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
![[asy] size(10cm); pair A, B, C, D, F; A = (0, tan(3 * pi / 7)); B = (1, 0); C = (-1, 0); F = rotate(90/7, A) * (A - (0, 2)); D = rotate(900/7, F) * A; draw(A -- B -- C -- cycle); draw(F -- D); draw(D -- B); label("$A$", A, N); label("$B$", B, E); label("$C$", C, W); label("$D$", D, W); label("$E$", F, E); [/asy]](http://latex.artofproblemsolving.com/e/9/e/e9e27ce12b66c38e61f3aeeeec6ac0760eebc7e9.png)
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 isoceles triangle are congruent.
Now we angle chase. 
, 
, 
, 
, 
, 
. Since 
as given by the problem, 
, so 
. Therefore, 
, and our desired angle is ![\[180-4\left(\frac{180}{7}\right) = \frac{540}{7}\]](http://latex.artofproblemsolving.com/1/c/8/1c8994eb36574f1458d2ee3824b34658f345eabc.png)
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 .
https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25
(Almost Mirrored)
See here for a video solution:
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 | ||
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