Difference between revisions of "2018 AMC 12A Problems/Problem 23"
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+ | For another way to find <math>\angle XMN</math>, note that <cmath>\angle XAM = 90 - \angle MXA = 90 - \frac{\angle AXP}{2} = 90 - \frac{\angle ATP}{2}= 90 - 44 = 46,</cmath> giving <math>\angle XMN = \angle XAG = 56 - 46 = 10</math> as desired. | ||
==Solution 3 (Nice, I Think?)== | ==Solution 3 (Nice, I Think?)== |
Revision as of 18:47, 18 January 2021
Contents
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 is the midpoint of and , or
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)
Note: Though this solution is excellent, setting makes life a tad bit easier ~ MathleteMA
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)
Sidenote
For another way to find , note that giving as desired.
Solution 3 (Nice, I Think?)
Let the bisector of intersect at We have so We claim that is parallel to this angle bisector, meaning that the acute angle formed by and is meaning that the answer is .
To prove this, let be the midpoint of where and are the points on and respectively, such that (The points given in this problem correspond to but the idea we're getting at is that will ultimately not matter.) Since and vary linearly with the locus of all points must be a line. Notice that so lies on this line. Let be the intersection of this line with (we know that this line will intersect and not because ). Notice that
Let Then and Thus, By the Angle Bisector Theorem, so Since is the midpoint of we also have Notice that:
Since the line containing all points must be parallel to This concludes the proof.
The critical insight to finding this solution is that the length probably shouldn't matter because a length ratio of or (as in the problem) is exceedingly unlikely to generate nice angles. This realization then motivates the idea of looking at all points similar to which then leads to looking at the most convenient such point (in this case, the one that lies on ).
(sujaykazi) Shoutout to Richard Yi and Mark Kong for working with me to discover the necessary insights to this problem!
Solution 4
Let the mid-point of be and the mid-point of be . Since and , we can conclude that . Similarly, we can conclude that . Construct and intersects at , which gives . Since , , we can find the value of , which is equal to . Since , which means , we can infer that . As we are required to give the acute angle formed, the final answer would be , which is . (Surefire2019)
Solution 5 (Simplest, I think)
Link , extend to so that . Then link and .
, is the middle point of and
is the middle line of
Notice that
As a result, , ,
Also,
As a result,
Therefore,
Since we are asked for the acute angle between the two lines, the answer to this problem is
~Solution by (Frank FYC)
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2018amc12a/473
~ dolphin7
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.