2023 AIME II Problems/Problem 12
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 1
Because is the midpoint of , following from the Stewart'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 on tells
As leads to
~bluesoul (supplemental note: ~Mathavi)
It's not actually immediately clear why this is the case. There are two ways to easily show this:
Notice that the problem statement tells us that point Q is EVERY piece of information in the problem statement is intentional, so we should try to use this to our benefit. None of the other solutions do, which is why they are more complicated than they need be.
Consider point Q' s.t. . Obviously, and are equal - we have perfect symmetry along line . Moreover, is a parallelogram as its diagonals bisect each other. Since point is unique, we know that . Thus is a parallelogram. Consider any quadrilateral whose diagonals intersect at s.t. and . We will prove that is .
(Note that in our problem, since and are not orthogonal, ( isn't isosceles) this is enough to show that is a parallelogram). -- By same base/same altitude, and
Therefore: Since , this reduces to
Let and . Then, by , and for some real . Then by LoC on and :
-- and is a parallelogram.
-- ( cannot be ; no negative sided polygons here!) and is a kite. . ~Mathavi
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 symmetric 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.