelomovskii@gmail.com, vvsss
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=k6hEFEVVzMI
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.