Difference between revisions of "2018 AMC 12A Problems/Problem 23"
(→Solution 2 (Overkill)) |
(→Solution 3 (Nice, I Think?)) |
||
Line 35: | Line 35: | ||
==Solution 3 (Nice, I Think?)== | ==Solution 3 (Nice, I Think?)== | ||
− | Consider the bisector of <math>\angle ATP.</math> This angle makes an <math>80^{\circ}</math> angle with <math>AP.</math> We claim that <math>MN</math> is parallel to this angle bisector, meaning that the | + | Consider the bisector of <math>\angle ATP.</math> This angle makes an <math>80^{\circ}</math> angle with <math>AP.</math> We claim that <math>MN</math> is parallel to this angle bisector, meaning that the acute angle that <math>MN</math> makes with <math>AP</math> is <math>80^{\circ},</math> meaning that the answer is <math>\boxed{\textbf{(E)}}</math>. |
+ | |||
+ | (sujaykazi) | ||
+ | Shoutout to Richard Yi and Mark Kong for working with me to glean the necessary insights for this problem! | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=22|num-a=24}} | {{AMC12 box|year=2018|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:18, 14 February 2018
Contents
[hide]Problem
In and Points and lie on sides and respectively, so that Let and be the midpoints of segments and respectively. What is the degree measure of the acute angle formed by lines and
Solution
Let be the origin, and lie on the x axis.
We can find and
Then, we have and
Notice that the tangent of our desired points is the the absolute difference between the y coordinates of the two points divided by the absolute difference between the x coordinates of the two points.
This evaluates to Now, using sum to product identities, we have this equal to so the answer is (lifeisgood03)
Solution 2 (Overkill)
Note that , the midpoint of major arc on is the Miquel Point of (Because ). Then, since , this spiral similarity carries to . Thus, we have , so .
But, we have ; thus .
Then, as is the midpoint of the major arc, it lies on the perpendicular bisector of , so . Since we want the acute angle, we have , so the answer is .
(stronto)
Solution 3 (Nice, I Think?)
Consider the bisector of This angle makes an angle with We claim that is parallel to this angle bisector, meaning that the acute angle that makes with is meaning that the answer is .
(sujaykazi) Shoutout to Richard Yi and Mark Kong for working with me to glean the necessary insights for this problem!
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.