2020 AIME I Problems/Problem 1
Contents
Problem
In with point lies strictly between and on side and point lies strictly between and on side such that The degree measure of is where and are relatively prime positive integers. Find
Solution 1
If we set to , we can find all other angles through these two properties: 1. Angles in a triangle sum to . 2. The base angles of an isosceles triangle are congruent.
Now we angle chase. , , , , , . Since as given by the problem, , so . Therefore, , and our desired angle is for an answer of .
See here for a video solution: https://youtu.be/4e8Hk04Ax_E
Solution 2
Let be in degrees. . By Exterior Angle Theorem on triangle , . By Exterior Angle Theorem on triangle , . This tells us and . Thus and we want to get an answer of .
Solution 3 (Official MAA)
Let . Because is isosceles, . Then Because and are also isosceles, Because is isosceles, is also , so , and it follows that . The requested sum is .
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25 (Almost Mirrored)
See here for a video solution:
Solution 4 (writing equations)
graph soon
We write equations based on the triangle sum of angles theorem. There are angles that do not need variables as the less variables the better.
Then, using triangle sum of angles theorem, we find that
Now we just need to find the variables.
Notice how all the equations equal 180. We can use this to write
Simplifying, we get
Theres more. We are at a dead end right now because we forgot that the problem states that the triangle is isosceles. With this, we can write the equation
Substituting with , we get
With this, we get
And a final answer of .
Video Solution by OmegaLearn
https://youtu.be/O_o_-yjGrOU?t=333
~ pi_is_3.14
Video solution
https://youtu.be/mgRNqSDCvgM ~yofro
Solution without words
vladimir.shelomovskii@gmail.com, vvsss
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.